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Kelvi

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Concentration of the Ag $^+$ ions in a saturated solution of Ag $_2$ C $_2$ O $_4$ is $2.2 \times 10^{-4}$ mol L $^{-1}.$ Solubility product of Ag $_2$ C $_2$ O $_4$ is

medium

Equilibrium

2017

chemistry

$2.66 \times 10^{-12}$

$4.5 \times 10^{-11}$

$5.3 \times 10^{-12}$

Explanation

To find the solubility product $(K_{sp})$ of Ag $_2$ C $_2$

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$2.42 \times 10^{-8}$

O

_4,

we start with the dissolution equilibrium:

\\

Ag

_2

C

_2

O

_4 (s) \rightleftharpoons 2

Ag

^+ (aq) +

C

_2

O

_4^{2-} (aq) \\

Let

s

be the solubility of

Ag

_2

C

_2

O

_4

in mol/L.

\\

From the stoichiometry of the reaction, we have:

\\ [

Ag

^+] = 2s \\ [

C

_2

O

_4^{2-}] = s \\

Given that the concentration of

Ag

^+

ions is

2.2 \times 10^{-4}

mol/L, we have:

\\ 2s = 2.2 \times 10^{-4} \\

Solving for

s: \\ s = \frac{2.2 \times 10^{-4}}{2} = 1.1 \times 10^{-4}

mol/L

\\

Now, we can calculate the solubility product

K_{sp}: \\ K_{sp} = [

Ag

^+]^2 [

C

_2

O

_4^{2-}] \\

Substitute the concentrations:

\\ K_{sp} = (2.2 \times 10^{-4})^2 (1.1 \times 10^{-4}) \\

Calculate:

\\ K_{sp} = (4.84 \times 10^{-8}) (1.1 \times 10^{-4}) \\ K_{sp} = 5.324 \times 10^{-12} \\

Therefore, the solubility product of

Ag

_2

C

_2

O

_4

is approximately

5.3 \times 10^{-12}. \\

This corresponds to Option 3.