PH calculation-chem

Question

For a 0.01 M solution of formic acid at pH 4, calculate the concentrations of the conjugate base and acid. The pKa of formic acid is 3.77.

thank you

Answer 1

Formic acid = HCOOH or CH₂O₂

HCOOH + H₂O ==> H₃O⁺ + HCOO^-

....A.............B............CA............CB..........

0.01............................0...............0..........Initial

-x...............................+x............+x..........Change

0.01-x.........................x...............x...........Equilibrium

NOTE: The problem with the question as posed is that you are given the pH of 4 and the [HCOOH] as 0.01 M, but those two values do not agree with each other given the pKa of 3.77. I will ignore the pH value in the first calculation below, and solve for [H3O+] and [HCOO-], and then in the second calculation below, I will use the pH of 4 to calculate these concentrations, and you should see that they do not agree.

(1)

Ka = [H3O⁺][HCOO^-] / [HCOOH]

We can find the Ka from the pKa that is provided:

Ka = 1x10^-3.77 = 1.70x10^-4

1.70x10^-4 = (x)(x) / 0.01 - x

x² = 1.7x10^-6 - 1.7x10^-4x and solving for x using the quadratic equation we have...

x = 0.00122 M = [H₃O⁺] = [HCOO^-]

(2)

HCOOH + H₂O ==> H₃O⁺ + HCOO^-

pH = 4

[H3O⁺] = 1x10^-4 M and from the equilibrium expression, this is also the [HCOO^-]

[HCOO^-] = 1x10^-4 M