What is the Ka of a 1.9 ~ 10-2 M solution of carbonic acid (H2CO3)

with a pH of 3.88?

Ka = [ ? ] × 10!?)

Hep

Ka = 9.2x10⁻⁷

Explanation:

The equilibrium of carbonic acid in water is:

H₂CO₃ ⇄ HCO₃⁻ + H⁺

Where Ka is defined as:

Ka = [HCO₃⁻] [H⁺] / [H₂CO₃]

The equilibrium concentration of the species is:

[H₂CO₃] = 1.9x10⁻² - X

[HCO₃⁻] = X

[H⁺] = X

As pH is -log[H⁺]

3.88 = -log[H⁺]

1.318x10⁻⁴ = [H⁺] = X

Replacing:

[H₂CO₃] = 1.9x10⁻² - 1.318x10⁻⁴ = 1.8868x10⁻²

[HCO₃⁻] = 1.318x10⁻⁴

[H⁺] = 1.318x10⁻⁴

Replacing in ka equation:

Ka = [1.318x10⁻⁴] [1.318x10⁻⁴] / [1.8868x10⁻²]