Let the concentration of HCOOH be c
use:
pH = -log [H+]
2.39 = -log [H+]
[H+] = 4.074*10^-3 M
HCOOH dissociates as:
HCOOH -----> H+ + HCOO-
c 0 0
c-x x x
Ka = [H+][HCOO-]/[HCOOH]
Ka = x*x/(c-x)
1.8*10^-4 = 4.074*10^-3*4.074*10^-3/(c-4.074*10^-3)
c-4.074*10^-3 = 9.22*10^-2
c=9.627*10^-2 M
volume , V = 1*10^2 mL
= 0.1 L
use:
number of mol,
n = Molarity * Volume
= 9.627*10^-2*0.1
= 9.627*10^-3 mol
Molar mass of HCOOH,
MM = 2*MM(H) + 1*MM(C) + 2*MM(O)
= 2*1.008 + 1*12.01 + 2*16.0
= 46.026 g/mol
use:
mass of HCOOH,
m = number of mol * molar mass
= 9.627*10^-3 mol * 46.03 g/mol
= 0.4431 g
Answer: 0.443
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