Question

calculate the pH of the 120mL of a 0.12 M HCOONa,Ka=1.8*10^-5,solution are added to 50mL of a 0.05 M H2SO4 solution.Determine the pH of the sodium formate solution before and after the H2SO4 addition.
please someone help me!

Answer 1

I ASSUME this is a beginning class and they expect you to view H2SO4 as a dibasic acid; i.e., 0.05M x 2 = 0.1M in (H^+). Actually, that isn't right as (H^+) really is closer to o.059M and not 0.1. But we will go with the assumption. Let's call formic acid as HFm to save typing and formate as Fm^-
millimols Fm^- = mL x M = 120 x 0.12 = 14.4
mmols H2SO4 = 50 x 0.1 M = 5.0
.......Fm^- + H^+ ==> HFm
I.....14.4....0........0
Add..........5.0.......0
C.....-5.0..-5.0.......5.0
E.....9.4.....0........5.0
Plug the E line into the Henderson-Hasselbalch equation and solve for pH.

For part A, the pH of the formate by itself, that's a hydrolysis question.
.......Fm^- + HOH ==> HFm + OH^-
I....0.12..............0....0
C.....-x...............x....x
E..0.12-x..............x....x

Kb for Fm^- = (Kw/Ka for HFm) = (x)(x)/0.12-x
Solve for x = (O)H^-) and convert to pH.