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

To find the hydroxide ion concentration, we can use the pH to find the hydrogen ion concentration:

Given pH = 11.11,

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

Using the relation between $[H^+]$ and $[OH^-]$:

$K_w = [H^+][OH^-] = 1.00 \times 10^{-14}$

Thus,

$[OH^-] = \frac{K_w}{[H^+]} = \frac{1.00 \times 10^{-14}}{7.76 \times 10^{-12}} = 1.29 \times 10^{-3} \text{ M}$

From the equilibrium expression obtained in part a:

Substituting in the values:

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

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

We can rearrange the equation to find [H₃BO₃]:

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

Assuming $K$ value is known or determined, we would substitute it here to find [H₃BO₃]. For example, if $K$ = 6.01 × 10⁻⁴:

$[H_3BO_3] = \frac{(6.01 \times 10^{-4}) \times (0.100)}{(1.29 \times 10^{-3})} = X M$ (solve for X accordingly).