Given a diprotic acid, H 2 A, with two ionization constants of K a1 = 3.0 * 10 - 4 and K a2 = 4.0 * 10 - 11, calculate the pH for a 0.117 M solution of NaHA.

The pH for a 0.117 M solution of NaHA is 2.227

Explanation:

To solve the question we check the difference in the Ka values thus

Ka₁ / Ka₂ = 7500000 < 10⁸ so we are required to calculate each value as follows

We therefore have

H₂X→ H⁺¹+HX⁻¹ with Ka₁ = 3.0 * 10⁻⁴

Therefore

3.0 * 10⁻⁴ = (x²) / (0.117)

x² = 3.0 * 10⁻⁴ * 0.117 and x = 5.925 * 10⁻³ = [H⁺]

Similarly

Ka₂ = 4.0 * 10⁻¹¹

and

4.0 * 10⁻¹¹ = (x²) / (0.117)

x² = 0.117 * 4.0 * 10⁻¹¹

x = 2.16 * 10⁻⁶

Total H⁺ = 5.925 * 10⁻³+2.16 * 10⁻⁶ = 5.927 * 10⁻³

Since pH = - log of hydrogen ion concentration,

pH = - log 5.927 * 10⁻³ = 2.227