CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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%
loader
B
100%
loader
C
50%
loader
D
20%
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image