Calculate the pH for each of the following points in the titration of 50.0 mL of a 1.8 M H3PO3(aq) with 1.8 M KOH(aq).

Question

before KOH
after addition of 25, 50,75,100 ml KOH

Answer 1

The secret to these is to know where you are on the titration curve.
First, find the equivalence points. There are three of them.
Write the equation and balance it.
Mols H3PO4 = M x L = ?
mols KOH = mols H3PO4 that (look at the coefficients in the equation).
mols KOH = M KOH/L KOH. You know mols and M, solve for L.

Then at zero mL KOH you have 1.8M H3PO4. I'm sure you've done hundred o these.

Between zero and the first equivalence point use th Henderson-Hasselbalch equation.

At each eq pt use the hydrolysis of the salt. For the first one it will be
.........H2PO4^- + HOH =-=> H3PO3 + H2O
I.........0.09...............0........0
C.........-x.................x........x
E.........0.09-x..............x........x

Kb for H2PO4^- = (Kw/k1 for H3PO4) = (x)(x)/(0.09-x) and solve for x = OH^-. Convert to pH.

The 0.09 comes from this.
You had 50 mL x 1.8 M H3PO4 initially. It will take 50 mL of 1.8 M KOH for the first eq pt so the concn of the salt will be 1.8 x (50/100) = 0.09
Post your work if you get stuck.

Answer 2

your explanation was extremely poor.

"I'm sure you've done hundreds of these"

If we did, then we wouldn't be looking up how to do the problem.