7.49 Calculate the pH of the following solutions:

a) 2 g of TlOH dissolved in water to give 2 litre of solution.

b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.

c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.

d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

(a)

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

$\left[{\mathrm{TIOH}}_{\left(\mathrm{aq}\right)}\right]=\frac{2}{2}\mathrm{g}/\mathrm{L}\phantom{\rule{0ex}{0ex}}=\frac{2}{2}×\frac{1}{221}\mathrm{M}\phantom{\rule{0ex}{0ex}}=\frac{1}{211}\mathrm{M}\phantom{\rule{0ex}{0ex}}{\mathrm{TIOH}}_{\left(\mathrm{aq}\right)}\to {\mathrm{TI}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}\phantom{\rule{0ex}{0ex}}\left[{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}\right]=\left[{\mathrm{TIOH}}_{\left(\mathrm{aq}\right)}\right]=\frac{1}{211}\mathrm{M}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]\phantom{\rule{0ex}{0ex}}{10}^{-14}=\left[{\mathrm{H}}^{+}\right]\left(\frac{1}{221}\right)\phantom{\rule{0ex}{0ex}}221×{10}^{-14}=\left[{\mathrm{H}}^{+}\right]\phantom{\rule{0ex}{0ex}}⇒\mathrm{pH}=-\mathrm{log}\left[{\mathrm{H}}^{+}\right]=-\mathrm{log}\left(221×{10}^{-14}\right)\phantom{\rule{0ex}{0ex}}=-\mathrm{log}\left(2.21×{10}^{-12}\right)\phantom{\rule{0ex}{0ex}}=11.65$

(b)

For 0.3 g of $Ca{\left(\mathrm{OH}\right)}_{2}$ dissolved in water to give 500 mL of solution:

(c)

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

(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)

= (– 0.1335 + 2) = 1.866 = 1.87