Find the hydrogen ion, H\(^+\) concentration and hydroxide ion, OH\(^-\) concentration in 0.06 moldm\(^{-3}\) solution of H\(_2\)SO\(_4\).

Sulphuric acid (H₂SO₄) is a diprotic acid, meaning each molecule donates two hydrogen ions when it dissociates completely in water:

\[\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{H}^+(aq) + \text{SO}_4^{2-}(aq)\]

Step 1: Find [H⁺]

Since each mole of H₂SO₄ produces 2 moles of H⁺:

\[[\text{H}^+] = 2 \times 0.06 = 0.12 \text{ mol dm}^{-3} = 1.2 \times 10^{-1} \text{ mol dm}^{-3}\]

Step 2: Find [OH^-]

Using the ionic product of water at 25 °C:

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

\[[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-1}}\]

\[[\text{OH}^-] = \frac{1.0}{1.2} \times 10^{-14+1} = 0.833 \times 10^{-13} = 8.3 \times 10^{-14} \text{ mol dm}^{-3}\]

Therefore [H⁺] = \(1.2 \times 10^{-1}\) mol dm^-3 and [OH^-] = \(8.3 \times 10^{-14}\) mol dm^-3.

A common mistake is forgetting that sulphuric acid is diprotic and using [H⁺] = 0.06 instead of 0.12, which would give \(1.2 \times 10^{-2}\) instead of \(1.2 \times 10^{-1}\) and an incorrect OH^- concentration of \(8.3 \times 10^{-13}\).