Sciencemadness Discussion Board - Acids/bases: heats of neutralisation

Acids/bases: heats of neutralisation

blogfast25 - 31-7-2010 at 12:55

This is a follow up on the back end of this thread here:

http://www.sciencemadness.org/talk/viewthread.php?tid=14164#...

... where it was suggested by some that the neutralisation of a weak acid with a strong base would produce less enthalpy than the neutralisation of a strong acid with a strong base. I decided to put that assumption to the test.

So far we have the following data points:

1) The enthalpy of neutralisation of a strong acid with a strong base is approx. - 57.3 kJ/mol of water (literature).

2) The neutralisation of NaHCO3 with a strong acid is endothermic (literature).

3) The neutralisation of Na2CO3 with HCl appears experimentally neither much exothermic nor much endothermic.

4) The neutralisation enthalpy of acetic acid with NaOH was found experimentally to be about - 54.5 kJ/mol of water, so quite close to - 57.3 kJ/mol of water. The strength of the acetic acid (distilled vinegar) will be experimentally verified by titration tomorrow.

Some new data points are:

5) The neutralisation enthalpy of NH4Cl (a very weak acid: pKa = 9.25) with NaOH (strong base) was experimentally found to be - 5.1 kJ/mol of water and for (NH4)2SO4 (half a mol), - 5.0 kJ/mol of water, so much less than - 57.3 kJ/mol for strong bases with strong acids.

6) The neutralisation enthalpy of NH3 with HCl was experimentally found to be -9.7 kJ/mol of water, so also much lower than - 57.3 kJ/mol. The strength of both the NH3 and HCl solutions have been verified with acid/base titrometry.

All in all quite inconclusive, I'd say. It'd be interesting to include some other alkanoic acids like 1-propanoic, 1-butanoic etc because we know these decrease gradually in acid strength (Ka) as the aliphatic group increases in length...

[Edited on 31-7-2010 by blogfast25]

12AX7 - 31-7-2010 at 18:19

It should vary with pKa because pKa is an indirect measure of energy. The neutralization of pH ~14 with pH ~ 7 should have half as much energy as to pH ~0. And the neutralization of an alkyl metal (pKa ~ 50) in water certainly has more energy. I forget what the conversion is though.

Tim

blogfast25 - 1-8-2010 at 04:26

Hi Tim,

I made that point here:

http://www.sciencemadness.org/talk/viewthread.php?tid=14164#...

For low Ka (< 1), the DG (delta G) of:

HA + H2O ---> H3O+ + A-

... is positive. Assuming no entropic effects, DH (delta H) for the dissociation reaction is then also positive. But for the acetic acid the data point really doesn't fit that theory at all.

A bit more later...

[Edited on 1-8-2010 by blogfast25]

blogfast25 - 1-8-2010 at 11:59

The acidity of my grade of 'distilled vinegar' was confirmed to be 4.65 w% of acetic acid, for '4.5 % acidity' advertised on the bottle, so the molarity used in the calculation was correct.

I'm wondering whether there is some entropic effect that I'm overlooking:

in HA(aq) + H2O(l) ---> H3O+(aq) + A-(aq)

... which side is the most probable state, most chaotic, most entropic, left or right?

Nicodem - 1-8-2010 at 14:40

I would tend to believe that in acid/base reactions where non-fully dissociated species in water react it is the enthalpy of solvation changes that contributes most. For example:

CH3COOH(aq) + OH-(aq) <=> CH3COO-(aq) + H2O
Acetic acid is solvated via H-bond (mostly as donor) but I think acetate should be more solvated. Most of the acetic acid is not dissociated, so it takes some energy to break the O-H bond (pKa being the equilibrium constant). Perhaps the solvation changes compensate for the energy loss in O-H bond breaking. Hydroxyde is highly solvated in water so it takes some energy to desolvate it, but this energy is already included in the -57.3 kJ/mol enthalpy value.

CO32-(aq) + 2H+(aq) <=> CO2(g) + H2O
Carbonate is strongly solvated while CO2 is not (leaves the reaction as gas), so a lot of energy is consumed in desolvating the carbonate anion. I would not be surprised that all the energy from protonation is required for this, thus leaving the overall enthalpy close to zero. Protons are extremely tightly solvated, but again this energy is already included in the -57.3 kJ/mol enthalpy value.

NH4+ + OH-(aq) <=> NH3(aq) + H2O
The ammonium ion is surely more solvated than ammonia, so this change takes a lot of energy. Same with breaking the relatively strong N-H bond...

You can also look at this as entropic changes given that it is about changes in the overall (dis)order of the system, but from the perspective of individual ions it is all about how much energy it takes to break the solvation bonds with the solvent or how much energy is gained by forming such bonds. Obviously the overall reaction enthalphy is the sum of all these changes plus the enthalpy of the H+(aq) + OH-(aq) reaction.

Edit: Mostly changed the interpretation of the acetic acid + hydroxide reaction as the previous made no sense. I still think a lot does not make sense which is not that strange as it is mostly guesswork. The point being that the answer most likely is connected with the solvation enthalpies.

[Edited on 2/8/2010 by Nicodem]

blogfast25 - 2-8-2010 at 04:41

@ Nicodem:

Yes, solvation or other energies must play a part in weak acid/base neutralisations, as it can be shown easily that even in the case of the weakest possible monoprotonic acid 1 mol of water is formed per mol of acid neutralised. So by rights the neutralisation energy for H3O+ + OH- ---> 2 H2O of -57.3 kJ/mol of H2O must be liberated, whatever the monoprotonic acid. Endothermic effects must offset that.

The case of NH4+/NH3 is particularly confusing, at least to moi. Most ammonium salts I've tested are lightly endothermic when you dissolve them in water (at least the chloride and sulphate are). NH3 itself though is strongly exothermic. Wouldn't one on these grounds expect the neutralisation of NH4+ salts to be strongly exothermic? I mean, where does the dissolution energy of the NH3 go???

These things keep me awake at night! ;-)

woelen - 2-8-2010 at 05:41

I think that splitting the proton from NH4(+) ions takes a lot of energy, which compensates for the solvation energy, released by dissolving NH3 in water. On the other hand, dissolving NH3 in water is not that exothermic, I tried a few times and it is much less exothermic than e.g. dissolving gaseous HCl in water.

[Edited on 2-8-10 by woelen]

blogfast25 - 2-8-2010 at 08:15

Good point, Woelen.

To get this absolutely right we probably would to have construct something akin to the Born Haber Cycle but for watery reactions, taking into account all energy consumptions and releases.

As regards the exothermicity of dissolving NH3 in water, in one patent I found the released energy is recovered to use as energy for a stripper section. In the stripper section a process solution is thermally stripped of dissolved NH3... Can't be that low if it's worth recovering. The NH3 cycled between solutions, acting as a NaF solubility suppressant.

Nicodem - 3-8-2010 at 00:59

Quote: Originally posted by blogfast25

Yes, solvation or other energies must play a part in weak acid/base neutralisations, as it can be shown easily that even in the case of the weakest possible monoprotonic acid 1 mol of water is formed per mol of acid neutralised. So by rights the neutralisation energy for H3O+ + OH- ---> 2 H2O of -57.3 kJ/mol of H2O must be liberated, whatever the monoprotonic acid. Endothermic effects must offset that.

Quote:

The case of NH4+/NH3 is particularly confusing, at least to moi. Most ammonium salts I've tested are lightly endothermic when you dissolve them in water (at least the chloride and sulphate are). NH3 itself though is strongly exothermic. Wouldn't one on these grounds expect the neutralisation of NH4+ salts to be strongly exothermic? I mean, where does the dissolution energy of the NH3 go???

The enthalpy of dissolution is no direct indication of the solvation enthalpy, certainly not so in dissolving salts where the enthalpy is the difference between solvation and lattice energy. OK, the dissolution of NH3 is more directly connected with the solvation, but I think the most direct indication of the solvation energy would be the dilution enthalpy. Anyway, the solvation enthalpy of the formed NH3 goes in the overall enthalpy, thus it is consumed in the desolvation of the ammonium ions and N-H bond breaking (as Woelen noted above). Obviously this energy per self would not be enough, but since there is also the H2O "association" energy released, the overall reaction is what it is, a very slightly exothermic neutralisation.

blogfast25 - 3-8-2010 at 06:34

To be sure, a second reading of the reaction enthalpy of HAc (aq) + NaOH (aq) ---> NaAc (aq) + H2O (l) (acetic acid) was obtained as -55.5 kJ/mol, making the average value -55.0 kJ/mol.

