A3B2 is a sparingly soluble salt of molar mass M (g mol^-1) and solubility x g L^-1. The solubility

Q: Why is the concentration of $A^{2+}$ written as $3S$ and not just $S$?

Because of the stoichiometry of the salt $A_3B_2$. When 1 mole of the solid dissolves, it produces 3 moles of $A^{2+}$ ions. Therefore, if $S$ moles of the solid dissolve per liter, it will produce $3S$ moles of $A^{2+}$ ions per liter.

Q: Why do we raise the concentrations to the power of their coefficients in the $K_{sp}$ expression?

This comes directly from the Law of Mass Action. The equilibrium constant for any reaction is the product of the concentrations of the products raised to their stoichiometric coefficients, divided by the reactants. For sparingly soluble salts, the reactant is a pure solid, so its concentration is considered constant and incorporated into the $K_{sp}$ value.

Q: What is the difference between $x$ and $S$ in this specific question?

They both represent solubility, but in different units. $x$ is the solubility in grams per liter (g/L), which tells you the mass of salt dissolved. $S$ is the molar solubility in moles per liter (mol/L), which tells you the number of molecules (in moles) dissolved. $K_{sp}$ calculations strictly require molar solubility.

Q: Why do we ignore the solid $A_3B_2$ in the $K_{sp}$ formula?

The concentration of a pure solid is constant because its density and molar mass do not change regardless of how much solid is present. In equilibrium expressions, constant terms are merged into the equilibrium constant itself. Thus, $K_{sp}$ only includes the aqueous ions.

Q: Is there a shortcut formula for these types of salts?

Yes! For any general salt $A_x B_y$, the relationship between $K_{sp}$ and molar solubility $S$ is always $K_{sp} = x^x y^y S^{(x+y)}$. In this case, $x=3$ and $y=2$, so $K_{sp} = 3^3 \cdot 2^2 \cdot S^{(3+2)} = 27 \cdot 4 \cdot S^5 = 108S^5$.

Question

$A_3B_2$ is a sparingly soluble salt of molar mass $M$ (g mol$^{-1}$) and solubility $x$ g L$^{-1}$. The solubility product satisfies $K_{sp} = a \left(\frac{x}{M}\right)^5$. The value of $a$ is ..........

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are in a chemistry lab, and you drop a pinch of a sparingly soluble salt, $A_3B_2$, into a beaker of pure water. It doesn't vanish completely like table salt. Instead, a tiny, almost invisible fraction of it dissolves, breaking apart into its constituent ions.

This creates a delicate dance—a dynamic equilibrium between the solid chunk sitting at the bottom and the free-floating ions in the water. The balanced chemical equation for this dissociation is the foundation of our problem:

$$A_3B_2(s) ightleftharpoons 3A^{2+}(aq) + 2B^{3-}(aq)$$

Notice the stoichiometry here. For every single molecule of $A_3B_2$ that dissolves, it releases **three** $A^{2+}$ ions and **two** $B^{3-}$ ions into the solution.

### The Master Equation: Solubility Product
To quantify this equilibrium, we use the concept of Molar Solubility, denoted by $S$. This represents the maximum number of moles of the solid salt that can dissolve in one liter of water.

Based on our stoichiometry, if $S$ moles of $A_3B_2$ dissolve, the concentration of the resulting ions will be:
$$[A^{2+}] = 3S$$
$$[B^{3-}] = 2S$$

Now, we construct the Solubility Product Constant ($K_{sp}$) expression. Remember the golden rule: $K_{sp}$ is the product of the ion concentrations, each raised to the power of its stoichiometric coefficient.

$$K_{sp} = [A^{2+}]^3 [B^{3-}]^2$$

### Executing the Calculation
Let's substitute our equilibrium concentrations into the $K_{sp}$ expression. Don't rush through the algebra; this is where silly mistakes happen!

$$K_{sp} = (3S)^3 	imes (2S)^2$$
$$K_{sp} = (27S^3) 	imes (4S^2)$$
$$K_{sp} = 108S^5$$

We have a beautiful, clean expression for $K_{sp}$ in terms of molar solubility $S$. But wait, there is a catch! The question doesn't give us $S$ directly. It gives us the solubility $x$ in units of **grams per liter (g/L)** and the molar mass $M$ in **g/mol**.

We must convert this mass-based solubility into molar solubility. How do we convert grams to moles? We divide by the molar mass!

$$S = \frac{x}{M}$$

### The Final Reveal
Now, we bring it all together. Substitute this expression for $S$ back into our $K_{sp}$ equation:

$$K_{sp} = 108 \left(\frac{x}{M}ight)^5$$

The problem states that $K_{sp} = a \left(\frac{x}{M}ight)^5$. By simply comparing our derived equation with the given one, the answer reveals itself in all its glory.

**$	herefore a = 108$**

This problem is a classic test of keeping your units straight and respecting the stoichiometry of the dissociation. Master this, and you can conquer any solubility product question JEE throws at you!

Animated Solution for Chemistry - Ionic Equilibrium: $A_{3} B_{2}$ is a sparingly soluble salt of molar mass $M$ (g mol $^{- 1}$ ) and solubility $x$ g L $^{- 1}$ . The solubility product satisfies $K_{s p} = a (\frac{x}{M})^{5}$ . The value of $a$ is ..........

Enter Numerical Value:

Visualized Solution

$A_{3} B_{2}$ Dissociation Setup

Dissociation Equation

Molar Solubility ( $S$ )

Solubility Product ( $K_{s p}$ ) Expression

Calculating $K_{s p}$ in terms of $S$

Relating $S$ with $x$ and $M$

Final Comparison

The Way Forward

The Sigma Insight: Solubility Product and Common Ion Effect

Analyzing the Setup Imagine you are in a chemistry lab, and you drop a pinch of a sparingly soluble salt, $A_{3} B_{2}$ , into a beaker of pure water

The Master Equation

Executing the Calculation Let's substitute our equilibrium concentrations into the $K_{s p}$ expression

The Final Reveal Now, we bring it all together

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3​B2​ Dissociation Setup

Dissociation Equation

Molar Solubility ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S)

Solubility Product ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ksp​) Expression

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ksp​ in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S