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.
Show Hint
- B(OH)\(_3\) will precipitate before A(OH)\(_2\)
- A(OH)\(_2\) and B(OH)\(_3\) will precipitate together
- A(OH)\(_2\) will precipitate before B(OH)\(_3\)
- Both A(OH)\(_2\) and B(OH)\(_3\) do not show precipitation with NH\(_4\)OH
The Correct Option is A
Approach Solution - 1
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 (Ksp) of each hydroxide.
- 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-]:
\([OH^-] = \sqrt{9 \times 10^{-10}} = 3 \times 10^{-5} \text{ M}\)
- 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-]:
\([OH^-] = \sqrt[3]{27 \times 10^{-18}} = 3 \times 10^{-6} \text{ M}\)
- 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 Ksp is reached at a lower [OH-] concentration.
Conclusion: B(OH)3 will precipitate before A(OH)2.
Approach Solution -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)\(_2\).
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