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 determine which hydroxide will precipitate first when NH\(_4\)OH is added to a solution containing 1M A\(^{2+}\) and 1M B\(^{3+}\) ions, we need to compare their solubility product constants (K\(_{sp}\)).

Given:

• K\(_{sp}\)[A(OH)\(_2\)] = \(9 \times 10^{-10}\)
• K\(_{sp}\)[B(OH)\(_3\)] = \(27 \times 10^{-18}\)

For precipitation to occur, the ionic product of the hydroxide must exceed the K\(_{sp}\) value.

1. For A(OH)\(_2\), the precipitation condition is: \[ [A^{2+}][OH^-]^2 > K_{sp}[A(OH)_2] \]

Given [A\(^{2+}\)] = 1M, \([OH^-]^2 > 9 \times 10^{-10}\)

Simplifying: \[ [OH^-] > \sqrt{9 \times 10^{-10}} = 3 \times 10^{-5} \text{ M} \]

2. For B(OH)\(_3\), the precipitation condition is: \[ [B^{3+}][OH^-]^3 > K_{sp}[B(OH)_3] \]

Given [B\(^{3+}\)] = 1M, \([OH^-]^3 > 27 \times 10^{-18}\)

Simplifying: \[ [OH^-] > \sqrt[3]{27 \times 10^{-18}} = 3 \times 10^{-6} \text{ M} \]

Since \([OH^-] > 3 \times 10^{-6}\) M is required for B(OH)\(_3\) and \([OH^-] > 3 \times 10^{-5}\) M for A(OH)\(_2\), B(OH)\(_3\) will precipitate at a lower concentration of hydroxide ions.

Thus, B(OH)\(_3\) will precipitate before A(OH)\(_2\).