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What would be the pH of an aqueous solution of sulphuric acid which is \(5\times10^{–5}\) \(mol\) \(litre^{–1}\) in concentration?
Sulphuric acid dissociates in water as:
Each mole of sulphuric acid gives of \(H^+\) ions in the solution. One litre of \(H_2SO_4\) solution contains \(5\times10^{–5}\) moles of \(H_2SO_4\) which would give
\(2×5×10^{–5}=10×10^{–5}\) or
\(1.0×10^{–4}\) moles of \(H^+\) ion in one litre of the solution.
Therefore,
\([H^+]=1.0×10^{–4}\) \(mol\) \(litre^{–1}\)
\(pH=–log_{10}[H^+]\)
\(= –log_{10}10^{–4}\)
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