Question

In a solution buffered to pH 3.040 that contains hydrazoic acid, benzoic acid, propanoic acid, and formic acid, what percent of formic acid is protonated?

ka=1.8*10^4

Answer 1

Q. Formic acid K_a = 1.8 x 10^-4

pK_a = -log(K_a)

pK_a = -log(1.8 x 10^-4)

pK_a = 3.74

According to Henderson - Hasselbalch equation,

pH = pK_a + log([conjugate base] / [weak acid])

pH = pK_a + log([HCOO^-] / [HCOOH])

3.040 = 3.74 + log([HCOO^-] / [HCOOH])

log([HCOO^-] / [HCOOH]) = 3.040 - 3.74

log([HCOO^-] / [HCOOH]) = -0.70

[HCOO^-] / [HCOOH] = 10^-0.70

[HCOO^-] / [HCOOH] = 0.20

Adding 1 to both sides

([HCOO^-] / [HCOOH]) + 1 = 0.20 + 1

([HCOO^-] + [HCOOH]) / [HCOOH] = 1.20

Total concentration / [HCOOH] = 1.20

taking reciprocal,

[HCOOH] / total concentration = 1 / 1.20

[HCOOH] / total concentration = 0.835

percent of formic acid protonated = ([HCOOH] / total concentration) * 100

percent of formic acid protonated = (0.835) * 100

percent of formic acid protonated = 83.5 %