# Calculate the pH of the following solutions:

Question:

Calculate the pH of the following solutions:

a) $2 \mathrm{~g}$ of $\mathrm{TlOH}$ dissolved in water to give 2 litre of solution.

b) $0.3 \mathrm{~g}$ of $\mathrm{Ca}(\mathrm{OH})_{2}$ dissolved in water to give $500 \mathrm{~mL}$ of solution.

c) $0.3 \mathrm{~g}$ of $\mathrm{NaOH}$ dissolved in water to give $200 \mathrm{~mL}$ of solution.

d) $1 \mathrm{~mL}$ of $13.6 \mathrm{M} \mathrm{HCl}$ is diluted with water to give 1 litre of solution.

Solution:

(a) For $2 \mathrm{~g}$ of $\mathrm{TIOH}$ dissolved in water to give $2 \mathrm{~L}$ of solution:

$\left[\mathrm{TlOH}_{(a q)}\right]=\frac{2}{2} \mathrm{~g} / \mathrm{L}$

$=\frac{2}{2} \times \frac{1}{221} \mathrm{M}$

$=\frac{1}{221} \mathrm{M}$

$\mathrm{TlOH}_{(a q)} \longrightarrow \mathrm{Tl}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-}$

$\left[\mathrm{OH}_{(\text {oqf })}^{-}\right]=\left[\mathrm{TlOH}_{(\text {oq })}\right]=\frac{1}{221} \mathrm{M}$

$K_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]$

$10^{-14}=\left[\mathrm{H}^{+}\right]\left(\frac{1}{221}\right)$

$221 \times 10^{-14}=\left[\mathrm{H}^{+}\right]$

$\Rightarrow \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(221 \times 10^{-14}\right)$

$=-\log \left(2.21 \times 10^{-12}\right)$

$=11.65$

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

$\mathrm{Ca}(\mathrm{OH})_{2} \longrightarrow \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-}$

$\left[\mathrm{Ca}(\mathrm{OH})_{2}\right]=0.3 \times \frac{1000}{500}=0.6 \mathrm{M}$

$\left[\mathrm{OH}^{-}{ }_{a q q}\right]=2 \times\left[\mathrm{Ca}(\mathrm{OH})_{2 a q q}\right]=2 \times 0.6$

$=1.2 \mathrm{M}$

$\left[\mathrm{H}^{+}\right]=\frac{K_{w}}{\left[\mathrm{OH}_{a q}^{-}\right]}$

$=\frac{10-14}{1.2} \mathrm{M}$

$=0.833 \times 10^{-14}$

$\mathrm{pH}=-\log \left(0.833 \times 10^{-14}\right)$

$=-\log \left(8.33 \times 10^{-13}\right)$

$=(-0.902+13)$

$=12.098$

(c) For $0.3 \mathrm{~g}$ of $\mathrm{NaOH}$ dissolved in water to give $200 \mathrm{~mL}$ of solution:

$\mathrm{NaOH} \longrightarrow \mathrm{Na}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-}$

$[\mathrm{NaOH}]=0.3 \times \frac{1000}{200}=1.5 \mathrm{M}$

Then, $\left[\mathrm{H}^{+}\right]=\frac{10^{-14}}{1.5}$

$=6.66 \times 10^{-13}$

$\mathrm{pH}=-\log \left(6.66 \times 10^{-13}\right)$

$=12.18$

(d) For $1 \mathrm{~mL}$ of $13.6 \mathrm{M} \mathrm{HCl}$ diluted with water to give $1 \mathrm{~L}$ of solution:

$13.6 \times 1 \mathrm{~mL}=\mathrm{M}_{2} \times 1000 \mathrm{~mL}$

(Before dilution) (After dilution)

$13.6 \times 10^{-3}=\mathrm{M}_{2} \times 1 \mathrm{~L}$

$\mathrm{M}_{2}=1.36 \times 10^{-2}$

$\left[\mathrm{H}^{+}\right]=1.36 \times 10^{-2}$

$\mathrm{pH}=-\log \left(1.36 \times 10^{-2}\right)$

$=(-0.1335+2)$

$=1.866 \sim 1.87$