In 1 L saturated solution of AgCl -Ksp-1-6- 10-10- 0-01 mole of CuCl -Ksp-1- 10-6- is added- The resultant concentration of Ag- in the solution is 1-6- 10-x M- The value of x is -

K_sp(CuCl) gt; gt;K_sp(AgCl) Thus assume [Cl^ ] in solution comes from CuCl only [Cl^ ]=√(K_spCuCl)=10^ 3 M For AgCl [Ag^+][Cl^ ]=1.6× 10^ 10 [Ag^+]=1.6× 1 ...

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Byju's Answer

Standard XII

Chemistry

Solubility Product

In 1 L satura...

Question

In $1 L$ saturated solution of $A g C l [K_{s p} = 1.6 \times 10^{- 10}]$ , $0.01 m o l e o f C u C l [K_{s p} = 1 \times 10^{- 6}]$ is added. The resultant concentration of $A g^{+}$ in the solution is $1.6 \times 10^{- x} M$ . The value of $x$ is :

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Solution

$K_{s p} (C u C l) >> K_{s p} (A g C l)$
Thus assume $[C l^{-}]$ in solution comes from $C u C l$ only
$[C l^{-}] = \sqrt{K_{s p} C u C l} = 10^{- 3} M$
$For A g C l$
$[A g^{+}] [C l^{-}] = 1.6 \times 10^{- 10}$
$[A g^{+}] = 1.6 \times 10^{- 7}$
$x = 7$

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In 1 L saturated solution of AgCl [Ksp=1.6×10−10], 0.01 mole of CuCl [Ksp=1×10−6] is added. The resultant concentration of Ag+ in the solution is 1.6×10−x M. The value of x is :

Ksp(CuCl)>>Ksp(AgCl) Thus assume [Cl−] in solution comes from CuCl only [Cl−]=√KspCuCl=10−3 M For AgCl [Ag+][Cl−]=1.6×10−10 [Ag+]=1.6×10−7 x=7

In $1 L$ saturated solution of $A g C l [K_{s p} = 1.6 \times 10^{- 10}]$ , $0.01 m o l e o f C u C l [K_{s p} = 1 \times 10^{- 6}]$ is added. The resultant concentration of $A g^{+}$ in the solution is $1.6 \times 10^{- x} M$ . The value of $x$ is :

$K_{s p} (C u C l) >> K_{s p} (A g C l)$
Thus assume $[C l^{-}]$ in solution comes from $C u C l$ only
$[C l^{-}] = \sqrt{K_{s p} C u C l} = 10^{- 3} M$
$For A g C l$
$[A g^{+}] [C l^{-}] = 1.6 \times 10^{- 10}$
$[A g^{+}] = 1.6 \times 10^{- 7}$
$x = 7$