[ponyoponyo]

A 0.4509 L solution of 1.73 M sulfurous acid (ka1 = 1.50x10^-2, ka2 = 1.0x10^-7) is titrated with 1.450 M KOH. What is the pH when 0.9201 L of KOH has been added?

I know the answer is pH=7.39 but I don't know how to get it.

I know I have 0.78 mol H2SO3 and 1.334 mol OH- initially.
When they react there is 0 mol H2SO3, 0.554 mol OH-, 0.78 mol HA-.
I thought that the pH would be determined solely based on the OH- left in solution, but I guess not since that gave me a pH of 13.6.

I don't know what I'm doing wrong.

[Borek]

Treat it as a buffer solution.

[ponyoponyo]

I keep getting the wrong answers so I'm not sure if I'm setting it up right.

Should I set it up as

HSO₄^-    +    OH^-     -->     SO₄^2-    +    H₂O
0.57 M      0.97 M              0 M

[Borek]

Assume neutralization went to the end, calculate concentrations of acid and conjugate base, plug them into Henderson-Hasselbalch equation.

Note that it is about HSO₃^-/SO₃^2-, not about HSO₄^-/SO₄^2-.

[ponyoponyo]

I'd have 0.78 mol HSO3- and 1.33 mol OH-. And if neutralization went to the end, wouldn't the OH- react with all the HSO3- and leave free OH- in solution. Then I can't use H-H equation.

[Borek]

I'd have 0.78 mol HSO3- and 1.33 mol OH-. And if neutralization went to the end, wouldn't the OH- react with all the HSO3- and leave free OH- in solution. Then I can't use H-H equation.

That's not the case here, as there is not enough KOH to fully neutralize the acid. But in general if there is an excess of OH^- you should calculate pH just from this excess.

[ponyoponyo]

I don't get the exact answer but is this correct?


      H₂SO₃      +       OH-       -->       HSO3^-      +    H2O
I    0.57 M             0.97 M                    0 M
C   0.57 M             0.57 M                0.57 M
E       0 M             0.40 M                0.40 M



      HSO3^-      +       OH-       -->       SO3^2-      +    H2O
I     0.57 M            0.40 M                   0 M
C    0.40 M            0.40 M               0.40 M
E    0.17 M                 0 M               0.40 M



pH = pka2 + log [A²-] / [HA-]
pH = 7.4

[Borek]

I don't get the exact answer but is this correct?


      H₂SO₃      +       OH-       -->       HSO3^-      +    H2O
I    0.57 M             0.97 M                    0 M
C   0.57 M             0.57 M                0.57 M
E       0 M             0.40 M                0.40 M

Using ICE table for stoichiometry is rather unheard of, but if it helps you to follow what is happening... I don't get what is the 0.40M of HSO₃^-.

      HSO3^-      +       OH-       -->       SO3^2-      +    H2O
I     0.57 M            0.40 M                   0 M
C    0.40 M            0.40 M               0.40 M
E    0.17 M                 0 M               0.40 M

0.17 M and 0.40 M are correct, so I suppose 0.40 above was just a typo.

pH = pka2 + log [A²-] / [HA-]
pH = 7.4

What is the pK_a2 value you use?

I didn't get 7.4, I got 6.5 for pK_a2=6.91