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 $1.3$
(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), (b) and (c)

Solution:- (B) (a), (b) and (c)

(a) $\begin{array}{cccc}{H}_{2}S{O}_4& +& 2NaOH& ⟶N{a}_{2}S{O}_4+2H_{2}O\\ 400\times .1=40& & 400\times .1=40& \\ 20& & 0& \end{array}$

$\therefore \left[H^{+}\right]=\frac{20\times 2}{800}=\frac{1}{20}\Rightarrow pH=-\log \left(\frac{1}{20}\right)$

$\therefore pH=1.3$

So (a) is correct

(b) $\log \left(\frac{K{w}_2}{K{w}_1}\right)=\frac{\Delta H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

So, ionic product of water is temp. dependent hence (b) is correct.

(c) $K_a={10}^{-5},pH=5\Rightarrow \left[H^{+}\right]={10}^{-5}$

$K_a=\frac{c{\alpha }^2}{(1-\alpha )}\Rightarrow K_a=\frac{\left[H^{+}\right].\alpha }{(1-\alpha )}$

${10}^{-5}=\frac{{10}^{-5}.\alpha }{(1-\alpha )}\Rightarrow 1-\alpha =\alpha \Rightarrow \alpha =\frac{1}{2}=50\%$

so (c) is correct.

(d) Le-chatelier's principle is applicable to common-Ion effect so option (d) is wrong

$\therefore$ correct answer (2)