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

the answer is b. dilution