Question

How many moles of formic acid need to be dissolved in 1.00 L of water to make a solution with a pH of 1.97? The pKa for formic acid is 3.75.

Answer 1

use:

pKa = -log Ka

3.75 = -log Ka

Ka = 1.778*10^-4

use:

pH = -log [H+]

1.97 = -log [H+]

[H+] = 1.072*10^-2 M

formic acid is HCOOH

HCOOH    <—>   H+       +   HCOO-

C                   0           0       (initial)

C-x                   x           x       (at equilibrium)

since,

[H+] = 1.072*10^-2 M

x = 1.072*10^-2 M

Ka = [H+][HCOO-]/[HCOOH]

1.778*10^-4 = x*x / (C-x)

1.778*10^-4 = (1.072*10^-2)*(1.072*10^-2) / (C-(1.072*10^-2))

C-(1.072*10^-2) = 0.646

C = 0.657 M

C =number of mol / volume

0.657 M = n / 1 L

n = 0.657 mol

Answer: 0.657 mol