Q.If the solubility product of CuS is $6 \times 10^{-16}$, calculate the maximum molarity of CuS in aqueous solution.

The key idea is that for a sparingly soluble salt like CuS, the molar solubility is directly related to its solubility product $K_{sp}$.

1. CuS dissociates in water as:

   $\text{CuS(s)} \rightleftharpoons \text{Cu}^{2+}(\text{aq}) + \text{S}^{2-}(\text{aq})$

2. If the molar solubility of CuS is $s$ mol/L, then $[\text{Cu}^{2+}] = s$ and $[\text{S}^{2-}] = s$.

3. The solubility product expression is:

   $K_{sp} = [\text{Cu}^{2+}][\text{S}^{2-}] = s \times s = s^2$ …

The key to this problem is understanding what "maximum molarity" means in the context of a sparingly soluble salt. When you drop CuS into water, only a tiny amount dissolves. The dissolved CuS breaks apart completely into ions:

The solution becomes saturated when no more solid can dissolve. At that point, the product of the ion concentrations hits a fixed ceiling — the solubility product constant, $K_{sp}$. That ceiling is what we use to find the molar solubility.

Now, let's walk through the calculation.

1. Write the dissociation equilibrium. Every mole of CuS that dissolves gives one mole of $\text{Cu}^{2+}$ and one mole of $\text{S}^{2-}$. If the molar solubility is $s$ mol/L, then:

2. Write the $K_{sp}$ expression. From the equilibrium:

3. Plug in the given value. We know $K_{sp} = 6 \times 10^{-16}$. So:

4. Solve for $s$. Take the square root of both sides:

Since $\sqrt{6} \approx 2.449$, we get: …

Method: Solubility Product (Ksp) to Molar Solubility

Concept (Why this works)

For a sparingly soluble salt, the solubility product $K_{sp}$ is the equilibrium constant for its dissolution. At saturation, the product of ion concentrations (each raised to the power of its stoichiometric coefficient) equals $K_{sp}$. The maximum molarity of the salt that can stay dissolved is its molar solubility, $s$.

Steps

Step 1: Write the dissociation equation

CuS is a 1:1 (AB-type) salt -- one mole of CuS gives one mole of $\text{Cu}^{2+}$ and one mole of $\text{S}^{2-}$.

Step 2: Express ion concentrations in terms of molar solubility $s$

If $s$ mol/L of CuS dissolves:

Step 3: Write the $K_{sp}$ expression

Step 4: Substitute and solve …

Here are the common mistakes students make when solving this type of solubility product problem, along with how to avoid each.

1. Forgetting the Stoichiometry of Dissociation

The Mistake:

Students often write $K_{sp} = [\text{Cu}^{2+}][\text{S}^{2-}]$ and then set $[\text{Cu}^{2+}] = [\text{S}^{2-}] = s$, but then incorrectly write $K_{sp} = s^2$ without checking the dissociation equation.

Why it happens:

They memorise the formula $K_{sp} = s^2$ for all 1:1 salts, but CuS dissociates as:

Here, one mole of CuS gives one mole of each ion, so $s = [\text{Cu}^{2+}] = [\text{S}^{2-}]$ and $K_{sp} = s^2$ is actually correct for this case. The mistake is assuming this holds for every salt (e.g., for $\text{Ag}_2\text{CrO}_4$, $K_{sp} = 4s^3$).

How to avoid:

Always write the balanced dissociation equation first. Then express each ion concentration in terms of $s$ (molar solubility). Only then substitute into the $K_{sp}$ expression.

2. Confusing Molar Solubility with $K_{sp}$

The Mistake:

Students think $K_{sp}$ is the solubility, so they answer $6 \times 10^{-16}$ M directly.

Why it happens:

They see “solubility product” and “maximum molarity” and assume they are the same number.

How to avoid:

Remember:

• $K_{sp}$ = equilibrium constant (product of ion concentrations at saturation).
• Molar solubility ($s$) = concentration of the salt that dissolves (in mol/L).

For CuS:

So the correct calculation is:

Key result: The maximum molarity is $2.45 \times 10^{-8}$ M, not $6 \times 10^{-16}$ M.

3. Incorrect Square Root Calculation

The Mistake:

Taking $\sqrt{6 \times 10^{-16}}$ and writing $3 \times 10^{-8}$ (forgetting to take square root of 6) or $6 \times 10^{-8}$ (taking square root of exponent only).

Why it happens:

Rushing the arithmetic or not knowing that $\sqrt{a \times 10^b} = \sqrt{a} \times 10^{b/2}$.

How to avoid:

Break it down step by step:

Then approximate $\sqrt{6} \approx 2.45$.

Always check: the exponent should be halved (from $-16$ to $-8$).

4. Ignoring Units or Significant Figures

The Mistake:

Writing the answer as $2.45 \times 10^{-8}$ without units, or using too many/few decimal places.

Why it happens:

Carelessness in final presentation.

How to avoid:

• Always include units: M (mol/L) or mol L$^{-1}$. …