Question

# Solubility of $$AgCl$$ in $$0.2\ M\ NaCl$$ is $$x$$ and that in $$0.1\ M\ AgNO_{3}$$ is $$y$$ then which of the following is correct?

A
x=y
B
x>y
C
y>x
D
Cannot be predicted

Solution

## The correct option is B $$y>x$$In $$0.2M$$ $$NaCl$$, $$[Cl^-]=0.2$$$$(K_{sp})_{AgCl}=[Ag^+][Cl^-]$$$$\therefore (K_{sp})_{AgCl}=x\times 0.2$$$$\Rightarrow x=\cfrac {(K_{sp})_{AgCl}}{0.2}=5(K_{sp})_{AgCl}$$In $$0.1M$$ $$AgNO_3$$, $$[Ag^+]=0.1M$$$$\Rightarrow (K_{sp})_{AgCl}=0.1\times y$$$$\Rightarrow y=\cfrac {(K_{sp})_{AgCl}}{0.1}=10(K_{sp})_{AgCl}$$$$\therefore y >x$$Chemistry

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