Quote: Originally posted by Nicodem

I think what you keep neglecting is the energy required to deprotonate weak acids. HCl(aq), AcOH(aq) or NH4+ can not give the same amount of heat upon reaction with OH- even if you could somehow eliminate all enthalpy changes resulting from ions solvations and desolvations. The energy required for full dissociation (deprotonation energy) of these three acids is very different (given by pKa) and this energy must be detracted from the -57.3 kJ/mol value you give for the H2O "association". In the case of strong acids which are more or less fully dissociated in water this energy is close to zero, but with weak acids like acetic or ammonium it is certainly the major factor. I think that only in reactions such as the neutralisation of (hydrogen)carbonates (where the strongly solvated anion changes into a species that leaves the solution) is the solvation change is the prevalent factor explaining the overall enthalpy.

(Snip...)

The enthalpy of dissolution is no direct indication of the solvation enthalpy, certainly not so in dissolving salts where the enthalpy is the difference between solvation and lattice energy. OK, the dissolution of NH3 is more directly connected with the solvation, but I think the most direct indication of the solvation energy would be the dilution enthalpy. Anyway, the solvation enthalpy of the formed NH3 goes in the overall enthalpy, thus it is consumed in the desolvation of the ammonium ions and N-H bond breaking (as Woelen noted above). Obviously this energy per self would not be enough, but since there is also the H2O "association" energy released, the overall reaction is what it is, a very slightly exothermic neutralisation.

Regarding the first part I think Nicodem is being a little unfair to me, in the sense that I took the deprotonation reaction into account right from the start. But it doesn't explain everything IMHO. Let me reiterate my initial thoughts on the reaction enthalpies for the neutralisations of weak acids with strong bases:

1) Each reaction produces one mol of water (for monoprotonic acids) and thus - 57.3 kJ/mol of water MUST be released.

2) The Gibbs Free Energy of the deprotonation (index d) reaction (for weak acids i.e. K < 1): HA + H2O ---> H3O+ + A- is positive because:

ΔGd = -RTlnK and K < 1

For instance for acetic acid pKa = 4.76, Ka = 10^-4.76. This value has to be divided by the water constant of 55.55 mol/l, so K = 3.13 10^-7

and so isothermal ΔGd = 8.31 J/ K mol x 298 K x ln (3.13 10^-7) = + 37.1 kJ/mol

Of course ΔGd = ΔHd - TΔSd. We can't assume T ΔSd ≈ 0 because then ΔHd ≈ + 37.1 kJ/mol and that doesn't fit -55.0 = - 57.3 + 37.1. So either we assume ΔGd is mostly made of entropy or other enthalpies must play a part.

Incidentally, if for arguments sake we assumed NO other enthalpies play a part, it can the be shown that ΔSd = - 0.115 kJ/K mol, which would suggest the dissociated acid to be less disorderly than the non-dissociated form and that doesn't seem to make a lot of sense to me. Take KCl for instance: endothermic when dissolved because the state K+ (aq) + Cl- (aq) + H2O (l) is more disordered than the state KCl (s) + H2O (l).

So I do believe solvation energies play a part. But how precisely?

Incidentally, for the reaction NH4+ (aq) + OH- (aq) ---> NH3 (aq) + H2O (l)

with Ka = 10^-9.25 and K = 1.01 10^-11, @ 298 K, ΔGd = + 62.7 kJ/mol and that doesn't fit either.

[Edited on 3-8-2010 by blogfast25]

Nicodem - 3-8-2010 at 07:06

While reading your calculations it occurred to me that I was quite wrong about solvation enthalpy and that this phenomenon actually only explains the (bi)carbonate neutralization. The reason is in that the (de)solvation enthalpies are already included in the dissociation enthalpy! This becomes obvious if you for example read the dissociation equilibrium reaction:
AcOH(aq) + H2O <=> AcO-(aq) + H3O-(aq)

Obviously here the energy change between solvated acetic acid and solvated acetate anion is already included and these are also the starting and ending species in the neutralization of acetic acid.

(Bi)carbonates are an exception because the dissociation equilibrium reaction and the neutralization reaction are not complementary:
HCO3-(aq) + H2O <=> H2CO3(aq) + OH-(aq)

The subsequent release of CO2 is not included in the equilibrium described by pKa1 and pKa2, so the enthalpy of this missing reaction needs to be included in the neutralization enthalpy:
H2CO3(aq) <=> CO2(g) + H2O

If you think entropic factors in these reactions are large, then maybe you can take some time and do a few measurements of AcOH + NaOH reaction at 3 or 4 different temperatures so that we can calculate its entropy. Better waste a bit of time playing with your calorimeter than being unable to sleep.

blogfast25 - 3-8-2010 at 07:44

Quote: Originally posted by Nicodem

If you think entropic factors in these reactions are large, then maybe you can take some time and do a few measurements of AcOH + NaOH reaction at 3 or 4 different temperatures so that we can calculate its entropy. Better waste a bit of time playing with your calorimeter than being unable to sleep.

That's a nice idea. Perhaps for ammonium sulphate: for ammonium neutralisations the discrepancy seems to be the largest. Not to mention that I've plenty ammonium sulphate... Perhaps 0 C, 25 C, 50 and 70 C or for starters 0 C and 75 C, just to see.

Your concern for my insomnia is touching... : -) In all 'seriousness'. I just think these are interesting little problems worth investigating...

I do see one potential complication: the enthalpy of neutralisation (H3O+ + OH-) may be temperature dependent. Normally making that correction for small temperature differences can be done mathematically but what's the heat capacity of a mol of H3O+ (LOL)? Or one mol of OH-? It'd be cautious to measure the enthalpy of neutralisation at at least two temps, just to be sure...

[Edited on 3-8-2010 by blogfast25]

woelen - 3-8-2010 at 22:44

Use dilute solutions, at most 3 mol/l. Then the heat capacity of the solution roughly is equal to the heat capacity of water. Of course, your temperature changes are less and harder to obtain accurately, but at least you do not have to worry about problems with wrong heat capacity.

On the other hand, if you use measurements at different concentrations (e.g. 1.5 mol/, 3 mol/l, 6 mol/l) then you can determine the heat capacity of the other than water compounds in the solution. You can do that by measuring the change of temperature for a certain solution. If you take a solution with exactly twice the concentration, then you expect a change of temperature, exactly twice as large (of course you also must take into account the heat capacity of your vessel, so you might have somewhat less than 2 times). Any deviation can be used to compute the effect on heat capacity of the dissolved chemicals. Repeat measurements in order to reduce the effect of random (non-systematic) errors.

blogfast25 - 4-8-2010 at 03:59

Thanks for the advice, Woelen, but it didn't really teach me anything I didn't already know. I've been using concentrations around 1 M for most experiments. I've always assumed the heat capacity of the solution is Cp, water but I know more precise methods are available. Regarding replicas, repeating each experiment with a 1 l 'calorimeter' (and old thermos flask) would be costly but I do have about 3 repeats and they show the method is remarkably reproducible if care is taken... Some 20 experiments have been conducted so far and this afternoon 2 more will be conducted.

blogfast25 - 4-8-2010 at 05:23

First up, the cold neutralisation of 1 mol of NH4+ (1/2 mol (NH4)2SO4) with 1 mol NaOH.

As a side observation, dissolving 1/2 a mol of ammonium sulphate in 800 ml of water caused the solution to cool down by about 1 C, so negligibly endothermic.

By contrast, dissolving 1 mol of NaOH in 200 ml of water the temperature rise was about 52 C, corresponding to about 44 kJ/mol released in dissolution enthalpy!

The neutralisation was carried out at T = 1.6 C and the measured enthalpy of reaction ΔH = -6.7 kJ/mol of water which isn't significantly different from the RT value of -5.1 kJ/mol.

This experiment will now be repeated at about 70 - 80 C.

Second, the neutralisation enthalpy of NaOH + HCl at T = 71 C. I found a value of about - 55.8 kJ/mol of water, so not significantly different from the RT literature value of - 57.3 kJ/mol.

