Concentration of the Ag^(+) ions in a saturated solution of Ag_2C_2O_4 is 2.2×10^(-4) mol L^(-1). Solubility product of Ag_2C_2O_4 is

Correct answer: (D) 5.3×10^(-12). Ag_2C_2O_4s ⇌ 2Ag^(+)aq + C_2O42-aq 2s s K_sp=Ag+^2 C_2O42- Ag^(+)=2.2×10^(-4)mol/L ∴ C_2O42-=(2.2×10^(-4))/2 M=1.1×10^(-4)M ∴ K_sp=2.2×10-4^2 1.1×10^(-4) (mol/L)^3 =5.324×10^(-12) (mol/L)^3.

Q56 / 180 · NEET 2017 · Chemistry

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

A
$2.42 \times 10^{- 8}$
B
$2.66 \times 10^{- 12}$
C
$4.5 \times 10^{- 11}$
Correct answer
D
$5.3 \times 10^{- 12}$

Solution

\begin{matrix} {Ag}_{2} C_{2} O_{4} (s) & ⇌ & 2 {Ag}^{+} (aq) & + & C_{2} O_{4}^{2 -} (aq) \\ 2 s & s \end{matrix}

K_{sp} = {[{Ag}^{+}]}^{2} [C_{2} O_{4}^{2 -}]

[{Ag}^{+}] = 2 .2 \times 10^{- 4} mol / L

∴ [C_{2} O_{4}^{2 -}] = \frac{2 .2 \times 10^{- 4}}{2} M = 1 .1 \times 10^{- 4} M

∴ K_{sp} = {(2 .2 \times 10^{- 4})}^{2} (1 .1 \times 10^{- 4}) (mol / L)^{3}

= 5 .324 \times 10^{- 12} (mol / L)^{3}

Hence, the correct answer is option: 'D'.

Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.2×10-4 mol L-1. Solubility product of Ag2C2O4 is

Solution

Relevant Questions

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