The Ka for formic acid (HCHO2) is 1.8 x 10^-4. What is the pH for a 0.35?

ph-value

#1

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