!! LONG ANSWER!!
The pH of the solution is 2.63.
So, you're titrating sulfurous acid, H_2SO_3, a weak acid, with potassium hydroxide, KOH, a strong base.
The equations for the two dissociation equilibrium reactions that will be established in solution will be
H_2SO_(3(aq)) + H_2O_((l)) rightleftharpoons HSO_(3(aq))^(-) + H_3O_((aq))^(+), K_(a1) = 5.9 * 10^(-3)
HSO_(3(aq))^(-) + H_2O_((l)) rightleftharpoons SO_(3(aq))^(2-) + H_3O_((aq))^(+), K_(a2) = 6.0 * 10^(-8)
Now, because you have a significant difference in magnitude between K_(a1) and K_(a2), you won't get any SO_3^(2-) anions in your solution until after all of the sulfurous acid has been converted to hydrogen sulfite, HSO_3^(-).
In order for that to happen, you have to add enough moles of strong base to completely neutralize all of the sulfurous acid present in the initial solution.
If you don't add enough moles of strong base, the solution will essentially become a buffer, because you'll have both a weak acid, H_2SO_3, and its conjugate base, HSO_3^(-), present in solution.
Use the molarities of the two solutions to determine how many moles of each you add together
C = n/V => n = C * V
n_(H_2SO_3) = "0.10 M" * 70 * 10^(-3)"L" = "0.0070 moles " H_2SO_3
n_(KOH) = "0.10 M" * 50 * 10^(-3)"L" = "0.0050 moles " KOH
When you start addin the base, this is what will happen
H_2SO_(3(aq)) + KOH_((aq)) -> KHSO_(3(aq)) + H_2O_((l))
The net ionic equation will be
H_2SO_(3(aq)) + OH_((aq))^(-) -> HSO_(3(aq))^(-) + H_2O_((l))
You have less moles of KOH than you have of H_2SO_3, which means that the strong base will be consumed by this reaction.
The number of moles of H_2SO_3 will decrease by the same number of moles of base added. At the same time, you'll produce
some conjugate base,HSO_3^(-). So, you get
n_(KOH"remaining") = 0
n_(H_2SO_3"remaining") = 0.0070 - 0.0050 = "0.0020 moles"
n_(HSO_3^(-)"produced") = "0.0050 moles"
The final volume of the solution will be
V_"final" = V_(H_2SO_3) + V_(KOH)
V_"final" = 70 + 50 = "120 mL"
Your solution will now contain 0.0020 moles of weak acid and 0.0050 moles of conjugate base. Use the Henderson-Hasselbalch equation to determine the pH of the resulting solution
pH_"sol" = pK_(a1) + log((["conjugate base"])/(["weak acid"]))
Since you still have some sulfurous acid present in solution, the first equilibrium reaction will be established and you'll get
pK_(a1) = -log(K_(a1)) = -log(5.9 * 10^(-3)) = 2.23
The concentrations the weak acid and conjugate base will be
[H_2SO_3] = "0.0020 moles"/(120 * 10^(-3)"L") = "0.01667 M"
[HSO_3^(-)] = "0.0050 moles"/(120 * 10^(-3)"L") = "0.04167 M"
Now plug all your values into the H-H equation
pH_"sol" = 2.23 + log((0.04167cancel("M"))/(0.01667cancel("M"))) = color(green)(2.63)
