(a) \(10 \mathrm{~mL}\) of \(0.2 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2=10 \times 0.2\) millimoles \(=2\) millimoles of \(\mathrm{Ca}(\mathrm{OH})_2\)
\(25 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{HCl}=25 \times 0.1\) millimoles \(=2.5\)
millimoles of \(\mathrm{HCl}\)
\(\mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O}\)
\(\because \quad 1\) millimole of \(\mathrm{Ca}(\mathrm{OH})_2\) reacts with 2 millimoles of
\(\begin{aligned} \therefore \quad & 2.5 \text { millimol } \\ \text { of } \mathrm{Ca}(\mathrm{OH})_2 \end{aligned}\)
\(\therefore \quad \mathrm{Ca}(\mathrm{OH})_2\) left \(=2-1.25=0.75\) millimoles \((\mathrm{HCl}\) is
the limiting reactant)
Total volume of the solution \(=10+25 \mathrm{~mL}=35 \mathrm{~mL}\)
\(\therefore \quad\) Molarity of \(\mathrm{Ca}(\mathrm{OH})_2\) in the mixture solution \(=0\).
\(75 / 35 \mathrm{M}=0.0214 \mathrm{M}\)
\(\left[\mathrm{OH}^{-}\right]=2 \times 0.0214 \mathrm{M}=0.0428 \mathrm{M}\)
\(=4.28 \times 10^{-2}\)
\(\mathrm{pOH}=-\log \left(4.28 \times 10^{-2}\right)=2-0.6314=1.3686 \approx 1.37\)
\(\mathrm{pH}=14-1.37=12.63\)
(b) \(10 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4=10 \times 0.01 \mathrm{M} \mathrm{mol}\)
\(=0.1 \mathrm{M} \mathrm{mol}\)
\(10 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2=10 \times 0.01 \mathrm{M} \mathrm{mol}\)
\(=0.1 \mathrm{M} \mathrm{mol}\)
\(\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{SO}_4 \rightleftharpoons \mathrm{CaSO}_4+2 \mathrm{H}_2 \mathrm{O}\) 1 mole of \(\mathrm{Ca}(\mathrm{OH})_2\) reacts with 1 mole of \(\mathrm{H}_2 \mathrm{SO}_4\)
\(\therefore 0.1\) millimole of \(\mathrm{Ca}(\mathrm{OH})_2\) will react completely with \(0.1\) millimole of \(\mathrm{H}_2 \mathrm{SO}_4\). Hence, solution will be neutral with \(\mathrm{pH}=7,0\)
\(\begin{array}{ll}\text { (c) } & 10 \mathrm{~mL} \text { of } 0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4=1 \text { millimole } \\ & 10 \mathrm{~mL} \text { of } 0.1 \mathrm{M} \mathrm{KOH}^2=1 \text { millimole } \\ & 2 \mathrm{KOH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O} \\ & 1 \text { millimole of } \mathrm{KOH} \text { will react with } 0.5 \text { millimole of } \\ & \mathrm{H}_2 \mathrm{SO}_4 \\ \therefore \quad & \mathrm{H}_2 \mathrm{SO}_4 \text { left }=1-0.5=0.5 \text { millimole } \\ & \text { Volume of reaction mixture }=10+10=20 \mathrm{~mL} \\ \therefore \quad & \text { Molarity of } \mathrm{H}_2 \mathrm{SO}_4 \text { in the mixture solution } \\ & \frac{0.5}{20}=2.5 \times 10^{-2} \\ & {\left[\mathrm{H}^{+}\right]=2 \times 2.5 \times 10^{-2}=5 \times 10^{-2}} \\ & \mathrm{pH}=-\log \left(5 \times 10^{-2}\right)=2-0.6989=1.3011\end{array}\)