(a) Molar conc. of \(\mathrm{TlOH}\)
\(\begin{aligned}
&=2 \mathrm{~g} /(204+16+1) \mathrm{g} \mathrm{mol}^{-1} \times 1 / 2 \mathrm{~L} \\
&=4.52 \times 10^{-3} \mathrm{M} \\
&{\left[\mathrm{OH}^{-}\right]=[\mathrm{T} l \mathrm{OH}]=4.52 \times 10^{-3} \mathrm{M}} \\
&{\left[\mathrm{H}^{+}\right]=10^{-14} /\left(4.52 \times 10^{-3}\right)=2.21 \times 10^{-12} \mathrm{M}} \\
&\mathrm{pH}=-\log \left(2.21 \times 10^{-12}\right)=12-(0.3424)=11.655
\end{aligned}\)
(b) Molar conc. of \(\mathrm{Ca}(\mathrm{OH})_2\)
\(=0.3 /\left[(40+34) g \mathrm{~mol}^{-1}\right] \times 1 / 0.5 \mathrm{~L}\)
\(=8.10 \times 10^{-3} \mathrm{M}\)
\(\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \mathrm{Ca}^{2+}+2 \mathrm{OH}\)
\(\left[\mathrm{OH}^{-}\right]=2\left[\mathrm{Ca}(\mathrm{OH})_2\right]=2 \times\left(8.10 \times 10^{-3}\right) \mathrm{M}\)
\(=16.2 \times 10^{-3} \mathrm{M}\)
\(\mathrm{pOH}=-\log \left(16.2 \times 10^{-3}\right)=3-1.2101=1.79\)
\(\mathrm{pH}=14-1.79=12.21\)
(c) Molar conc. of \(\mathrm{NaOH}\)
\(=0.3 \mathrm{~g} / 40 \mathrm{~g} \mathrm{~mol}^{-1} \times 1 / 0.2 \mathrm{~L}\)
\(=3.75 \times 10^{-2} \mathrm{M}\)
\(\mathrm{pOH}=-\log \left(3.75 \times 10^{-2}\right)\)
\(=2-0.0574=1.426\)
\(\mathrm{pH}=14-1.426=12.57\)
(d) \(\mathrm{M}_1 \mathrm{~V}_1=\mathrm{M}_2 \mathrm{~V}_2\)
\(\therefore 13.6 \mathrm{M} \times 1 \mathrm{~mL}=\mathrm{M}_2 \times 1000 \mathrm{~mL}\)
\(\therefore \mathrm{M}_2=1.36 \times 10^{-2} \mathrm{M}\)
\({\left[\mathrm{H}^{+}\right]=[\mathrm{HCl}]=1.36 \times 10^{-2} \mathrm{M} }\)
\(\mathrm{pH}=-\log \left(1.36 \times 10^{-2}\right)\)
\(=2-0.1335=1.8665\)