Boric acid (H₃BO₃) is a weak acid - VCE - SSCE Chemistry - Question 3 - 2002 - Paper 1

To calculate the hydrogen ion concentration, use the pH value:

$pH = -\log[H^+]$

Substituting the given pH:

$11.11 = -\log[H^+]$

This yields:

$[H^+] = 10^{-11.11} \approx 7.76 \times 10^{-12} \text{ M}$

Next, use the relationship between hydrogen and hydroxide ions:

$K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$

Where:

$[OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{7.76 \times 10^{-12}} \approx 1.29 \times 10^{-3} \text{ M}$

Using the equilibrium expression derived earlier:

$K = \frac{[OH^-][H_3BO_3]}{[B(OH)_4^-]}$

We know the values:

• $[B(OH)_4^-] = 0.100 \text{ M}$
• $[OH^-] = 1.29 \times 10^{-3} \text{ M}$

Substituting into the equilibrium expression:

$K = \frac{(1.29 \times 10^{-3})[H_3BO_3]}{0.100}$

And since we don't have K from the question but can derive it from the concentrations,

$K = 6.01 \times 10^{-10}$

Thus rearranging gives:

$[H_3BO_3] = \frac{K \times [B(OH)_4^-]}{[OH^-]}$

Substituting in we find:

$[H_3BO_3] \approx 1.29 \times 10^{-3} \text{ M}$