Question

Calculate $pH$ of following mixtures-
(i) $10mL$ of $0.2MCa(OH{)}_2+25mL$ of $0.1MHCl$.
(ii) $10mL$ of $0.01M{H}_{2}S{O}_4+10mL$ of $0.01MCa(OH{)}_2$
(iii) $10mL$ of $0.1M{H}_{2}S{O}_4+10mL$ of $0.1MKOH$.

Answer 1

(a) Moles of $H^{+}=\frac{25\times 0.1}{1000}=0.0025$ mol

Moles of $O{H}^{-}=\frac{10\times 0.2\times 1}{1000}=0.004$ mol

Thus excess of $O{H}^{-}=0.0015$ mol

$\left[O{H}^{-}\right]=\frac{0.0015}{35\times {10}^{-3}}$ mol/L=$0.0428$

$pOH=-\log \left[OH\right]=1.36$

$pH=12.63$

(b) Moles of $H^{+}=\frac{10\times 0.01\times 2}{1000}=\frac{0.2mol}{1000}$

Moles of $O{H}^{-}=\frac{10\times 0.01\times 2}{1000}=\frac{0.2mol}{1000}$

$\because$ there is neither an excess of $H^{+}$ or $O{H}^{-}$,

So the $so{l}^n$ is neutral. Hence $pH=7$

(c) Moles of $H^{+}=\frac{2\times 10\times 0.1}{1000}=0.002mol$

Moles of $O{H}^{-}=\frac{10\times 0.1}{1000}=0.001mol$

excess of $H^{+}=0.001mol$

$\therefore \left[H^{+}\right]=\frac{0.001}{20\times {10}^{-3}}=\frac{10}{20\times {10}^{-3}}$

$=0.05$

$\therefore pH=-\log \left(0.05\right)$

$=1.3$