I think this is due to the two temperature corrections that would normally have to be applied to go from ΔH, 298 to ΔH, T cancel each other because the heat capacity of H3O+ (aq) + OH- (aq) + H2O (l) is essentially the same as that of H2O (l).

I may repeat this experiment at near O C.

Nicodem - 4-8-2010 at 13:05

Blogfast, it is nice that you take so much time for these measurments and it is interesting to follow your results, but with all the effort invested I can't do otherwise but to remind you that the proper way to measure enthalpies is to use an electric heating element incorporated in the calorimeter/thermos to introduce heat (I^2×R×t) until the same ΔT is obtained as obtained in the reaction. Just measuring the ΔT with a thermometer and then using water's Cp is certainly not properly scientific. Though, at least you get similar errors by using similar approximations, but the numbers, even if good for comparison, are not reliable and can not be compared with literature values.

blogfast25 - 5-8-2010 at 07:12

That's a rather curious comment, Nicodem. "Not properly scientific"? Very sloppy formulation: if I had a penny for every undergrad student that had determined the neutralisation enthalpy or the latent heat of ice in a chemistry of physics practicum, using the method I use, I'd have a tidy sum. Your comment also potentially rubbishes much of what many experimenters on SM are doing. To judge the 'scientificness' of a method solely by its accuracy is contentious: Bang!, goes the Hadron collider experiment...

Measuring the enthalpies by means of coulocalorimety bypasses the problem of Cp,H2O not being temperature independent but only a good operator with good kit will get that method to work more precisely than the one I use.

It is not a question of "Just measuring the ΔT with a thermometer and then using water's Cp" either. Let me explain with an example, carried out this morning: the neutralisation of 0.5 mol NaOH in 500 g of water with 0.5 mol of HCl in 500 g of water. In fact no ΔT is determined.

The calorific value k of the flask had previously independently determined to be 86 J/K.

The NaOH solution loaded into the flask and closed, while I prepared the HCl solution from standardised HCl stock, was swirled for about 2 minutes.

Temp. of the NaOH sol. + flask = 24.4 C (closed flask)
Temp. of the HCl sol. = 17.8 C
After mixing and stabilising (closed flask) = 28.2 C

Enthalpy content of NaOH + flask, relative to O C; H1 = (4.1813 x 500 + 86) x 24.4 = 53.1 kJ
Enthalpy content of HCl, relative to 0 C; H2 = 4.1813 x 500 x 17.8 = 37.2 kJ
Enthalpy of content and flask after mixing and reaction, relative to 0 C; H3 = (4.1813 x 1000 + 86) x 28.2 = 120.3 kJ

Reaction enthalpy ΔH = H3 - H2 - H1 = 30.0 kJ, estimated neutralisation enthalpy = - 60.0 kJ/mol of water, for -57.3 kJ/mol listed. Yesterday I obtained - 62.7 kJ/mol but for one mol and using quite concentrated HCl.

Yes, it does rely on constant Cp but in most experiments the actual temperature does orbit around 298 K, bar a few exceptions.

I think more interesting criticisms would include:

* measurement of mass by volume
* neglect of heat capacity of solutes
* quality of used reagents
* possibly the use of too concentrated solutions, thereby introducing inadvertently dilution enthalpies.

Instead of being a bit all over the place, I will now actually concentrate on the systems HAc + NaOH and HCl + NaOH, addressing some of the concerns in that list...

Nicodem - 5-8-2010 at 09:46

Well, sorry, I did not mean to badly criticize your method. Actually I think you are doing a wonderful job. What I wanted to say, was that your way of measuring the enthalpy might be good for preliminary experiments, but I do not think it is a good idea to invest so much time and effort into measurements, but no effort in using a more reliable method. Your way of doing it does indeed give relatively good results, but how sure are we the +/- few kJ/mol are good enough? For example, are you sure your measurements will give a good enough data trend to calculate the entropy? It might end up with you doing days of experiments just to realize the data is too scattered to draw a enthalpy vs. temperature plot.
Incorporating a heating element inside your thermos can be easy, so perhaps you should consider doing it. All you need is a multimeter for measuring the current and potential across a resistor/heater. You can use an electronics resistor for a heater (just insulate the wires). I never heard of undergraduate students using anything else in their practice but this kind of control. This is the method we used at the university when I was a student. And that is how we measured the calorimeter constant at different temperatures.

Here are a few references that you might find interesting:
http://dx.doi.org/10.1016/0021-9614(77)90198-7
http://dx.doi.org/10.1021/ja01122a008
http://dx.doi.org/10.1021/ed045p57
http://dx.doi.org/10.1039/CT8885300865 (very old!)
http://dx.doi.org/10.1039/CT8875100593 (very old!)

I only read the abstracts, so I'm unsure if you can find anything interesting there (also, I do not have access to the first one).

blogfast25 - 5-8-2010 at 11:42

Nicodem:

Yes, nothing I can disagree with there. But I'll pass on the amperocalorimetric determinations for now. I'll check out these links, thanks...

What I think is needed is an experiment design that concentrates on determining the difference between the heats of neutralisation of the systems HAc + NaOH and HCl + NaOH, in my particular conditions, rather than try and determine absolute values for the enthalpies. After all it's the 'missing energy' that concerns me here. And signal to noise ratio of course needs to be determined, at least roughly. If that difference could be established with roughly known accuracy then repeating the design at different temperatures should reveal whether entropic effects are in play or not.

blogfast25 - 11-8-2010 at 07:16

The following experiment was carried out at RT (between 20 and 21 C initial temperature):

Two standardised solutions of 0.2 M nominal were prepared, one HAc (acetic acid), one HCl.

One solution of 0.2 M NaOH nominal but about 5 % stronger than the acid solutions was prepared.

An amount, precisely weighed, of about 200 g of the acid solutions was reacted with a slight excess of precisely weighed NaOH solution (also about 200 g). The precise amounts of acid neutralised where thus known for each run. These amount were 0.03 - 0.04 mol.

Temperatures were determined as outlined above and the heats of neutralisation were obtained. Three runs were recorded for each acid. The results are as follows:

1) For the heat of neutralisation of 0.2 M HCl with 0.2 M NaOH the values -56.4; - 56.1; -58.5 kJ/mol were obtained with an average of -57.0 kJ/mol and SD = 1.6 kJ/mol.
2) For the heat of neutralisation of 0.2 M HAc with 0.2 M NaOH the values -42.1; - 38.4; -39.8 kJ/mol were obtained with an average of -40.1 kJ/mol and SD = 1.9 kJ/mol.

Note that there's more spread in values than I anticipated and that for HCl + NaOH I'm now quite close to the literature listed value of -57.3 kJ/mol. The real value of -57.3 kJ/mol can probably only be determined by determining the limit for conc. ---> 0.

The value for HAc + NaOH is considerably less than previously obtained in more concentrated solutions (0.8 M) and the more reliable value is the one in more diluted conditions (I feel) of -40.1 kJ/mol.

Pooling together both SDs (they are not significantly different), the relative difference between the heats of neutralisation of HCl and HAc is -57.0 - (-40.1) = -16.9 kJ/mol +/- 3.5 kJ/mol.

That difference is higher than previously estimated but still doesn't account for the Free Energy change for the deprotonation of HAc, estimated to be +37.1 kJ/mol, at least not if we consider most of that Free Energy to be enthalpy (but we don't know that, of course).

