This is a typical basic salt equilibrium problem:
NaCHO2(aq) → Na+(aq) + CHO2–(aq)
CHO2–(aq) + H2O(l) → ← HCHO2(aq) + OH–(aq)
Kb = [HCHO2]e[OH–]e [CHO2–]e = 1.0×10–14/1.8×10–4 = 5.6×10–11
Initial0.050 0 0
Change–x +x +x
Equilibrium0.050 – x x x
Approximate? 0.050/5.6×10–11 = 8.9×108 Yes
5.6×10–11 = x2/0.050
x = 1.7×10–6 = [OH–]e
pOH = –log[OH–]e = –log(1.7×10–6) = 5.77
pH = 14.00 – pOH = 14.00 – 5.77 = 8.23