Question

Which of the following happens when NH\(_4\)OH is added gradually to the solution containing 1M A\(^{2+}\) and 1M B\(^{3+}\) ions? Given: K\(_{sp}\)[A(OH)\(_2\)] = 9 \(\times\) 10\(^\text{-10}\) and K\(_{sp}\)[B(OH)\(_3\)] = 27 \(\times\) 10\(^\text{-18}\) at 298 K.

• A. B(OH)\(_3\) will precipitate before A(OH)\(_2\)
• B. A(OH)\(_2\) and B(OH)\(_3\) will precipitate together
• C. A(OH)\(_2\) will precipitate before B(OH)\(_3\)
• D. Both A(OH)\(_2\) and B(OH)\(_3\) do not show precipitation with NH\(_4\)OH

Hint:

When adding NH\(_4\)OH to a solution containing metal ions, the ion that reaches its precipitation limit first (due to its lower required OH\(^-\) concentration) will precipitate first.

Answer 1

The Correct Option is A

To ascertain the precipitation order when NH₄OH is added to a solution containing 1M A²⁺ and 1M B³⁺ ions, the hydroxide ion concentration ([OH^-]) required to saturate the solubility product (K_sp) for each hydroxide must be computed.

1. Commence with the dissociation of A(OH)₂: \(A(OH)_2 \leftrightarrow A^{2+} + 2OH^-\)  

The solubility product expression is:

\(K_{sp}[A(OH)_2] = [A^{2+}][OH^-]^2\)

Given K_sp[A(OH)₂] = 9 × 10^-10 and assuming equilibrium concentrations at the onset of precipitation, with [A²⁺] = 1 M, we determine:

\(9 \times 10^{-10} = 1 \times [OH^-]^2\)

Solving for [OH^-]:

\([OH^-] = \sqrt{9 \times 10^{-10}} = 3 \times 10^{-5} \text{ M}\)

1. Next, examine the dissociation of B(OH)₃:  \(B(OH)_3 \leftrightarrow B^{3+} + 3OH^-\)

The solubility product expression is:

\(K_{sp}[B(OH)_3] = [B^{3+}][OH^-]^3\)

Given K_sp[B(OH)₃] = 27 × 10^-18 and with [B³⁺] = 1 M, we find:

\(27 \times 10^{-18} = 1 \times [OH^-]^3\)

Solving for [OH^-]:

\([OH^-] = \sqrt[3]{27 \times 10^{-18}} = 3 \times 10^{-6} \text{ M}\)

1. Comparison: The OH^- concentration required for B(OH)₃ precipitation is 3 × 10^-6 M, which is less than the 3 × 10^-5 M needed for A(OH)₂. Consequently, B(OH)₃ will precipitate first as its K_sp is achieved at a lower [OH^-].

Conclusion: B(OH)₃ precipitates prior to A(OH)₂.