Question 67 of 69intermediate🔧 ApplyNumerical3 marks

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp=6.3×10−18K_{sp} = 6.3 \times 10^{-18}).

Correct Answer

5.02×10−95.02 \times 10^{-9} M

Exercise: EXERCISES | Q: 6.71 | (Page 46)
For More Understanding

Explanation

To prevent the precipitation of iron sulphide, the ionic product of the constituent ions must not exceed the solubility product constant (KspK_{sp}). When two equimolar solutions are mixed in equal volumes, the concentration of each solution is halved due to dilution. By equating the ionic product to the KspK_{sp}, we can find the maximum initial concentration required to just avoid precipitation.

Solution Steps

  1. Step 1: Let the maximum concentration of the equimolar solutions of ferrous sulphate (FeSO4\text{FeSO}_4) and sodium sulphide (Na2S\text{Na}_2\text{S}) be cc M.

  2. Step 2: When these two solutions are mixed in equal volumes, the total volume doubles, causing the concentration of each salt to become half. Therefore, [Fe2+]=c2[\text{Fe}^{2+}] = \frac{c}{2} and [S2−]=c2[\text{S}^{2-}] = \frac{c}{2}.

  3. Step 3: For no precipitation of iron sulphide (FeS\text{FeS}), the ionic product (QQ) must be less than or equal to the solubility product (KspK_{sp}). Thus, [Fe2+][S2−]≤Ksp[\text{Fe}^{2+}][\text{S}^{2-}] \le K_{sp}.

  4. Step 4: Substituting the concentrations into the inequality: (c2)×(c2)≤6.3×10−18\left(\frac{c}{2}\right) \times \left(\frac{c}{2}\right) \le 6.3 \times 10^{-18}, which simplifies to c24≤6.3×10−18\frac{c^2}{4} \le 6.3 \times 10^{-18}.

  5. Step 5: Solving for cc, we get c2≤25.2×10−18c^2 \le 25.2 \times 10^{-18}. Taking the square root of both sides gives c≤25.2×10−18=5.02×10−9c \le \sqrt{25.2 \times 10^{-18}} = 5.02 \times 10^{-9} M. Hence, the maximum concentration is 5.02×10−95.02 \times 10^{-9} M.

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