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Offline pcubis

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pH question
« on: May 15, 2007, 10:32:53 PM »
I am making a mixture with various salts (ZnCl2, MnCl2, BaCl2, etc.).  After adding all the salts, the pH was around 6.  In order to increase the solubility, I wanted to lower the pH of the solution with nitric acid and then try to raise the pH back up to around 7-7.5 with NaOH. 

Learning from my mistake => I added 30 mL of nitric acid to a 3 liter solution and that resulted in a pH of 1.5.  So after the addition of 3 g of NaOH, I could not bring the pH to above 2.  (basically a trial and error approach  :-\ )

I was wondering if there was a way to calculate how much nitric acid (mL) to add to the mixture to lower the pH to 2 and how much NaOH (mg) is needed to be added to raise the pH back up to 7-7.5 without the guessing game. 

Thank you in advance for any help.   

Offline DevaDevil

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Re: pH question
« Reply #1 on: May 16, 2007, 01:36:02 AM »
If you are using only Chloride salts you can ignore the "buffer" behavior of the salt. (as HCl is a strong acid and the chloride ions will not recombine to HCl in any noticable number)

basically the pH calculation will become very easy if you are using a strong acid (like nitric acid) to lower the pH. Assume full disprotonation of the acid upon solvation.

pH = -10log[H+]

so this will give you the value of the protons (or hydronium if you prefer that) needed.
claculate the concentration back to moles, and you have the number of moles of acid you need to add. The amount of ml then is simply: moles needed / Molarity of your nitric acid.

(assumed concentration started at pH = 7; if the pH is initially lower, calculate the number of moles of protons already present in solution with this same formula, and subtract those from the number you need to have in the end to be precise, but since the pH is on a logarithmic scale the difference between starting at pH 7 or 6 will not be large)


To increase the pH again you need to add basically as many moles of NaOH as you added of acid before.

Offline pcubis

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Re: pH question
« Reply #2 on: May 16, 2007, 10:00:25 PM »
Thank you for the explanation! :)


If you are using only Chloride salts you can ignore the "buffer" behavior of the salt. (as HCl is a strong acid and the chloride ions will not recombine to HCl in any noticable number)

basically the pH calculation will become very easy if you are using a strong acid (like nitric acid) to lower the pH. Assume full disprotonation of the acid upon solvation.

pH = -10log[H+]

so this will give you the value of the protons (or hydronium if you prefer that) needed.
claculate the concentration back to moles, and you have the number of moles of acid you need to add. The amount of ml then is simply: moles needed / Molarity of your nitric acid.

(assumed concentration started at pH = 7; if the pH is initially lower, calculate the number of moles of protons already present in solution with this same formula, and subtract those from the number you need to have in the end to be precise, but since the pH is on a logarithmic scale the difference between starting at pH 7 or 6 will not be large)


To increase the pH again you need to add basically as many moles of NaOH as you added of acid before.

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