Question

# $$17.4\%$$ $$K_2SO_4$$ 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 $$K_2SO_4$$ in aq. solution?[Note: Molarity of both the solution are equal]

A
40%
B
100%
C
50%
D
20%

Solution

## The correct option is C $$50\%$$For $$K_2SO_4$$,             $$K_2SO_4 \longrightarrow 2K^++SO_4^{2-}$$At eqm,   $$1-\alpha$$             $$2\alpha$$           $$\alpha$$       [$$\alpha$$= degree of dissociation of $$K_2SO_4$$]$$\therefore$$ Vant Hoff factor $$(i)=\cfrac {1-\alpha+\alpha+2\alpha}{1}=1+2\alpha$$Let the concentration be $$C$$ of $$K_2SO_4$$$$\therefore$$ Concentration at equilibrium $$= C(1+2\alpha)$$For $$NaOH$$,              $$NaOH \longrightarrow Na^++OH^-$$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 $$K_2SO_4$$ in aqueous solution is $$50$$% .Chemistry

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