As Nicodem suggested I will now rerun some of the experiments at higher temperature, as significant changes in the results would indicate that TΔS may play some part. From preliminary results with HCl + NaOH reported above, as well as from theory, I will assume the neutralisation of HCl + NaOH isn't temperature sensitive.

kmno4 - 11-8-2010 at 09:13

Your value of heat of neutralisation of CH3COOH with NaOH solution is in error.
I do not know what, but you do something wrongly.
Literature values for neutralisation of ~0,3 M solutions are:
NaOH - HCl : 57,5 kJ/mol
NaOH - CH3COOH : 56,1 kJ/mol
Values found on some web page are 57,1 and 55,9 kJ/mol and also here http://dx.doi.org/10.1021/ja01378a011 (worth of reading)
BTW.
Thermodynamic data for reaction
CH3COOH(aq) -> CH3COO(-)(aq) + H3O(+)(aq)
are (in kcal/mol, rewritten from book by G.Kortum) :
ΔG = 6,49 ( = -RTlnKa) ; ΔH = -0,09 ; TΔS = - 6,58

[Edited on 11-8-2010 by kmno4]

entropy51 - 11-8-2010 at 10:45

blogfast25, how are you standardizing your solutions of acid and base? What standard acids and bases are you using for comparison? What indicator are you using for your titrations? Are you checking the pH in some way after your calorimetry to verify that you had equivalent amounts of acid and base? Thanks!

blogfast25 - 11-8-2010 at 12:29

Quote: Originally posted by kmno4

Your value of heat of neutralisation of CH3COOH with NaOH solution is in error.
I do not know what, but you do something wrongly.
Literature values for neutralisation of ~0,3 M solutions are:
NaOH - HCl : 57,5 kJ/mol
NaOH - CH3COOH : 56,1 kJ/mol
Values found on some web page are 57,1 and 55,9 kJ/mol and also here http://dx.doi.org/10.1021/ja01378a011 (worth of reading)
BTW.
Thermodynamic data for reaction
CH3COOH(aq) -> CH3COO(-)(aq) + H3O(+)(aq)
are (in kcal/mol, rewritten from book by G.Kortum) :
ΔG = 6,49 ( = -RTlnKa) ; ΔH = -0,09 ; TΔS = - 6,58

[Edited on 11-8-2010 by kmno4]

If you look a little higher up, I found two vales for HAc neutralisation of -54.5 and -55.5 kJ/mol, using 0.8 M HAc. But the latest results, at 0.2 M were much lower. That part will now be repeated as I can see nothing particularly wrong with the raw data.

As regards the calculation of ΔG by G.Kortum, are you sure you're copying correctly? That part has been elaborated on higher up: ΔG = -RT lnK, not Ka, with K = [H3O+][Ac-]/[HAc][H2O] for the reaction:

CH3COOH(aq) + H2O(l) -> CH3COO(-)(aq) + H3O(+)(aq)

For acid/base calculations [H2O] can be incorporated in Ka but for thermocalcs you need the real thing: K, not Ka.

I'm intrigued about the values for ΔH and TΔS, where does he get them from?

Quote: Originally posted by entropy51

blogfast25, how are you standardizing your solutions of acid and base? What standard acids and bases are you using for comparison? What indicator are you using for your titrations? Are you checking the pH in some way after your calorimetry to verify that you had equivalent amounts of acid and base? Thanks!

A fundamental standard was created as a solution from a known quantity of pure anhydrous Na2CO3 (washing soda, filtered and very carefully recrystallised, then dehydrated for 1 h @ 110 C (till constant weight)). Used immediately to prepare the standard solution, 0.1 N with known titre, using a calibrated measuring flask.

Against it is titrated a HCl solution, using a 0.1 N HCl solution as titrant and determining the last end point by means of pH meter (direct pH readings, continuously throughout titration) to determine solution's titre. The pH jump is used (throughout all titrations) to determine end point. It is very sharp in all cases. I use a 25 ml volumetric burette.

Against the standardised HCl solution is then titrated a 0.1 N NaOH solution, using pH meter as end point determinator.

Standardised NaOH and HCl solutions are the used to calibrate HAc and HCl 0.2 M solutions, using the pH meter.

I accept there are more accurate ways of doing this but having no pure oxalic acid and no commercially standardised solutions, I have to 'make do'. Suggestions are most welcome.

Final pH of neutralisations is always checked: a small excess of base is used to ensure full neutralisation of the acid.

As it happens I prepared a new sample of pure anhydrous Na2CO3 today and will verify the titre of the master solutions of NaOH and HCl tomorrow.

[Edited on 11-8-2010 by blogfast25]

entropy51 - 11-8-2010 at 12:44

Quote: Originally posted by blogfast25

I accept there are more accurate ways of doing this but having no pure oxalic acid and no commercially standardised solutions, I have to 'make do'. Suggestions are most welcome.

It sounds as if you're being fairly rigorous in your standardizations! The only suggestion I would make is that I try to compare to more than a single standard, and when I do this I often find that the Na2CO3 is the outlier.

I recrystallize technical grade oxalic acid several times for a standard. It seems to work better for me than Na2CO3, but that may just be me.

I also use accurate dilutions of the constant boiling HCl which I find to probably be the most accurate standard for me. This one is not too difficult if you have distillation equipment and an accurate balance.

Keep up the good work!

blogfast25 - 11-8-2010 at 12:49

entropy51:

I have had some problems with Na2CO3 in the past, particularly with turbid solutions which turned even more turbid overnight! I seem to have that under control now.

Explain your constant boiling HCl method, please...

blogfast25 - 11-8-2010 at 13:09

Also, when using oxalic acid, do you use the dihydrate or anhydrous as a standard? If the latter how do you dehydrate? One source mentions possible brown or black spots if you get the dehydration at just below 100 C (IIRW) wrong...

entropy51 - 11-8-2010 at 14:42

The constant boiling HCl method is in the literature. Essentially a somewhat diluted HCl solution is distilled until a constant boiling point is reached and that fraction is collected. The BP depends on the barometric pressure and there are tables for finding the exact concentration of HCl in the distillate. A weighed amount of the distillate is diluted to produce a standard solution. If Google doesn't turn it up, I can find a citation for you. I'd be surprised if it hadn't been posted on the forum already.

I used the oxalic acid dihydrate. IIRC I dried the recrystallized oxalic acid in a vacuum dessicator because it can indeed decompose if heated too much. But so can Na2CO3. If the Na2CO3 method works for you that is good, but I would cross check it against another primary standard just to be sure. At one point I did a detailed comparison of the methods until I found agreement between at least two methods and that point I began to trust the results.

Beware of Na2CO3 contamination of your NaOH. Apparently a good method, which I use, is to make a 50% solution of NaOH. Over time the Na2CO3 will precipitate out of this and the clear supernatant can be diluted and standardized.

blogfast25 - 12-8-2010 at 04:04

entropy51:

Yes, the HCl azeotrope method re-occurred to me last night. It's within my capability envelope.

Na2CO3 is a tricky one. Hydrolysis of the carbonate is the main danger. I'm revising a new procedure based on decomp. of bicar (rather than starting from commercial washing soda) and new recryst. procedure of the thus formed Na2CO3. I'll keep you posted.

Carbonate in NaOH is a worry: precautions are taken but you can't do much against stuff that's already in the product. I will definitely try the 50 % method, almost immediately actually.

[Edited on 12-8-2010 by blogfast25]

kmno4 - 12-8-2010 at 06:05

Value ΔG = 6,49 kcal/mol strictly corresponds to thermodynamic constant K=1,754x10-5 for acetic acid
( ΔG= -RTlnKa) and you can calculate it by yourself.
K contains activities and for diluted solutions aH2O=1.
Parts ΔH and ΔS are calculated on the basis of measurements of temperature coefficient of K .
Unfortunately this is possible only for weak acids, for which K is small and independent (almost) on concentration.
Author of my book gives reference to unavailable electronically article:
http://adsabs.harvard.edu/abs/1949ZNatA...4..171B
for more datails.
BTW.
Na2CO3 can be estimated in NaOH by double titration of the same sample: first with phenolphthalein and next with methyl orange.

blogfast25 - 12-8-2010 at 06:45

A new 0.2 M HAc solution was prepared and standardised.

This afternoon, three runs for the heat of neutralisation of that solution with 0.2 M gave -55.6; -56.4; -56.9 kJ/mol, average ΔH = -56.3 kJ/mol, SD = 0.7 kJ/mol. That's again much in line with previous values and literature. I believe the previous faulty previous runs to be caused by an unknown standardising error, or at least that sounds like the most plausible explanation.

Assuming kmno4 is correct, then the mystery is essentially solved: if Ka is indeed the thermodynamical K for aH2O = 1. The heat of neutralisation of a weak acid like HAc isn't much different from that of a strong acid and most of the deprotonation Free Energy manifests itself as entropy.

I will try and confirm that by running three neutralisation tests with HAc at near zero C, at 50 C and at 75 C, a total of four estimates.

What's still quite amazing is that for very, very weak acids like NH4+ the actual neutralisation enthalpy is practically zero: there the entropic effect must be even higher...

