botanistbee
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pH help
Hey SM, Long time lurker here, finally decided to make an account.
I have a ph problem that I was wondering if you all could help me with. I am trying to neutralize some very alkali mud with a 1 molar HCl solution so
I get a specific final volume. I know the pH of the mud slurry which is consistently 9.2 in a 60% wt/wt slurry with water which has a pH of 7.8. I
know the dirt here contains some calcium carbonates and some phosphates.
I want to neutralize a specific dirt sample size (say 2.0kg) to a pH of 7.0 while ensuring a 60%wt/wt slurry. I realize this will be a system of
equations of some kind but I am having trouble setting it up.
Again here is what I know:
pH of 60%wt/wt slurry is 9.2
pH of water used to make slurry is 7.8
Desired pH is 7.0
Using 1 molar HCl as modifier
Final slurry concentration must be 60%
dirt sample size is 2.0 kg
I appreciate any help in advanced, I have spent several days on this problem and have gotten nowhere.
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Metacelsus
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Without knowing the precise composition of the mud, this will be impossible to calculate. Instead, I suggest doing a small-scale test. For example,
take 3 gram portions of mud, and to each portion add 2 grams of hydrochloric acid solution of varying concentration, and measure the final pH. This
will enable you to experimentally determine the required acid concentration.
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blogfast25
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Metacelsus is right.
Alternatively, dry the mud to determine its water content. The amount of HCl needed could then be estimated but would still have to be experimentally
verified.
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aga
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Does the Mud start off as a 60w% mud/water mixture ?
If so, then HCl (max 37w% concentration) might be the wrong thing to be using, because you'll be adding water along with the acid.
Something cheap and powdered, maybe vitamin C would be more practical.
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blogfast25
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Quote: Originally posted by aga  | Does the Mud start off as a 60w% mud/water mixture ?
If so, then HCl (max 37w% concentration) might be the wrong thing to be using, because you'll be adding water along with the acid.
Something cheap and powdered, maybe vitamin C would be more practical. |
| Quote: | | I am trying to neutralize some very alkali mud with a 1 molar HCl solution so I get a specific final volume. |
At pH = 9, [OH-] = 10-5 mol/L
Even if the mud was 100 % water, to neutralise 1 L of mud would take only 0.01 ml (millilitre) of 1 M HCl.
[Edited on 18-5-2016 by blogfast25]
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aga
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Such is the Power of maths.
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Metacelsus
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| Quote: Originally posted by blogfast25 |
At pH = 9, [OH-] = 10-5 mol/L
Even if the mud was 100 % water, to neutralise 1 L of mud would take only 0.01 ml (millilitre) of 1 M HCl.
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The mud is almost certainly a buffered system, so it will take much more acid than that to bring it to pH 7.
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aga
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Why would it be buffered ?
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AJKOER
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Here is an idea in place of HCl, used a powdered metal (like Al or possibly Zn), which may avoid some of the carbonate buffering issue.
Another choice is powdered FeO working with air and water producing Fe2O3.xH2O and H+. This is more of a green solution ( actually reddish brown). The
equation concerning the action of O2 on a ferrous salt in water is given by:
O2(aq) + 4Fe2+ + 6H2O ↔ 4FeOOH(s) + 8H+
Source: "Air Oxidation of Ferrous Iron in Water" by Ahmet Alıcılar, Göksel Meriç, Fatih Akkurt and Olcay Şendil, link: https://www.google.com/url?sa=t&source=web&rct=j&...
[Edited on 18-5-2016 by AJKOER]
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Metacelsus
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Because:
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blogfast25
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My point was not an exact calculation (that's not possible w/o more compositional data), rather it was to show how little alkaline pH = 9 really is.
Unless it's high ionic strength buffer you won't need much acid to correct that bit of alkalinity.
Ultimately only experimentation can show, of course...
[Edited on 19-5-2016 by blogfast25]
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aga
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| Quote: Originally posted by blogfast25 |
At pH = 9, [OH-] = 10-5 mol/L
Even if the mud was 100 % water, to neutralise 1 L of mud would take only 0.01 ml (millilitre) of 1 M HCl.
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Just so i can be sure i got this straight ...
pH + pOH = 14, so at pH 9, pOH = 5
p being -log()
So if there's a pH of 9 and he wants it at 7, there's a difference of 9.9*10<sup>-6</sup> mols of OH<sup>-</sup> to react.
Certainly helps to know that even a Drop of 1 [M] HCl would be a LARGE amount instead of starting off with a bucket full of acid !
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blogfast25
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| Quote: Originally posted by aga |
Just so i can be sure i got this straight ...
pH + pOH = 14, so at pH 9, pOH = 5
p being -log()
So if there's a pH of 9 and he wants it at 7, there's a difference of 9.9*10<sup>-6</sup> mols of OH<sup>-</sup> to react.
Certainly helps to know that even a Drop of 1 [M] HCl would be a LARGE amount instead of starting off with a bucket full of acid !
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Yes. 10-5 - 10-7
My point was to show just how dilute in OH- a pH = 9 really is.
So when someone says, "OMG, pH = 11, sooooooo alkaline!", that person is almost certainly not a chemist: 1011-14 = 0.001 M of
OH-.
Even pH = 14 is only 1 M OH-.
[Edited on 19-5-2016 by blogfast25]
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aga
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So, how would a really realy basic or acidic solution be measured ?
Presumably not on the pH scale ?
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