(a)
From data booklet : \(pK_a = 4.87 \Rightarrow K_a = 10^{-4.87} = 1.35 \times 10^{-5}\).
For a weak monoprotic acid of concentration \(c = 1.00 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}\), using the usual approximation \(x \ll c\):
\[ [\mathrm{H}^+] \approx \sqrt{K_a c} = \sqrt{(1.35 \times 10^{-5})(1.00 \times 10^{-3})} = \sqrt{1.35 \times 10^{-8}} = 1.16 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}. \] \[ \boxed{\,\mathrm{pH} = -\log_{10}[\mathrm{H}^+] = -\log_{10}(1.16 \times 10^{-4}) \approx \mathbf{3.94}\,} \] (Accept the exact quadratic solution: \[ x=\frac{-K_a+\sqrt{K_a^2+4K_a c}}{2}=1.096\times 10^{-4}\ \Rightarrow\ \mathrm{pH}=3.96. \]) (M1) A1 A1
[3 marks]
(b)Features of the sketch (weak acid titrated with strong base, same concentrations):
- S-shaped curve; initial pH near the weak-acid value (\(\approx 3.9\text{–}4.0\)).
- Half-equivalence at \(12.5\ \mathrm{cm^3}\) with \(\mathrm{pH} \approx pK_a \approx 4.87\).
- Equivalence point at \(25\ \mathrm{cm^3}\) with \(\mathrm{pH} > 7\) (basic due to propanoate hydrolysis).
- As \(V_\text{NaOH}\to 50\ \mathrm{cm^3}\), solution approaches the base pH (\([\mathrm{OH^-}]=0.001\ \mathrm{M}\Rightarrow \mathrm{pOH}=3,\ \mathrm{pH}\approx 11\)).
A1 A1 A1[3 marks] 