Oooops... I've just noticed that kmno4's values are in kcal/mol, not kJ/mol:

Quote: Originally posted by kmno4

Thermodynamic data for reaction
CH3COOH(aq) -> CH3COO(-)(aq) + H3O(+)(aq)
are (in kcal/mol, rewritten from book by G.Kortum) :
ΔG = 6,49 ( = -RTlnKa) ; ΔH = -0,09 ; TΔS = - 6,58

Converting:

ΔG = +27.1 kJ/mol ( = -RTlnKa) ; ΔH = -0.38 kJ/mol ; TΔS = - 27.5 kJ/mol

Due to the small value of ΔH this changes very little...

[Edited on 12-8-2010 by blogfast25]

kmno4 - 12-8-2010 at 08:36

Yes - the book I mentioned ( translated) is “Lehrbuch der Elektrochemie” by G. Kortüm from ~1960 and in older references cited there they use kcal.

I have noticed this:

6) The neutralisation enthalpy of NH3 with HCl was experimentally found to be -9.7 kJ/mol of water, so also much lower than - 57.3 kJ/mol. The strength of both the NH3 and HCl solutions have been verified with acid/base titrometry.
Is this your experiment ?
From another book this is 12,72 kcal/mol (25 C, 0,3 M).
BTW.
There should be only small heat effect in reaction NH4(+) with NaOH.
It is because neutralisation of CH3COOH gives almost the same effect as neutralisation of HCl.
It would correspond to reaction NaCl + NaOH (zero heat).
BTW 2.
For NH4(+) acid:
ΔG=12,56 kcal/mol; ΔH=12,4 kcal/mol; TΔS=-0,16 kcal/mol

Lambda-Eyde - 12-8-2010 at 08:39

Quote: Originally posted by blogfast25

Also, when using oxalic acid, do you use the dihydrate or anhydrous as a standard? If the latter how do you dehydrate? One source mentions possible brown or black spots if you get the dehydration at just below 100 C (IIRW) wrong...

http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv1...

blogfast25 - 12-8-2010 at 09:03

Quote: Originally posted by kmno4

6) The neutralisation enthalpy of NH3 with HCl was experimentally found to be -9.7 kJ/mol of water, so also much lower than - 57.3 kJ/mol. The strength of both the NH3 and HCl solutions have been verified with acid/base titrometry.
Is this your experiment ?

Yes, that was my experiment. I wouldn't put my hand in the fire on that value as it's a single data point, obtained at 0.5 M, with a slightly different method. But it's not far from the value you quote.

For NH4+ and NaOH I found about DH = -5.0 kJ/mol.

All in all, it seems that entropy is explaining a lot here... Good, I like a bit of disorder ;-)

[Edited on 12-8-2010 by blogfast25]

blogfast25 - 12-8-2010 at 09:09

Lambda-Eyde:

That was actually the chemtext I was referring to...

blogfast25 - 14-8-2010 at 10:40

Method for primary base standard from washing soda

Firstly, some fundamental data about Na2CO3's interesting solubility behaviour can be found here:

http://www.genchem.com/properties.asp

Don't forget to click on the Fig. 2-1 hyperlink, it's the phase diagram for Na2CO3/H2O

Useful factoid: the Na2CO3 content of a mixture of A parts Na2CO3.10 H2O and B parts of water can be calculated from:

% (anh.) Na2CO3 = 37.1 x A / (0.629 x A + B)

1. 200 g of washing soda were dissolved into 200 ml of tap water (about 23 %, based on Na2CO3.10 H2O) and brought to about 80 C quickly. Do NOT boil to avoid loss of CO2 (hydrolysis). Filter till clear. I preheat my filter to about 100 C to avoid any crystalisation on the filter. In my case there's a fine, white powder left on the filter which I presume is mainly NaHCO3. I will verify that.

2. Allow the filtrate to cool or force cool it, then cool further on ice bath. Crystals of relatively pure Na2CO3.10 H2O form and grow visibly. I stopped cooling at about 14 C which gave a nice crop of wholesome crystals.

3. Decant off the supernatant liquid and collect the crystals on a clean tea strainer or ceramic filter. I use a SS fine mesh (+/- 0.5 mm) strainer. Allow to drip, then wash with small amounts of ice cold deionised water (DIW). These crystals are undoubtedly already sufficiently pure for much analysis and synthesis, so set some aside.

4. For the second crytallisation, we take advantage of the fact that soda's maximum solubility is 33.2 % at about 35 C and drops steadily towards 0 C.

Because the slush contains so much liquid water, not much needs to be added. I've had success with adding 1, 2 or even 0 parts of DIW to 10 parts of Na2CO3.10 H2O/water slush. I think 2 parts probably works best. Heat that sludge au bain marie to about 35 C, making sure all crystals dissolve. Allow to cool or force cool on ice bath. Due to purity, often the solution is reluctant to crystallise and you may need to add a crystal from the first crop: the solution then beautifully crystallises all out at once.

5. Repeat 3. Allow time to drip to almost dryness. Dehydrate to constant weight in oven at 110 C. Use a plastic (PP) or silicone cup for this: anhydrous sodium carbonate sticks like mad to glass. Use immediately. I do believe it can be stored in CaCl2 desiccator for some months.

6. Quality test: dissolve some in DIW, a clear solution has to be obtained, without turbidity or insoluble bits. I've become so paranoid about this I leave the test solution standing overnight for a final check.

[Edited on 14-8-2010 by blogfast25]

entropy51 - 14-8-2010 at 11:12

blogfast25, when you oven dried the decahydrate to constant weight, did it lose the theoretical amount of water?

blogfast25 - 15-8-2010 at 04:39

Quote: Originally posted by entropy51

blogfast25, when you oven dried the decahydrate to constant weight, did it lose the theoretical amount of water?

In a certain sense the decahydrate doesn't really exist: the sludge from the second crystallisation is impossible to dry without the some of the crystals losing some water. I see this on all utensils used in contact with the Na2CO3.10 H2O solutions: apart from some clear, well formed crystals forming, in air quite a bit of them effloresce. The commercial 'washing soda' also looks like a mixture of decahydrate and lower hydrates and commercial products usually mention 'min. x % sodium carbonate decahydrate', never 'pure sodium decahydrate'. Holleman confirms the effloresence.

Pure Na2CO3.10 H2O may possibly obtained by gentle drying at low temp. under mild vacuum? It's a shame because the decahydrate has a much higher gravitational lever, better for primary standards..

So the decahydrate second crop is Na2CO3.10H2O/liquid water sludge. This is then oven dehydrated. Ergo I cannot know whether the obtained weight loss is in accordance with the theoretical weight loss.

[Edited on 15-8-2010 by blogfast25]

blogfast25 - 15-8-2010 at 08:58

Well, well. At first glance the white insoluble matter found in my washing soda doesn't appear to be NaHCO3.

I took 6 g of the still wet stuff and reacted it with 10 ml 20 % HCl. Much effervescence AND considerable heat evolves but not everything dissolves. The heat is strange: it would seem to exclude NaHCO3 and CaCO3 (I checked: a bit of limestone with 20 % HCl doesn't generate much heat on dissolving in HCl).

The solution was filtered and diluted a little and strong (5 M) NaOH was added: a white, gelatinous precipitate dropped out, quite a bit of it too. The alkali metals wouldn't do that and Ca(OH)2 isn't particularly gelatinous as a precipitate, IIRW. It was filtered and washed slightly, then scraped of the filter. Strong Na2CO3 was added to the filtrate, which didn't respond.

To the precipitate was then added strong Na2CO3 and the solution was heated to test the displacement Ca(OH)2 (s) + Na2CO3 (aq) --> CaCO3 (s) + 2 NaOH (aq) but apart from the gelatinous precipitate dispersing through the solution, nothing much else happens.

I'm now drying some of the white stuff for some further testing...

watson.fawkes - 15-8-2010 at 09:41

Quote: Originally posted by blogfast25

Well, well. At first glance the white insoluble matter found in my washing soda doesn't appear to be NaHCO3.

In the US at least, much washing soda is derived from trona, so you might consider mineral-type contaminants such as borates, silicates, sulfates etc.

arsphenamine - 15-8-2010 at 11:07

Quote: Originally posted by watson.fawkes

In the US at least, much washing soda is derived from trona, so you might consider mineral-type contaminants such as borates, silicates, sulfates etc.

