The hydrogen carbonate ion (HCO₃⁻) can act both as an acid and as a base - VCE - SSCE Chemistry - Question 5 - 2006 - Paper 1

When acting as a base, HCO₃⁻ accepts a proton (H⁺) from water. The chemical equation is:

$HCO_3^-(aq) + H_2O(l) ⇌ CO_2(g) + H_3O^+(aq)$

To find the pH of a 0.50 M solution of hypochlorous acid, we use the formula for the acidity constant:

$K_a = \frac{[H^+][ClO^-]}{[HOCl]}$

Let x be the concentration of H⁺ and ClO⁻ ions produced:

$K_a = 3.0 × 10^{-7} = \frac{x^2}{0.50 - x}$

Assuming x is small compared to 0.50, we can simplify this to:

$3.0 × 10^{-7} ≈ \frac{x^2}{0.50}$

Solving for x gives:

$x^2 ≈ 3.0 × 10^{-7} × 0.50$ $x^2 ≈ 1.5 × 10^{-7}$ $x = \sqrt{1.5 × 10^{-7}} ≈ 1.22 × 10^{-4} M$

Then, the pH is:

$pH = -\log[H^+] = -\log(1.22 × 10^{-4}) ≈ 3.91$

Given the pOH is 3:

$pOH = 3$

The relationship between pOH and hydroxide ion concentration ([OH⁻]) is:

$[OH^-] = 10^{-pOH} = 10^{-3} = 0.001 M$

Using the relationship between pH and pOH:

$pH + pOH = 14$

Calculating pH:

$pH = 14 - 3 = 11$

Now, for the hydrogen ion concentration ([H⁺]):

$[H^+] = 10^{-pH} = 10^{-11} M$