In 1 L saturated solution of AgCl KspAgCl - 1-6 - 10-10- 0-1 mol of CuCl Ksp CuCl - 1-0 - 10-6 is added- The resultant concentration of Ag- in the solution is 1-6 - 10-x- Calculate the value of x-

The solubility product of CuCl is 1.0 × 10^ 6 . #160; #160; #160; #160; #160; #160; #160; CuCl(s) → Cu^+ + Cl^ initially ...

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Byju's Answer

Standard XII

Chemistry

Solubility Product

In 1 L satu...

Question

In $1 L$ saturated solution of $A g C l (K_{s p} A g C l = 1.6 \times 10^{- 10})$ , 0.1 mol of $C u C l (K_{s p} C u C l = 1.0 \times 10^{- 6})$ is added. The resultant concentration of $A g^{\oplus}$ in the solution is $1.6 \times 10^{- x}$ . Calculate the value of $x$ .

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Solution

The solubility product of $C u C l$ is $1.0 \times 10^{- 6}$ .

$C u C l (s) \to C u^{+} + C l^{-}$
initially 0.1M 0 0
At eqm. 0.1-x x x

$K_{s p, C u C l} = [C u^{+}] [C l^{-}]$

$K_{s p, C u C l} = x \times x$

The chloride ion concentration will be, $[C l^{-}] = x = \sqrt{{K_{s p}}_{, C u C l}} = \sqrt{1.0 \times 10^{- 6}} = 1.0 \times 10^{- 3} M$

$\Rightarrow K_{s p, A g C l} = [A g^{+}] [C l^{-}]$

The silver ion concentration is, $[A g^{+}] = \frac{K_{s p, A g C l}}{[C l^{-}]} = \frac{1.6 \times 10^{- 10}}{1.0 \times 10^{- 3}} = 1.6 \times 10^{- 7} M = 1.6 \times 10^{- x} M$
Hence, the value of $x$ is $7$ .

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In 1 L saturated solution of AgCl (KspAgCl=1.6×10−10), 0.1 mol of CuCl (KspCuCl=1.0×10−6) is added. The resultant concentration of Ag⊕ in the solution is 1.6×10−x. Calculate the value of x.

In $1 L$ saturated solution of $A g C l (K_{s p} A g C l = 1.6 \times 10^{- 10})$ , 0.1 mol of $C u C l (K_{s p} C u C l = 1.0 \times 10^{- 6})$ is added. The resultant concentration of $A g^{\oplus}$ in the solution is $1.6 \times 10^{- x}$ . Calculate the value of $x$ .