I would guess primarily silicates,
similar to sodium silicate (aka "water glass") which
has been used to clarify turbid waste water.

Particularly, it precipitates metals from solution.

blogfast25 - 15-8-2010 at 12:06

watson and arsphenamine:

Sure but what I' seeing isn't really consistent with those...

watson.fawkes - 15-8-2010 at 12:54

Quote: Originally posted by blogfast25

Sure but what I' seeing isn't really consistent with those...

Shale is another impurity of trona, but I think that's even less likely. Worth a guess, I suppose. The only other thing I've got is that if you've got output from a Solvay process, there's a different class of impurities.

[edit] P.S. It's also conceivable that your washing soda isn't even notionally pure, that it's been compounded for laundry. I mention this with the full disclosure that occasionally I have forgotten to check that something is plugged in.

[Edited on 15-8-2010 by watson.fawkes]

blogfast25 - 16-8-2010 at 06:37

Quote: Originally posted by watson.fawkes

Quote: Originally posted by blogfast25

Sure but what I' seeing isn't really consistent with those...

It's possible. It's also possible it's straight Solvay soda. It's only 1 £/1 kg in retail.

About half of the impurity dissolves in strong HCl, the other half doesn't. I'm now investigating the first half...

blogfast25 - 16-8-2010 at 12:02

The packet states 'min. 30 % decahydrate', which is definitely referring to the (poor) state of hydration. Endearingly it also states:' no petrochemicals', so in all likelihood it's not compounded.

Some of the dried water insoluble residue was treated with strong HCl and about half of it does dissolve in it with effervescence, the other half doesn't. Filtering and diluting a bit, the filtrate was then split into 2 equal portions. One was neutralised with strong Na2CO3: no precipitation even at pH = 11. The other was neutralised with strong NaOH and a whitish, gelatinous precipitate forms. I'll check the latter for amphoterism and complexation with NH3.

kmno4 - 17-8-2010 at 09:16

Quote: Originally posted by blogfast25

Quote: Originally posted by kmno4

This is rather very big difference between your -9,7kJ/mol and
literature -12,72 kcal/mol (=53,2 kJ/mol) value.
BTW
For reaction in kind of "salt1+salt2 -> something" you can estimate enthalpy of rection from available tables of thermodynamic data for aquaions and pure substances. In this way you can predict (to some extent) heat of rections: NH4(+) + OH(-) -> NH3 + H2O; CO3(2-) + 2H(+) -> CO2 + H2O and many other.

blogfast25 - 17-8-2010 at 11:44

Quote: Originally posted by kmno4

I'm inclined to stand by my value, as I remember the experiment yielding a very low ΔT. But I'm willing to repeat it at 0.5 M. It will take a few days because of the need to restandardise NH3 and other commitments. I'll also check the raw data of the experiment.

blogfast25 - 17-8-2010 at 11:58

As regards the white, water-insoluble residue in my washing soda, it's almost certainly MgO or Mg(OH)2, added probably as an anti-caking agent. I isolated some more and washed it carefully. It dissolves effortlessly in 1 M H2SO4, with mild turbidity and generation of considerable heat but no bubbles. That's been filtered and the salt will be crystallised tomorrow.

The stuff was also tested with very strong NaOH: it's not amfoteric. It doesn't react with strong NH3 either.

A cation that doesn't form an insoluble chloride, nor insoluble sulphate, nor insoluble carbonate but does precipitate a white, gelatinous hydroxide (presumed)? That looks very much like Mg2+...

blogfast25 - 18-8-2010 at 08:07

Well, using a new bottle or standardised NH3, I now get a value of -64.7 kJ/mol for NH3 + HCl at about 0.2 M! So that's over-shooting but shows the previous value of -5.0 is erroneous (although I can't figure out why from the raw data).

The value for NH4+ + NaOH will also have to be repeated.

But the value of -64.7 kJ/mol was obtained in slightly dodgy circumstances: it turned out that the concentration of NH3 in the latest bottle of 'household NH3' is only 0.37 w% (0.216 M)! The previous bottle was bad enough at 4 w% but this takes the biscuit... Dry distillation of garden grade ammonium sulphate with NaOH, here I come!

My local hardware store stocks some citric acid monohydrate ('no additives' it says), so I'll see if that's any use as a primary acid standard. Interesting solubility (and other) data on citric acid here (third search result from top, a *.DOC download):

http://www.google.co.uk/#hl=en&q=citric+acid+solubility&...

The solids content of my washing soda turns out to be about 38.6 w%, so fairly close to the 'theoretical' value of 37.1 w%.

[Edited on 18-8-2010 by blogfast25]

blogfast25 - 20-8-2010 at 12:41

Well, my first efforts at recrystallising citric acid monohydrate haven't been very successful. The raw product is a neat, white, well formed crystalline product, seemingly of 'good quality'.

Citric acid is extremely water soluble (over 90 w% @ 100 C and still over 50 w% @ 10 C) and forms syrupy solutions. Cooling these down to RT (say from 80 C saturated) yields no crystals but allowing cooling at about 5 C overnight yielded a nice crop. The problem is that the supernatant liquid is still highly concentrated. Washing with cold DIW is possible but drying the wet crystals (of presumably monohydrate) is hard: at 50 C more syrupy solution is formed between the crystals and the wash water and as the syrup is highly concentrated, driving off water is nigh impossible at atmospheric conditions.

I will titrate a known quantity of (raw) citric acid against a secondary NaOH standard, using the first two equivalence points, just to see.

The white residue in my washing soda is definitely MgO or Mg(OH)2. Dissolved in H2SO4 and filtered I obtained nice, big clear crystals. After washing and redissolving in DIW, adding Na2CO3 yields some fizz and upon heating the white, basic carbonate Mg(OH)2.3MgCO3.3 H2O (acc. Holleman) precipitates from the solution.

The only test left is for MgNH4PO4.6H2O but I've no H3PO4 at hand.

blogfast25 - 23-8-2010 at 08:28

The value of the reaction heat of NH4+ +OH- ---> NH3 + H2O has been confirmed at -5.2 kJ/mol. Using 400 ml of standardised NH4+ (about 0.4 M), 400 ml containing a slight excess of NaOH, the temps. were: TNaOH = 22.3 C, TNH4+ = 17.9 C, Tend = 20.3 C. Mols of NH4+ neutralised = 0.172 mol. ΔH = -5.2 kJ/mol tallies perfectly with three other values of around -5.0 kJ/mol obtained with NH4Cl at 20 C, (NH4)2SO4 at 20 and at 0 C and now this one. There's just no heat in it.

Music to my ears because I'll be dry distilling (NH4)2SO4 with NaOH and can do without runaway heats...

peach - 23-8-2010 at 15:31

This is an interesting discussion, and I will be scratching my head about it also.

I'm sure I remember talking about something similar in A-Level chemistry, but alas, I'm a biology boy over a chemist and so some of it has drifted into the darkness.

Given the odd measurements you're getting and the impurities you've found in the materials, it may be worth stopping and harvesting together something you can be more sure of (as per the long delay in that reply appearing in the AlCl3 thread, which will be followed by some more delays as I do the same again). If you're putting in so much effort, you'll undoubtedly want a solid result.

I have some questions regarding the practical side of things. I'll bold them to make them more immediately separate.

What resolution is the pH meter, and is it a pool kind or a lab kind? Is it temperature compensating?

I have one that displays to 0.001. That's useless for everyday neutralizations where I can pretty much get what I want by eye (without an outside indicator), but it does show up some of the more hidden effects of pH. For example, I can neutralize to 7, then sit and watch the meter and I can see it drifting back up or down to where it started because of all the zeros. Quite often, it can drift a fair way. I can accelerate to a true pH by watching the drift rate (which won't appear on the .1 meters for a long while).

A prime example of this is cannabis growers (and the guys growing those big watery tomatoes) using hydroponic feed bins. If the solution isn't buffered, they can set the pH (well below where it should be), walk away, and it'll be back over 7 the next day or two as the pH is still stabilizing a long time later. It can happen over an entire week, so using tiny solution volumes and molarities would help with that, and make the calorific measurements harder; ah, nature.

If your meter is reading to .1 and it's the pool type, that effect may be hiding from your on looking gaze, causing you to think it's fully neutralized prior to it being so.

