In: Chemistry
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.
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