Question

Calculate the pH for the following weak acid.

A solution of HCOOH has 0.15M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium?

Answer 1

The answer to your question is pH = 2.28

Step-by-step explanation:

Chemical reaction

HCOOH(ac) + H₂O ⇔ H₃O⁺ + HCOO⁻

I 0.15 -- 0 0

C - x -- +x +x

F 0.15 - x -- x x

Write the equation of equilibrium

`ka = ([H3O][HCOO])/([HCOOH])`

Substitution

`1.8 x 10^(-4) = ([x][x])/(0.15 - x)`

But 0.15 - x ≈ 0.15

`1.8 x 10^(-4) = (x^(2) )/(0.15)`

Solve for x

x² = (0.15)(1.8 x 10⁻⁴)

Simplification

x² = 0.000027

Result

x = 0.0052

pH = -log [H₃O⁺]

Substitution

pH = - log [0.0052]

Simplification and result

pH = 2.28