When I'm going for gold on the pH, it needs to be in an Erlenmyer with a big stir bar (I'm talking 75mm, the biggest that'll fit) running at full speed and it needs to sit for ten minutes between additions as I approach where I want it; I'll leave it for half an hour sometimes. The meter functions as a datalogger. When I'm done giving myself cancer from whatever the hell is now lurking in the AlCl3 orange solvent of the flask, I could plot a graph of it happening.

I refer to it as 'bouncing'.

I'm guessing you've calibrated it against a solid buffer. The pool ones often only have one calibration point, calibrate at or close to your end point (of coarse). If it only has one point, leaving it at the end point to check the stock solutions will skew the reading it gives for those. NurdRage on youtube has a video all about accounting for the skew between points on a wonky meter.

Accelerating heat loss

Remember not only does it take energy to warm the liner of the thermos up (and the water it's holding), it's being lost through the thermal bridges. And the rate will increase as you experiment at higher temperatures, which could squiffy up the tolerance bands. So you may wish to check the bridging by holding the thermos contents at certain temperatures and watching the energy it's taking. As Nicodem suggests, that could be as simple as a resistive heater left running for a while with a multimeter on it.

Pure materials

I have bottles of glacial and oxalic acid hanging around the house. The oxalic is a mite treatment for the bees, the acetic I bought ages ago and haven't used much. The former is only 3 or 6% I think, in sugar solution made up with distilled water (so that'd need cleaning up). The glacial I think is fairly clean, it does what it's name suggests.

If you're concerned about purities, how much are you talking about using? If it's not a lot, I can put some in the post free, for the good of science.

I'm sure I could pick up some pure materials from the lab supplier for some paypaling to cover the cost. I doubt someone bothered about enthalpies, entropies, neutralizations, the finer points and in need of technical grade materials is baking crack. I may be about to order up some more things (for my gelatinous mess experiments), so now may be your chance if you're having problems finding pure materials.

I have a 0.1mg balance that was just recalibrated by Avery Weigh Tronixs (it came out of Air Product's cryogenics lab). If you're having measuring issues and want the tolerances tighter, and are willing to paypal some £ for the materials I'll have to add to the order, I can premeasure stocks of them and print you receipts on the weights for dilution.

NurdRage also has a video about manganese dioxide he bought from eGimp, in which he dissolves the manganese, has a look under the microscope, and it's cut with sand. Tisk tisk...

I'm sorry if I'm teaching you how to suck eggs here, I'm not sorry for trying to help.

[Edited on 23-8-2010 by peach]

blogfast25 - 24-8-2010 at 05:27

That's a long post, John, I won't be able to address every single point but I'll try.

Firstly, my enthalpic measurements are of course approximate. I do my best but have my limitations: it's a roomy potting shed, not a state-of-the-art conditioned laboratory!

On pH measurements, I use a regularly calibrated (two points) 'pot grower' model which has served me excellently over the years. 0.1 pH point resolution. Max. temp. 50 C, but I never use it above 25 C. I treat it like the Queen bee!

For acid base titrations, it's the pH jump that indicates the end point, not some absolute value. Not perfect but more precise than indicators. Sometimes I use both. For titrations I haven't done before a 'master pH v. volume' curve is recorded to actually see the pH jump clearly and graphically. Currently I'm having bizarre problems with OTC citric acid. Need to confirm or I'll have egg on my face...

For the neutralisation experiments, pH doesn't even come into it. I standardise the acid solution, use a precise volume, then neutralise with a small but known excess of alkali. pH is only checked at the end to confirm it is higher than the theoretical end point of the acid (pH of the conjugated base).

For acid/base standard for the moment I use very carefully twice recrystallised and dehydrated soda. I've perfected that, even if I say so myself. I don't have purified oxalic acid. If you have some to spare, by all means send me some. Use the U2U to arrange something.

Scales to 0.1 mg? In my dreams! These things from Sartorius cost up to £5,000! I'm investing in a 0.001 g scales. But currently I manage 4 significant digits, that's OK for my purposes. A general lab upgrade is also in the works, considerable investment...

Thermos flask: heat losses are severely limited by the small temperature difference between in and out of the flask: rarely does the temperature in the flask exceed 25 C. But I get your point. And take the greatest care possible (okay, that's probably an exaggeration ;-) )

Manganese oxide and other oxides: beware of the 'Potters' (pottery grades), many are cut with silica (maybe just sand, of course), from my experience with various thermites/analysis. For glazing purposes a bit of the old silica is of course a bonus!

My pride and joy from the last years:

http://www.popsci.com/node/30347

Now I'm off to distill some ammonia. The things you've gotta do!

blogfast25 - 24-8-2010 at 12:14

Just finished the 'dry' distillation of ammonia from 0.5 mol garden grade ammonium sulphate and 1 mol household NaOH. Just mixed the crystals and pellets with a minimum of water into an old vinegar bottle, connected with silicone tubing (RC aircraft fuel lines!) to another vinegar bottle with about 0.5 L of iced DIW.

Gas starts flowing immediately and the reactor was then heated on steam bath. Amazing how the bubbles of NH3 get gobbled up by the cold water, with no bubbles surviving to the surface of the scrubber water! After 15 min gas evolution more or less stopped.

Resulting solution of pH about 11.8 and titrated as 2.56 w% NH3, which is about 1.5 M. Yield was thus about 75 %, not bad.

Interesting factoid: the silicone tube sits on a bit of brass tubing (yes, RC aircraft fuel line) that sticks through the cork and into the reactor. That bit of brass, exposed to moist NH3 had gone blue, presumably due to Cu2+ ammonia complex formation...

About half a mol of this ammonia will now be used for a neutralisation experiment.

peach - 24-8-2010 at 21:52

You seem to have those things I brought up under control.

The only thing I'd say is that the jump is indeed important, but it also depends on the solution volumes and concentrations being used. I've seen pH's jump, sit still, then drift a good way from where they're supposed to be as the solutions continue to react over ten minutes or more, for a few hundred ml's. With litres, and tens of litres, it can still be doing it days or a week plus later, even with gentle changes. Trying to acidify the garden at present, it's taking a year plus.

I have seen and used that RC silicon tubing before, it's nice stuff considering it's so readily available.

The brass thing doesn't surprise me at all. Whenever things start getting reactive, gas wise, things happen.

2007, 20kT of ammonia make a break for it at the Seward plant

The rusty stainless scissors from another post, left poking into the HCl(g) generator overnight, after it was empty;

The sink was splotless, prior to the HCl(g) being around, there are green patches and rust appearing all over it. The new chromed tap is going green at the stem;

blogfast25 - 25-8-2010 at 05:43

Regarding pH jumps, the ammonia titrations yesterday were carried out on undiluted, weighed samples (about 2 g) of the 1.5 M NH3, then diluted a little for the pH probe. Fine end points, no problem. But every time I read the endpoint at 5.0 to 4.3ish, the 4.5 drifted back up to 5 and beyond! Stirring made it go back down. I suspect NH3 gas in the headspace being absorbed near end point.

Well, my sink isn't quite as bad as yours but then it really is The Kitchen Sink!

Silicone tubing is fine form for impromptu, al fresco type 'apparati' but it's piss poor with acids: falls a part to almost pure silica very quickly.

Whenever things start getting reactive, gas wise, things happen.

If you can get it to react!

blogfast25 - 25-8-2010 at 08:33

OK. Another single data point for the neutralisation of NH4OH (aq) + HCl (aq) --> NH4Cl (aq) + H2O (l)

330.2 g of a 2.56 w% solution of NH3 (0.496 mol NH3) reacted with 330 ml of a 20.4 w% HCl solution (about 10 % HCl excess). TNH3 (in flask) = 18.5 C, THCl = 21.0 C, Tend (in flask) = 29.9 C.

Estimated value of ΔH = -58.5 kJ/mol of NH3. So a bit higher than the value for NaOH + HCl but considering my own observed standard deviations on very similar measurements, here for ammonia the value isn't substantially different from the actual neutralisation enthalpy for 1/2 H3O+ (aq) + 1/2 OH- (aq) ---> H2O (l).

I think I might yet try and include one more measurement, for the neutralisation of disodium citrate (say, Na2HCt): HCt 2- (aq) + OH- (aq) ---> Ct 3- (aq) + H2O (l). The third dissociation content of citric acid is very small...

