A 635 mL solution of HBr is titrated with 0.51 M KOH. If it takes 1569 mL of the base solution to reach the equivalence point, what is the pH when only 141 mL of the base has been added to the solution?
HBr + KOH ==> KBr + H2O
mols KOH = M x L = 1.569 x 0.51 approx 0.8.
mols HBr = 0.8 (look at the coefficients in the balanced equation; 1 mol HBr to 1 mol KOH.)
M HBr = mols HBr/L HBr = approx 0.8/0.635 = approx 12.6M
mols HBr initially = 0.8
mols KOH added with 141 mL = M x L = 0.51 x 0.141 = about 0.07
mols HBr remaining = -.8-0.7 = about 0.7.
Then (OH^-) = about 0.7
pOH = -log (OH^-)
and pH + pOH = pKw = 14.
You know pKw and pOH, solve for pH. Post your work if you get stuck. NOTE: I've estimated above. You must recalculate each step.