Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: AL on September 15, 2011, 09:08:18 PM
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I am trying to figure out the amount of 12.8M NaOH required to bring water from pH 7.2 to pH 10. Below is what I have done so far, but I am not sure if this is the right direction. I appreciate any help.
@ pH 7.2,
[H+] = 6.31*10-8
[OH-] = 1.6*10-4
@ pH 10
[H+] = 10-10
[OH-] = 10-4
I subtracted the [OH-] at pH 10 from pH 7.2
[OH-]10 - [OH-]7.2 = 9.98*10-5
Took that number and divided by concentration of NaOH to get the amount of 12.8M NaOH needed to change the pH of one liter of water.
9.98*10-5/12.8M = 7.8*10-6 liters needed.
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Looks OK to me.
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Still you don't have considered about the medium. you have assumed the system as a solution of strong acid. If there is any un dissociated weak acid, or dissociated weak base ( I mean the buffering reaction) your solution will not be correct. The only thing you have to confirm is that, your assumption.