(a)

For 2g of TIOH dissolved in water to give 2 L of the solution:

$\left[{\mathrm{TIOH}}_{\left(\mathrm{aq}\right)}\right]=\frac{2}{2}\mathrm{g}/\mathrm{L}$
$=\frac{2}{2}\times \frac{1}{221}\mathrm{M}$
$=\frac{1}{211}\mathrm{M}$
${\mathrm{TIOH}}_{\left(\mathrm{aq}\right)}\to {\mathrm{TI}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}$
$\left[{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}\right]=\left[{\mathrm{TIOH}}_{\left(\mathrm{aq}\right)}\right]=\frac{1}{211}\mathrm{M}$
${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$
${10}^{-14}=\left[{\mathrm{H}}^{+}\right]\left(\frac{1}{221}\right)$
$221\times {10}^{-14}=\left[{\mathrm{H}}^{+}\right]$
$\Rightarrow \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=-\log (221\times {10}^{-14})$
$=-\log (2.21\times {10}^{-12})$
$=11.65$

(b)

For 0.3 g of $Ca{\left(\mathrm{OH}\right)}_2$ dissolved in water to give 500 mL of solution:

$$\begin{gathered} \mathrm{Ca}{\left(\mathrm{OH}\right)}_2=0.3\times \frac{1000}{500}=0.6\mathrm{M} \\ \left[{\mathrm{OH}}_{\mathrm{aq}}^{-}\right]=2\times \left[\mathrm{Ca}{\left(\mathrm{OH}\right)}_{2\mathrm{aq}}\right]=2\times 0.6 \\ = 1.2 \mathrm{M} \\ \left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_w}{\left[{\mathrm{OH}}_{\mathrm{aq}-}^{}\right]} \\ =\frac{{10}^{-14}}{1.2}\mathrm{M} \\ =0.833\times {10}^{-14} \\ \mathrm{pH}=-\log (0.833\times {10}^{-14}) \\ =-\log (8.33\times {10}^{-13}) \\ =(-0.902+13) \\ =12.098 \end{gathered}$$

(c)

For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

$$\begin{gathered} \mathrm{NaOH}\to {\mathrm{Na}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-} \\ \left[\mathrm{NaOH}\right]=0.3\times \frac{1000}{200}=1.5\mathrm{M} \\ \left[{\mathrm{OH}}_{\mathrm{aq}}^{-}\right]=1.5\mathrm{M} \\ \mathrm{Then},\left[{\mathrm{H}}^{+}\right]=\frac{{10}^{-1\backslash 4}}{1.5} \\ = 6.66\times {10}^{-13} \\ \mathrm{pH}=-\log (6.66\times {10}^{-13}) \\ =12.18 \end{gathered}$$

(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of the solution:

13.6 × 1 mL = ${\mathrm{M}}_2$ × 1000 mL

(Before dilution) (After dilution)

$13.6 \times {10}^{–3} = {\mathrm{M}}_2 \times 1\mathrm{L} {\mathrm{M}}_2$

= $1.36 \times {10}^{-2} [\mathrm{H}+] = 1.36 \times {10}^{-2} \mathrm{pH} = – \log (1.36 \times {10}^{-2})$

= (– 0.1335 + 2) = 1.866 = 1.87