Solubility product constant of Ag _{2} Cr O _{4} and AgBr are 1.1 × 10^{-12} and 5.0 × 10^{-13} re

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6-2.Equilibrium-II (Ionic Equilibrium)

medium

The solubility product constant of $Ag _{2} Cr O _{4}$ and $AgBr$ are $1.1 \times 10^{-12}$ and $5.0 \times 10^{-13}$ respectively. Calculate the ratio of the molarities of the molarities of their saturated solutions.

Option A

Option B

Option C

Option D

Let $s$ be the solubility of $A g_{2} C r O_{4}$

Then, $A g_{2} C r O_{4} \longleftrightarrow A g^{2+}+2 C r O_{4}$

$K_{s p}=(2 s)^{2} \cdot s=4 s^{3}$

$1.1 \times 10^{-12}=4 s^{3}$

$s=6.5 \times 10^{-5} \,M$

Let $s$ ' be the solubility of $AgBr$.

$AgBr _{(s)} \longleftrightarrow Ag ^{+}+ Br ^{-}$

$K_{s p}=s^{\prime 2}=5.0 \times 10^{-13}$

$\therefore s^{\prime}=7.07 \times 10^{-7} \,M$

Therefore, the ratio of the molarities of their saturatecd solution is

$\frac{s}{s^{\prime}}=\frac{6.5 \times 10^{-5} \,M }{7.07 \times 10^{-7}\, M }=91.9$

Standard 11

Chemistry

medium

(AIIMS-2002)

hard

(IIT-1997)

medium

(AIPMT-2002)

medium

medium