Two salts A 2 X and MX have the same value of solubility product of 4 . 0 × 10 - 12 . The ratio of their molar solubilizes i.e. S A 2 X S ( MX ) = ________. (Round off to the Nearest Integer).

For A 2 X A 2 X → 2 A + + X 2 - 2 S 1 S 1 K sp = 4 S 1 3 = 4 × 10 - 12 S 1 = 10 - 4 for MX MX → M + + X - S 2 S 2 K sp = S 2 2 = 4 × 10 - 12 S 2 = 2 × 10 - 6 So S A 2 x S MX = 10 - 4 2 × 10 - 6 = 50

Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductMedium2 minPYQ_2021

ChemistryMediumstatement

Two salts $A_{2} X$ and $MX$ have the same value of solubility product of $4.0 \times 10^{- 12}$ . The ratio of their molar solubilizes i.e. $\frac{S (A_{2} X)}{S (MX)} =$ ________.

(Round off to the Nearest Integer).

Answer:

Solution:

For $A_{2} X$

$\begin{matrix} A_{2} X & \to & 2 A^{+} & + & X^{2 -} \\ 2 S_{1} & S_{1} \end{matrix}$ $K_{sp} = 4 S_{1}^{3} = 4 \times 10^{- 12}$

$S_{1} = 10^{- 4}$

for $MX$

$\begin{matrix} MX & \to & M^{+} & + & X^{-} \\ S_{2} & S_{2} \end{matrix}$

$K_{sp} = S_{2}^{2} = 4 \times 10^{- 12}$

$S_{2} = 2 \times 10^{- 6}$

So $\frac{S_{A_{2} x}}{S_{MX}} = \frac{10^{- 4}}{2 \times 10^{- 6}} = 50$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2021

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