Question

Consider the following statements
(a) The pH of a mixture containing $400mL$ of $0.1M{H}_{2}S{O}_4$ and $400mL$ of $0.1MNaOH$ will be approximately $\mathrm{1.3.}$
(b) Ionic product of water is temperature dependent.
(c) A monobasic acid with $K_a=10^{-5}$ has a $pH=5$. The degree of dissociation of this acid is $50\%$.
(d) The Le Chatelier's principle is not applicable to common-ion effect. the correct statement are :

• A. (a), (b) and (d)
• B. (a), (b) and (c)
• C. (a) and (b)
• D. (b) and (c)

Answer 1

Answer: (B)

(a) $H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$
$400\times .1=40400\times .1=40$
20 0
$\therefore \left[H^{+}\right]=\frac{20\times 2}{800}=\frac{1}{20}\Rightarrow pH=-\log \left(\frac{1}{20}\right)$
$\therefore pH=1.3$ so (a) is correct
(b) $\log \left(\frac{K{w}_2}{K{w}_1}\right)=\frac{\Delta H}{2.303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
so ionic product of water is temp. dependent hence (b) is correct.
(c) $K_a=10^{-5},pH=5\Rightarrow \left[H^{+}\right]=10^{-5}$
$K_a=\frac{c{\alpha }^2}{\left(1-\alpha \right)}\Rightarrow K_a=\frac{\left[H^{+}\right].\alpha }{\left(1-\alpha \right)}$
$\therefore 10^{-5}=\frac{10^{-5}.\alpha }{\left(1-\alpha \right)}\Rightarrow 1-\alpha =\alpha \Rightarrow \alpha =\frac{1}{2}=50\%$
so (c) is correct.
(d) Le-chatelier's principle is applicable to common -Ion effect so option (d) is wrong