How to solve Ksp to Solubility (S) conversions in NEET

Numerical/Formula

1. TITLE

How to solve Ksp to Solubility (S) conversions in NEET — Step-by-step

PROBLEM TYPE INTRO

This category of numerical involves finding the molar solubility ($S$) of sparingly soluble salts from their solubility product constant ($K_{sp}$). We will use the general formula $K_{sp} = x^x y^y S^{(x+y)}$. A typical mistake students make is forgetting to apply the stoichiometric coefficients both as multipliers for $S$ and as exponential powers.

MCQ

Question:
The solubility product constant ($K_{sp}$) of a sparingly soluble salt $A_2X_3$ in pure water is $1.1 \times 10^{-23}$ at 298 K. Assuming that neither kind of ion reacts with water, what is the molar solubility ($S$) of the salt?

(a) $1.0 \times 10^{-5} \text{ mol L}^{-1}$
(b) $2.56 \times 10^{-5} \text{ mol L}^{-1}$
(c) $1.8 \times 10^{-5} \text{ mol L}^{-1}$
(d) $3.3 \times 10^{-12} \text{ mol L}^{-1}$

STEP-BY-STEP SOLUTION

SOLUTION

Given:
* $K_{sp}$ of $A_2X_3$ = $1.1 \times 10^{-23}$
* Temperature = $298 \text{ K}$

To Find:
* Molar solubility ($S$) in $\text{mol L}^{-1}$

Formula:
$K_{sp} = x^x y^y S^{(x+y)}$
Use this formula to directly relate the solubility product constant to the molar solubility for any general salt $M_xX_y$.

Step 1: Write the dissociation equation and identify coefficients
For the salt $A_2X_3$, the dissociation in water is:
$A_2X_3(s) \rightleftharpoons 2A^{3+}(aq) + 3X^{2-}(aq)$
Here, $x = 2$ and $y = 3$.

Step 2: Substitute into the general $K_{sp}$ formula
If $S$ is the solubility, the equilibrium concentrations are $[A^{3+}] = 2S$ and $[X^{2-}] = 3S$.
$K_{sp} = (2S)^2 (3S)^3$
$K_{sp} = (4S^2) (27S^3) = 108S^5$

Step 3: Calculate the value of S
$108S^5 = 1.1 \times 10^{-23}$
$S^5 = \frac{1.1 \times 10^{-23}}{108}$
$S^5 = 0.01018 \times 10^{-23} \approx 1.0 \times 10^{-25}$
$S = (1.0 \times 10^{-25})^{1/5}$
$S = 1.0 \times 10^{-5} \text{ mol L}^{-1}$

Answer: $1.0 \times 10^{-5} \text{ mol L}^{-1}$ (Option a)

WHERE STUDENTS GO WRONG

❌ Mistake 1: Forgetting to apply stoichiometric coefficients to the concentration and the power, leading to the incorrect expression $K_{sp} = S^5$ (results in Option b).
❌ Mistake 2: Adding the exponents instead of raising them correctly (e.g., $(2S)(3S) = 6S^5$), leading to incorrect base multipliers (results in Option c).
❌ Mistake 3: Assuming the simplest $AB$ type formula $K_{sp} = S^2$ regardless of the salt's stoichiometry, calculating $\sqrt{K_{sp}}$ (results in Option d).

QUICK FORMULA REFERENCE

Formulas Used:
* Formula 1 ($AB$ salts): $K_{sp} = S^2$ — Use when the salt produces 2 total ions (e.g., $AgCl$, $BaSO_4$).
* Formula 2 ($AB_2$ or $A_2B$ salts): $K_{sp} = 4S^3$ — Use when the salt produces 3 total ions (e.g., $Ni(OH)2$, $CaF_2$).
* Formula 3 ($A_2B_3$ salts): $K{sp} = 108S^5${sp} = 108S^5$ — Use when the salt produces 5 total ions (e.g., $A_2X_3$).

NCERT REFERENCE

Based on: NCERT Class 11, Chapter 6, Problem 6.26

CTA

More solved numericals → Equilibrium Complete Guide
Practice problems → [Topic Test: 6-Equilibrium]

How to solve Ksp to Solubility (S) conversions in NEET — Step-by-step