HL Extension
Logarithmic pH Definitions & Kw
The logarithmic definitions link \([\text{H}^+]\) and \([\text{OH}^-]\) via the ion product constant of water \(K_{\text{w}}\):
\[\text{pH} = -\log_{10}[\text{H}^+], \quad \text{pOH} = -\log_{10}[\text{OH}^-], \quad K_{\text{w}} = [\text{H}^+][\text{OH}^-] = 1.00 \times 10^{-14} \text{ at } 298\text{ K}\]Key Equations
Worked Examples
Worked Example
pH of 0.10 M HCl
Problem: Calculate the pH of \(0.10\text{ mol dm}^{-3}\text{ HCl}\).
\[[\text{H}^+] = 0.10\text{ M} \implies \text{pH} = -\log_{10}(0.10) = 1.00\]
Worked Example
pH of 0.0044 M Ca(OH)₂ (Diprotic Base)
Problem: Calculate the pH of \(0.0044\text{ mol dm}^{-3}\text{ Ca(OH)}_2\) at \(298\text{ K}\).
\[[\text{OH}^-] = 2 \times 0.0044 = 0.0088\text{ M} \implies \text{pOH} = -\log_{10}(0.0088) = 2.06\] \[\text{pH} = 14.00 - 2.06 = 11.94\]
Examiner Trap
Polybasic Bases and Non-Standard Temperatures
- Polybasic bases: For \(\text{Ca(OH)}_2\) or \(\text{Ba(OH)}_2\), remember to multiply base concentration by 2 to get \([\text{OH}^-]\).
- Temperature matters: If the problem states \(T = 373\text{ K}\) (\(100^\circ\text{C}\)) where \(K_{\text{w}} = 5.13 \times 10^{-13}\), \(\text{pH} + \text{pOH} = 12.29\), not \(14.00\).
AQA GCSE & IB Chemistry
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