Question

Calculate the pH of the following solutions:
(i) 2.21 g of T $lOH$ dissolved in water to give 2 litre of solution. (Assume T$l$OH to be a strong base)
(ii) 0.49% w/v $H_{2}S{O}_4$ solution

Answer 1

i) $2.2$g of $TlOH=\frac{2.2}{221.39}mol=9.9\times {10}^{-3}$

$\therefore$ concentration of $TlOH=$$\frac{no.ofmoles}{volumeofsolution}.N$

$=\frac{9.9\times {10}^{-3}}{2}N=4.95\times {10}^{-3}$

$\therefore$ concentration of ${OH}^{-}=4.95\times {10}^{-3}N$ ($TlOH$ strong base)

$\therefore$ $pOH=-log\left[{OH}^{(-)}\right]=2.3$ $\therefore$ $prH=14-pOH=11.7$

(ii) $0.49$% w/v $H_2{SO}_4$ i.e. $0.49$gm of $H_2{SO}_4$ is dissolved in $100$ml $H_2{SO}_4$.

$\therefore$ concentration of $H_2{SO}_4$$=\frac{0.49\times 1000}{98\times 100}=0.05\left(M\right)$

$H_2{SO}_4\rightleftharpoons {2H}^{+}+{SO}_4^{2-}$

$\therefore$ concentration of $H^{+},\left[H^{+}\right]=0.05\times 2N$

$=0.1N$

$\therefore$ $pH=-log\left[H^{+}\right]=-log\left(0.1\right)=1$