Chemical And Ionic Equilibrium Result Question 4
4. Consider the following statements.
I. The pH of a mixture containing 400 mL of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ and $400$ mL of $0.1$ $ M$ $ NaOH$ will be approximately $1.3$ .
II. Ionic product of water is temperature dependent.
III. A monobasic acid with $K_a=10^{-5}$ has a $\mathrm{pH}=5$. The degree of dissociation of this acid is $50 %$.
IV. The Le-Chatelier’s principle is not applicable to common-ion effect.
The correct statements are
(2019 Main, 10 April I)
(a) I, II and IV
(b) II and III
(c) I and II
(d) I, II and III
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Answer:
Correct Answer: 4. ( d )
Solution:
- The explanation of given statements are as follows:
In statement (I), millimoles of $\mathrm{H}^{+}=400 \times 0.1 \times 2=80$
Millimoles of $\mathrm{OH}^{-}=400 \times 0.1=40$ (Limiting reagent)
$\therefore$ Millimoles of $\mathrm{H}^{+}$ left $=80-40=40$
$\begin{aligned} & {\left[\mathrm{H}^{+}\right]=\frac{40}{400+400}=\frac{40}{800} M=\frac{1}{20} M} \\ & \Rightarrow \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(\frac{1}{20}\right) \\ & =-\log 1+\log 2+\log 10 \\ & =-0+0.301+1 \\ & \Rightarrow 1.30 \\ & \end{aligned}$
Hence, the option (a) is correct.
In statement (II), ionic product of $\mathrm{H}_2 \mathrm{O}$ is temperature dependent.
$K_w=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \approx 10^{-14}(\mathrm{~mol} / \mathrm{L})^2 \text { at } 25^{\circ} \mathrm{C}$
With increase in temperature, dissociation of $\mathrm{H}_2 \mathrm{O}$ units into $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ions will also increase. As a result, the value of ionic product, $\left[\mathrm{H}^{+}\right] \times\left[\mathrm{OH}^{-}\right]$will be increased. e.g.
| Temperature | $K_w (mol/L)^2$ |
|---|---|
| $5^{\circ} \mathrm{C}$ | $0.186 \times 10^{-14}$ |
| $25^{\circ} \mathrm{C}$ | $1.008 \times 10^{-14}$ |
| $45^{\circ} \mathrm{C}$ | $4.074 \times 10^{-14}$ |
Hence, the option (b) is correct.
In statement (III), for a weak monobasic acid $HA$
$\begin{array}{r} \mathrm{H} A \rightleftharpoons \mathrm{H}^{\oplus}+A^{\ominus} \\ (1-\alpha) C \mathrm{M} \quad \alpha C \mathrm{M} \quad \alpha C \mathrm{M} \end{array}$
$\Rightarrow \mathrm{pH}$ of the solution is 5 , i.e.
${\left[\mathrm{H}^{+}\right]=10^{-5} \mathrm{M}=\alpha C} $
$\Rightarrow K_a=\frac{\alpha C \times \alpha C}{(1-\alpha) C}=\frac{10^{-5} \times \alpha}{1-\alpha} $
$\Rightarrow 10^{-5}=\frac{10^{-5} \times \alpha}{1-\alpha} $
$\Rightarrow \alpha=0.5 $
$\Rightarrow \alpha %=50$
Hence, the option (c) is correct.
In statement (IV), Le-Chatelier’s principle is applicable to common ion effect. Because, in presence of common ion (given) by strong electrolyte (say, $\left.\mathrm{Na}^{+} \bar{A}\right)$, the product of the concentration terms in RHS increases. For the weaker electrolyte, $\mathrm{H} A$ (say) the equilibrium shifts to the LHS, $\mathrm{H} A \rightleftharpoons \mathrm{H}^{\oplus}+A^{\ominus}$.
As a result dissociation of $\mathrm{H} A$ gets suppressed. Hence, the option (d) is incorrect.
BETA