Four hours later the temperature inside the flask had dropped to 28.5 C, more of a drop than I expected...

[Edited on 25-8-2010 by blogfast25]

blogfast25 - 26-8-2010 at 07:52

The trouble with citric acid monohydrate...

A week or so ago I created a master 'pH v volume' titration curve for titration of citric acid with standardised NaOH 0.1 M. I chose a citric acid concentration of about 0.05 M, hoping to titrate the first two equivalence points corresponding to pKa1 = 3.15 and pKa2 = 4.77. I dissolved about 7 g (accurately weighed) citric acid monohydrate (CAMH) in 750.0 ml of water.

On titration, the first part of the curve between pH = 2.5 and pH = 6.0 is almost linear and uneventful. The sharp first end point then appeared at pH about 9.

Problem: I used 20.0 ml of the CAMH solution but needed about 30 ml, just to reach the first equivalence point! That indicated a molarity of the CAMH solution of 0.15 M, NOT 0.05 M. The titration was repeated a few times, with very similar results.
I checked my calcs and weigh-up and couldn't find fault with it and decided to leave well alone for a couple of days.

Well, today I made a 0.1 M solution of CAMH: 15.76 g of CAMH in 750.0 ml of water. The MW of CAMH is 210.14 g/mol, so (15.76 g x 1 mol / 210.14 g) / 0.75 L = 0.1000 M (assuming the product is 100 % pure of course).

Titrated with 0.1 M NaOH (t = 0.981) just one titration of 20.0 ml of CAMH solution required... 59.9 ml of NaOH titrant solution to reach the first equivalence point!

This would indicate the CAMH solution was 0.294 M, not 0.1000 M, so again 3 times stronger than planned!

I know some will think I'm getting my normalities mixed up with my molarities but that's not the case: I rarely use normality anyway.

And a 0.1 M CAMH solution would require about the same amount of 0.1 M NaOH to neutralise the first CAMH proton anyway:

NaOH + H3Ct ---> NaH2Ct + H2O

I will titrate 20/3 or about 7 ml, to see if I can see all three equivalence points, or at least the first two...

Totally baffled...

Update:

I'm beginning to wonder whether the fact that the three pKa values of 3.15; 4.77 and 6.40 are really quite close together has something to do with it. The end-point for 0.1 M solutions are approx. 8.5; about 9 and 9.7! Titrating 5 ml of CAHM I ran right past number 2 and 3!

Compared to e.g. H3PO4: 2.148; 7.198 and 12.375 the CAMH values are really close together. Am I titrating all three protons together???

And looking closer at the pH curve it's fairly clear that's what's happening. The highest pH point is 10.6 but it looks very much part of the pH jump for equivalence point 1. Yet 10.6 seems already past EQ2 and EQ3! Bugger! The three protons come from fairly three fairly isolated acid groups and thus are about equally eager, unlike H3PO4...

[Edited on 26-8-2010 by blogfast25]

blogfast25 - 20-3-2011 at 10:50

Coulometric calorimetry (aka Electical Compensation Calorimetry)…

Months ago I bought a simple coulometric calorimeter off eBay, an educational tool by Phillip Harris (so old the documentation mentions a telex number but no email!)

It consists basically of a 200 ml thermos flask (Dewar bottle) with a 12 V - 21 W light bulb fitted so that it can be immersed into the flask. A second hole in the rubber bung allows for inserting a thermometer. Basic setup:

Connecting the light bulb to a 12 V DC power supply and either a watt meter or an amp meter and a volt meter allows the power output P = I . U (P = power (W), I in current (A), U = voltage (V)) to be measured. My two identical box standard multimeters:

I use a 12 V stabilised power source that once belonged to a laptop or a computer game (max. P = 48 W)

Measurement of reaction heats in watery medium (and at atmospheric pressure) can then be made as follows. Assume the calorimeter contains m gram of solution, with a heat capacity of Cp and that the calorimeter has a heat capacity of k, then if Q Joules of heat were released by a reaction, Q = m Cp ΔT + k ΔT = (m Cp + k) ΔT.

If we assume that for small values of ΔT, (m Cp + k) = constant or d(m Cp + k)/dT = 0 then if we add more heat (Q’) to the calorimeter by running current through the light bulb for a known period of time, we know that Q’ = (m Cp + k) ΔT’, with Q’ = I U Δt (Δt time of running current). It follows that Q = Q’ (ΔT/ ΔT’) or Q = I V Δt (ΔT/ ΔT’). It can be arranged so that ΔT=ΔT’ but that isn’t strictly necessary.

And if the precise number of moles reacted n is known the Standard Enthalpy of Reaction ΔH at RT can be estimated from ΔH = - Q / n.

Preliminary tests showed the immersed bulb to run at about 12.20 V and 1.70 A, which is indeed about 21 W.

1) Heat of neutralisation of ammonia solution (NH3 (aq)) with citric acid solution:

Using a standardised commercial solution of ammonia, three runs were carried out, each in a total volume of about 150 ml (75 g ammonia solution and 75 g citric acid solution) and neutralising a precisely known amount of NH3 with a small excess of the triprotic citric acid. Values obtained were - 45.3; - 42.9; - 44.3 kJ/mol with average of - 44.2 kJ/mol. So again, despite the combination of a weak acid AND a weak base, not far removed from the accepted wisdom of -57.3 kJ/mol for strong acid + strong base…

2) Heat of neutralisation of NaOH solution with HCl acid solution:

The first obtained values were much too high and re-standardising showed the HCl solution used was stronger than previously established. The HCl solution was then meticulously re-standardised and measurements repeated. Except, at the end of the first faulty run the bulb had clapped out! It turned out to be a box standard car bulb (for the reversing light), so that was replaced and sealed back into place with good quality silicone sealant.

Then using the newly standardised solutions two runs were made yielding values of - 52.4 kJ/mol and - 54.2 kJ/mol, both somewhat short and not better than the values obtained with the simpler method described above.

blogfast25 - 21-3-2011 at 13:41

Two more runs with HCl + NaOH, this time with stronger solutions (2 M HCl, standardised and slightly stronger NaOH, 75 g of both solution makes for 0.1422 mol of HCl neutralised per run) and slightly longer standing times (before temp. reading). This way the ΔT values has three significant digits (just over 10 K). Also Q’ was made a littler higher (4’ heating instead of previously 3’).

The two values obtained are - 52.8 kJ/mol and - 54.4 kJ/mol, average - 53.6 kJ/mol, still about 2.1 kJ/mol short of the - 55.7 kJ/mol.

Here’s a resource that provides plenty of ideas for the calibration/standardisation of various methods of calorimetry (pdf):

http://www.iupac.org/publications/pac/2001/pdf/7310x1625.pdf

Most are in the ΔH = +/- 20 kJ/mol range, leading to temperature rises that are too small to be accurately measured with my +/- 0.1 C thermocouple. Of particular interest could have been the standard molar heat of dissolution at 298 K of KCl(cr), certified by NIST to be + 17.584 kJ/mol (https://www-s.nist.gov/srmors/certificates/1655.pdf?CFID=139...) to a molality of m = 0.111 mol/kg but the estimated ΔT is only about 0.46 K (17584 x 0.111 / (1000 * 4.19) = 0.46 K)

blogfast25 - 22-3-2011 at 10:02

Another run, this time with home brewed HCl. The commercial HCl (23 w%) I’m using is yellow with Fe3+ contamination. To eliminate it as a possible cause of error, some 90 ml of approx. 2 M HCl were concocted using a basic HCl generator. 15 g technical CaCl2, 50 g H2SO4 (94.5 w%) in the right hand bottle (on steam bath) and 90 ml of deionised water in the left receiver bottle:

After standardising the obtained HCl solution, yield of HCl based on:

CaCl2 (s) + 2 H2SO4 (l) === > 2 HCl (g) + Ca(HSO4)2 (s)

… was found to be about 78 %.

Using this solution, a calorimetric experiment was then conducted using 0.1872 mol HCl and a small excess of NaOH, in a total mass of about 150 g of solutions.

One single run gave a value of - 53.7 kJ/mol heat of neutralisation of HCl solution with NaOH solution. This more or less excludes the iron contamination as a major source of error, as expected.