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 the order of precipitation when NH₄OH is added to a solution containing 1M A²⁺ and 1M B³⁺ ions, we need to calculate the concentration of hydroxide ions [OH^-] required to reach the solubility product (K_sp) of each hydroxide.

1. First, consider 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 \(K_{sp}[A(OH)_2] = 9 \times 10^{-10}\) and assuming equilibrium concentrations are approached as precipitation starts, with \([A^{2+}] = 1 \text{ M}\), we find:

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

Solve for [OH^-]:

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

1. Now, consider 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 \(K_{sp}[B(OH)_3] = 27 \times 10^{-18}\) and with \([B^{3+}] = 1 \text{ M}\), we find:

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

Solve for [OH^-]:

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

1. Comparison: The concentration of OH^- needed for B(OH)₃ to precipitate is \(3 \times 10^{-6} \text{ M}\), which is lower than the \(3 \times 10^{-5} \text{ M}\) required for A(OH)₂. This indicates B(OH)₃ will precipitate first as its K_sp is reached at a lower [OH^-] concentration.

Conclusion: B(OH)₃ will precipitate before A(OH)₂.

Answer 2

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)\(_2\).