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[*] posted on 10-10-2005 at 05:41
Anyone Knows Analytical Chemistry?


Hello

This question is a bet between us and our chemistry professor if any of us could answer it before the next lecture she/he will get 40 out of 100.

The question is : Calculate the pH value for HCl M = 10power -8 ( how do you write superscript anyway?)

When i try to solve it , i get a pH of 8!!! which is WRONG




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bereal511
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[*] posted on 10-10-2005 at 06:25


That's really strange...I always thought -log(pH would still work in this situation.

Perhaps you can use the K<sub>a of HCl and find your pH from there? It would probably end up being 8 anyway. so I'm not quite sure.

[Edited on 10-10-2005 by bereal511]




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[*] posted on 10-10-2005 at 06:52


Ph2....but best to read this.....solo

http://www.purchon.com/chemistry/ph.htm




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[*] posted on 10-10-2005 at 07:31


Your pH formula is an approximation which assumes that the concentration of H+ exceeds 10^-7M (common dissociation of water in fact).

With lower concentrations you need to use the quadratic equation, which can be derived by writing the mass and charge balance and solving that for H+.




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[*] posted on 10-10-2005 at 07:55


I agree with solo and the vulture but will give a few more hints.

1. 10^-8 molar contribution of H+ from the HCl is pretty minor as pure water contributes 10^-7 molar H+. So you know it will be very, very, slightly acid, i.e., very, very slightly below pH 7.

2. H+ from HCl is 10^-8 as it is a strong acid and dissociates completely.

3. Kw = 10^-14 = [H+][OH-].

4. HOH -> equal parts H+ and OH- on disocciation.

Solve relations simultaneously. And as the vulture has stated a quadratic equation will be involved.




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[*] posted on 10-10-2005 at 10:54


I am an analytical chemist, or at least that's what I'm paid for so I hope I know some analytical chemistry. I don't think I would ever have to do that calculation. It's a totally impractical matter, who would want to know the pH of M/10^8 HCl?
Not least, when you expose the solution to air it will pick up CO2 and drift to about pH5
I can't help wondering if you should share the 40% with the people who posted before me.
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[*] posted on 10-10-2005 at 16:23


I'm with vulture! My analytical professor tried to pull that on us last year. You must take into effect that the water dissociation constant plays a huge role here.

2H2O <----> H3O+ + OH-

the total Kw is 10^-14. The concentration of H3O+ is 10^-7. This is larger than the H+ concentration in your HCl so it must be considered. What is the pH of H2O? -log[10^-7] = 6

(10^-7)+(10^-8) = 10^-6

-log[10^-6] = pH 5.96

:P
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Magpie
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[*] posted on 10-10-2005 at 18:42


Of course you are right unionized but one could say the same thing for water. We must assume that this is an academic, not a practical question.

Countrychemist:

1. pH of pure H2O = -log [H+] = -(-7) = 7, not 6.

2. Last two equations are not only irrelvant but mathematically incorrect.




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[*] posted on 11-10-2005 at 11:42


In case someone's still stuck, the quadratic equation should look like this:

([H<sup>+</sup>])<sup>2</sup> - C<sub>acid</sub> [H<sup>+</sup>] - K<sub>w</sub> = 0

Where [H<sup>+</sup>] is the total contribution of H<sup>+</sup> from both water and HCl.
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[*] posted on 11-10-2005 at 22:43


Thanks to all of you

Sadly i am still stuck with this math things.

I don't get vulture , if H+ exceeds then thats a good thing right? as ionized H increases the acidity increases.

In HCl case when we calculate the pH we don't count its ionization constant as it is compeletly ionized so the H+ concentration will be its molar concentration... and we can calculate its pH through the negative logarithm... i don't why we should consider H2O ionization constant? why we use that horrible thing called quadratic equation?.

Phel : lets say that we are using your equation ... you are saying that H+ is ........ so it will be 10^-7*10^-8?.

Unionized : as Magpie said thats just an academic sort of question don't get angry ... and they deserve to have share in the 40% if i shall get it.




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[*] posted on 12-10-2005 at 04:02


I've tried to resist adding to this academic debate. But here goes
adding 10-8 HCl (if nothing happened to the Kw equilibrium) woud give a pH of 6.9 but at such low concentrations water will act as a buffer and I suspect the pH will remain as near to 7 as makes no difference. Of course (as has been mentioned) test some distilled water and you will find the pH is 5.5.
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[*] posted on 12-10-2005 at 07:58


Oh no I've forgot how to use logs.
log (10^-7+10^-8) = 7 - log(1.1)
or approx 7 - (0.1/2.3) = 6.96, but the pH will be even closer to 7 than this.
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[*] posted on 12-10-2005 at 09:08


I really want to get my share of these 40 points, so let's just do the math:

1. Kw = 10^-14 = [OH-][H+] eqn 1
2. pH = -log [H+] eqn 2

HOH -> OH- + H+

3. Let x = [H+] from water
4. But [OH-] also = x
4. [H+] from HCl = 10^-8
5. Using eqn 1: (x+10^-8)(x) = 10^-14

This is the quadratic equation.

Solving gives x = 0.95125x10^10-7, -1.05125x10^-7.

Adding HCl's 10^-8 contribution gives [H+] = 1.05125x10^-7, -0.95125x10^-7

The second root is clearly extraneous, so [H+] = 1.05125x10^-7. pH then = -log [H+] = 7-0.0217 = 6.9783.

I hope I have not made any math errors. ;)




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[*] posted on 12-10-2005 at 09:56


If you want to make approximations you can dispense with the quadratic
[H+][OH-] = 10^-14 (before)
[10^-8 + 10^-7 - d][10^-7 - d] = 10^-14 (after acid addition)
ignoring small terms
(2d).10^-7 = 10^-15
so d = 5x10^-9
[H+]= 10^-7 + 10^-8 - 5x10^-9
so pH=6.9788
(ok a little error in the 4th d.p.)
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[*] posted on 12-10-2005 at 11:48


Yes Quibbler your method is better. BTW I may have an error in the 4th decimal place also. ;)

(How can you do such detailed work in the Caribbean? I would want to be out sailing or skin diving. :D)




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[*] posted on 12-10-2005 at 12:14


I almost got it ... because that is a very low concentration we consider H+ from water ... but what is d in the last equation? and why we subtract it from the product of H+ from water and acid?

All i want is to understand...




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[*] posted on 12-10-2005 at 14:38


Calculation of pH of strong acid.



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[*] posted on 14-10-2005 at 04:13


Well pH is really -log(activity H+).
activity H+ = conc x activity coefficient
at a conc of 1M (HCl) activity coeff. = 0.811 so pH=-log(1x0.881) = 0.091 (not zero). At lower conc the activity coeff. can be calculated using the Debye-Huckel limiting law.

pH=-log[H+]+0.509x[H+]^½

so [H+] = 0.01 pH=2.051
[H+] = 0.001 pH=3.016
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[*] posted on 16-10-2005 at 06:55


THANKS TO EVERYONE HERE!

I showed him the answer and it was right but the problem that he wants the answer in another method (which he will discuss in the next lecture) so he is not giving me my 40% .........

Though i am not getting the 40% , he told me that he will help me in the oral exam...thats GREAT...
Thanks again




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[*] posted on 17-10-2005 at 13:55


water can act as an acid or base under anhydrous conditions.
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[*] posted on 22-10-2005 at 12:00


Under anhydrous conditions water isn't there and can't act as anything.

Just out of idle curiosity, did you think that your statment was on-topic?
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