The specific conductance of saturated solution of AgCl is found to be $1.86 \times 10^{- 6} o h m^{- 1} c m^{- 1}$ and that of water is $6 \times 10^{- 8} o h m^{- 1} c m^{- 1}$ . The solubility of AgCl given that $\land_{A g C l}^{\circ}$ = $137.2 o h m^{- 1} c m^{2} e q^{- 1}$ is

Q. What will be the solubility of

A g C l

in a

0.1 M

K C l

solution?

(K_{s p} A g C l = 1.20 \times 10^{- 10})

Q. Which is/are the correct statement(s) about the solubility of

A g C l (s)

?
Given :

K_{s p (A g C l)} = 10^{- 10}; K_{f} [A g (N H_{3})_{2}^{+}] = 10^{8}

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If solubility of AgCl is 0.2 M solution of AgNO3 is represented as y then find the value of y. (Given: Ksp(AgCl)=10−10)

The correct option is B 5×10−10MTh expression for the solubility product for AgCl is Ksp=[Ag+][Cl−]But Ksp=1×10−10,[Ag+]=0.2M and [Cl−]=x×10−10=y Substitute values in the above expression.1×10−10=0.2×x×10−10.Hence, x=10.2=5Hence, the value of y is 5×10−10

If solubility of $A g C l$ is $0.2 M$ solution of $A g N O_{3}$ is represented as $y$ then find the value of $y$ . (Given: $K_{s p (A g C l)} = 10^{- 10}$ )