When Ksp = 9.4x10^-6 (given missing in your question)
and Ksp = [X+][C2H3O2-]
so by substitution:
9.4x10^-6 = 3.5x10^-3 * [C2H3O2-]
∴ [C2H3O2-] =0.0027
then we need to get Ka we can get it from this formula:
Pka = - ㏒Ka
4.76 = -㏒Ka
∴Ka = 1.7x10^-5
Ka = [H+][C2H3O2-] /[HC2H3O2]
so by substitution:
1.7x10^-5 = [H+]*0.0027 / 1
∴[H+] = 0.0063 m
So finally we can get PH from this formula:
PH = - ㏒[H+]
= -㏒ 0.0063
∴ PH = 2.2
