SATHEE: Equilibrium - Result Question 68

Equilibrium - Result Question 68

####71. Concentration of the $A g^{+}$ ions in a saturated solution of $A g_{2} C_{2} O_{4}$ is $2.2 \times 10^{- 4} m o l L^{- 1}$ . Solubility product of $A g_{2} C_{2} O_{4}$ is :- [2017]

(a) $2.66 \times 10^{- 12}$

(b) $4.5 \times 10^{- 11}$

(d) $2.42 \times 10^{- 8}$

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Solution:

$A g_{2} C_{2} O_{4} (s) \to \underset{2 s}{2 A g^{+}} (a q) + C_{2} O_{4}^{2 -} (a q)$

$K_{s p} = [A g^{+}]^{2} [C_{2} O_{4}^{2 -}]$

$[A g^{+}] = 2.2 \times 10^{- 4} M$

Given that:

$∴$ Concentration of $C_{2} O_{4}^{2 -}$ ions,

$[C_{2} O_{4}^{2 -}] = \frac{2.2 \times 10^{- 4}}{2} M = 1.1 \times 10^{- 4} M$

$\begin{aligned} (72.) & ∴ K_{s p} & = (2.2 \times 10^{- 4})^{2} (1.1 \times 10^{- 4}) \\ = 5.324 \times 10^{- 12} \end{aligned}$

(b) $M Y \to∣ M^{+} + Y^{-}$

$K_{s p} = s^{2} = 6.2 \times 10^{- 13}$

$s = \sqrt{{6.2}^{'} 10^{- 13}}$

$s = 7.87 \times 10^{- 7} m o l L^{- 1}$

$N Y_{3} \to∣ N^{3 +} + 3 Y^{-}$

$K_{s p} = s \times (3 s)^{3} = 27 s^{4} = 6.2 \times 10^{- 13}$

$s = (\frac{6.2 \times 10^{- 13}}{27})^{1 / 4}$

$s = 3.89 \times 10^{- 4} m o l L^{- 1}$

$∴$ molar solubility of $N Y_{3}$ is more than MY in water.

Equilibrium - Result Question 68

Solution:

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Equilibrium - Result Question 68

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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