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

When HCO₃⁻ acts as a base, it accepts a proton from water, forming CO₂ and H₃O⁺:

$HCO₃⁻(aq) + H₂O(l) \rightleftharpoons CO₂(g) + H₃O⁺(aq)$

Using the expression for the acidity constant, we have:

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

Assuming x is the concentration of H⁺ and ClO⁻ produced,

$K_a = \frac{x^2}{0.50 - x} = 3.0 \times 10^{-5}$

For simplicity, assuming x is small compared to 0.50, we have:

$3.0 \times 10^{-5} = \frac{x^2}{0.50}$

Solving for x gives:

$x^2 = 3.0 \times 10^{-5} \times 0.50$ $x^2 = 1.5 \times 10^{-5}$ $x = \\sqrt{1.5 \times 10^{-5}} \approx 3.87 \times 10^{-3}$

Therefore, the pH is:

$pH = -\log(3.87 \times 10^{-3}) \approx 2.41$

Using the relation:

$pH + pOH = 14$

With a given pOH of 3:

$pH = 14 - 3 = 11$

Then to find [OH⁻]:

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

Using the pH we calculated:

$[H^+] = 10^{-pH} = 10^{-11} = 1.0 \times 10^{-11} M$