# Homework Help: PH of a titration

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1. Jul 15, 2015

### davev

1. The problem statement, all variables and given/known data
A 0.551 L solution of 1.37 M sulfurous acid (Ka1 = 1.5e-2 and Ka2 = 1.0e-7) is titrated with 1.65 M NaOH. What will the pH of the solution be when 0.7549 L of the NaOH has been added? The answer is 7.27.

2. Relevant equations
-log(H+)=pH
pH = pKa + log(salt/acid)

3. The attempt at a solution
I find the moles of the weak acid and strong base. After neutralization the strong base is left over. I find the concentration of the remaining moles of base. Then use -log(OH)=pOH. Then 14-pOH = pH. I keep getting 13.95. This is wrong, and I don't know how to get the right answer.

2. Jul 17, 2015

### epenguin

We'd really have to guess from the (lack of) information you give of your calculations. However my guess is you haven't taken account of the fact the acid is dibasic.

Anyway you are treating it like a stoichiometric calculation. It isn't, it's a calculation of an equilibrium, more exactly a pH-equilibrium, subject of the largest number of problems in this section of this site. You haven't used the relevant equation which you quote.

If there is no mistake in the question and given answer, since the pH is just above the pKa2 of 7, the number of moles of base added should be just over about 3/2 × moles of suphurous acid - the first deprotonation essentially complete and you have a bit less than half HSO3- and a bit over half SO32-. This question is really just about the equilibrium between those two species and H+ .

Last edited: Jul 17, 2015
3. Aug 6, 2015

### davev

How do I solve this problem in a way that would make sense given what I know at an introductory college chemistry level? Sorry, I just don't understand how to incorporate multiple Ka values even with the equation listed.

4. Aug 6, 2015

### Staff: Mentor

The difference between pKa values is so large (>3) you can safely assume first proton was completely neutralized before the neutralization of the second started. Then assume the neutralization of the second proton was stoichiometric, and use this information to find pH from the ratio of A2- and HA- concentrations.