Question

17.4% (w/V) $K_{2}S{O}_4$ solution at ${27}^{o}C$ is isotonic to 4% (w/V) $NaOH$ solution at same temperature. If $NaOH$ is at 100% ionised, what is the percentage of the ionisation of $K_{2}S{O}_4$ in aqueous solution?

• A. 20%
• B. 60%
• C. 100%
• D. 50%

Answer 1

Answer: (D)

50%

As we know that,

$\pi =iMRT$

$\because M=\frac{n}{V}=\frac{w}{mV}$

$\therefore \pi =\frac{\left(\frac{w}{V}\right)iRT}{m}$

For the reaction-

$\underset{\text{At equilibrium}}{\underset{\text{Initially}}{}}\underset{1-\alpha }{\underset{1}{K_{2}S{O}_4}}\rightleftharpoons \underset{2\alpha }{\underset{0}{2K^{+}}}+\underset{\alpha }{\underset{0}{{S{O}_4}^{2-}}}$

Total moles at equilibrium $=(1-\alpha )+2\alpha +\alpha =1+2\alpha$

Therefore,

$i=\frac{1+2\alpha }{1}=1+2\alpha$

Molecular weight of $K_{2}S{O}_4\left(m\right)=174g$

Therefore,

${\pi }_1=\frac{17.4\times 1000\times R\times T}{174\times 100}(1+2\alpha )=Rt(1+2\alpha )$

Now for the reaction-

$\underset{\text{At equilibrium}}{\underset{\text{Initially}}{}}\underset{1-\alpha }{\underset{1}{NaOH}}\rightleftharpoons \underset{\alpha }{\underset{0}{{Na}^{+}}}+\underset{\alpha }{\underset{0}{{OH}^{-}}}$

Total moles at equilibrium $=(1-\alpha )+\alpha +\alpha =1+\alpha$

Therefore,

$i=\frac{1+\alpha }{1}=1+\alpha$

$\because \alpha =1\left(\text{Given}\right)$

$\therefore i=1+1=2$

Molecular weight of $NaOH\left(m\right)=40g$

Therefore,

${\pi }_2=\frac{4\times 1000\times R\times T}{40\times 100}\left(2\right)=2RT$

$\because \alpha =1$

Given that both the solution are isotonic.

Therefore,

$2RT=Rt(1+2\alpha )$

$2=1+2\alpha$

$\Rightarrow \alpha =\frac{1}{2}=0.5$

Thus the percentage of the ionization of $K_{2}S{O}_4$ in aqueous solution is $50\%$.

Hence the correct answer is $\left(D\right)50\%$.