Question

4) Formic acid (

Answer 1

Ka of HCOOH = 1.8 x 10^-4
Ka = [H+][HCOO-] / [HCOOH]
molar mass of HCOONa = 68.0 g/mol
0.856 g HCOONa x (1 mol / 68.0 g) = 0.0126 mol HCOONa

Assume that adding 0.856 g HCOONa doesn't change the total volume.

M of HCOO- from the HCOONa
M = mol / L = 0.0126 mol / 0.200 L = 0.0629 M


Now, for the dissociation of the HCOOH.
Let x = dissociation of HCOOH
Then [H+] = x
And [HCOO-] = x + 0.0629
[HCOOH] = 0.759 -x
Ka = [H+][HCOO-] / [HCOOH]
1.8 x 10^-4 = (x)(x + 0.0629) / (0.759 -x)
Assume x << 0.759 and
assume x << 0.0629, then
1.8 x 10^-4 = x^2
x = 0.0134 M
x is about 21% of 0.0629, so that assumption is not valid.
Redo assuming only x << 0.759, then
1.8 x 10^-4 = (x)(x + 0.0629) / (0.759)
x^2 + 0.0629 x = 0.759 (1.8 x 10^-4) = 1.4 x 10^-4
x^2 + 0.0629 x - (1.4 x 10^-4) = 0
Solve for x using the quadratic equation:
x = -0.0629 +/- [(0.0629)^2 - 4(1)(-1.4 x 10^-4)]^.50
x = -.0629 +/- 0.0672 = 0.0043 (for the positive root)
The assumption that x << 0.759 is valid.
x = 0.0043 M = [H+]


The reverse reaction occurs (shifts left) when the HCOONa is added because the system tries to use up some of the excess HCOO- that is added (LeChatelier's principle).