I tried solving this problem but keep getting wrong answer. Chemistry problem

Question

A 25.0 mL sample of 0.10 M sodium acetate is titrated with 0.10 M HCl(aq). What is the pH after the addition of 35.0 mL of HCl(aq)? (Kb of CH3CO2-= 5.6e10-10

I got 4.75 as the answer but the correct answer is 1.78

Answer 1

Because Acetic acid is a weak acid, all of the acetate will react with HCl, and I won't be using K_b since the products of the reaction between Acetate and the water are negligible.

Keep in mind that concentration changes when other solutions are added.

The net ionic formula for the reaction between NaC₂H₃O₂ and HCl is:

C₂H₃O₂^-(aq) + H⁺(aq) --> HC₂H₃O₂(aq).

You can determine the new molarity of each solution with the formula M₁V₁=M₂V_2,rewritten as M₂=M₁V₁/V₂

[HCl] = 0.10 M * 35ml/60ml = 0.05833M

[C₂H₃O₂-] = 0.10 M * 25ml/60ml = 0.04167M

Because the ratio of H+ to C₂H₃O₂- is 1:1. and there is more HCl than NaC₂H₃O₂, there is excess H+. Which will have a molarity of 0.05833M-0.04167M=0.01666M

Then, we calculate the pH

pH = -log(0.01666) = 1.77824, rounded to sigfigs as 1.78