Question

$17.4\%$ $K_{2}S{O}_4$ solution at ${27}^o$C is isotonic with $4\%$ $NaOH$ solution at the same temperature. If $NaOH$ is $100\%$ ionised. What is the degree of ionisation of $K_{2}S{O}_4$ in aq. solution?

[Note: Molarity of both the solution are equal]

• A. $40\%$
• B. $100\%$
• C. $50\%$
• D. $20\%$

Answer 1

The correct option is C $50\%$

For $K_{2}S{O}_4$,

$K_{2}S{O}_4⟶2K^{+}+S{O}_4^{2-}$

At eqm, $1-\alpha$ $2\alpha$ $\alpha$ [$\alpha$= degree of dissociation of $K_{2}S{O}_4$]

$\therefore$ Vant Hoff factor $\left(i\right)=\frac{1-\alpha +\alpha +2\alpha }{1}=1+2\alpha$

Let the concentration be $C$ of $K_{2}S{O}_4$

$\therefore$ Concentration at equilibrium $=C(1+2\alpha )$

For $NaOH$,

$NaOH⟶N{a}^{+}+O{H}^{-}$

At eqm, $1-\beta$ $\beta$ $\beta$

$i=1-\beta +\beta +\beta =1+\beta$

$\therefore$ Let the concentration also be $C$ (since both solution are isotonic)

$\therefore$ Concentration at eqm= $C(1+\beta )$

Both solution are isotonic

$\therefore C(1+2\alpha )=C(1+\beta )$

$\Rightarrow 1+2\alpha =1+\beta =1+1$ [$\because \beta =1$ i.e. $100$% ionized]

$\Rightarrow 1+2\alpha =2$

$\Rightarrow \alpha =0.5$

$\therefore$ Degree of ionisation of $K_{2}S{O}_4$ in aqueous solution is $50$% .