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

Condition for precipitation \( Q_{ip}>K_{sp} \) For A(OH)\(_2\): \[ [A^{2+}][OH^-]^2>9 \times 10^{-10} \] \[ [A^{2+}] = 1M \] \[ \Rightarrow [OH^-]>3 \times 10^{-5} \, M \] For B(OH)\(_3\): \[ [B^{3+}][OH^-]^3>27 \times 10^{-18} \] \[ [B^{3+}] = 1M \] \[ \Rightarrow [OH^-]>3 \times 10^{-6} \, M \] So, B(OH)\(_3\) will precipitate before A(OH)\(_2\).