The Ka for formic acid (HCHO2) is 1.8 x 10^-4. What is the pH for a 0.35 aqueous solution of sodium formate (NaCHO2)?
Answer:
Ka = [H+][CHOO-]/[CHOOH]
CHOO- + H2O = CHOOH + OH-
Kb = [OH-][CHOOH]/[CHOO-]
Ka*Kb = [H+][OH-] = K_w = 1E-14
Kb = 1E-14/Ka = 1e-14/1.8e-4 = 5.55556 E-11
Kb = [x][x]/[0.35 -x] = 5.55556 E-11; x = [OH-] =[CHOOH]
Suppose 0.35 - x ~ 0.35
Then x^2 /0.35 = 5.55556 E -11
x = 4.40959 E-6 M = [OH-]
pH = -log[H+] = 14 - pOH = 14 + log [4.40959 E-6]
pH = 8.644