Question

Calculate the \(K_{\mathrm{sp}}\) of the following compounds, given their molar solubilities. (a) TlBr: \(1.9 \times 10^{-3}\) (b) \(\mathrm{Ni}(\mathrm{OH})_{2}: 5.2 \times 10^{-6}\) (c) \(\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}: 3.7 \times 10^{-8}\)

Short answer

Question: Calculate the solubility product constant (\(K_{\mathrm{sp}}\)) for the following compounds using their molar solubilities: a) TlBr with a molar solubility of \(1.9 \times 10^{-3}\) b) Ni(OH)2 with a molar solubility of \(5.2 \times 10^{-6}\) c) Cu3(AsO4)2 with a molar solubility of \(3.7 \times 10^{-8}\) Answer: The solubility product constants for the given compounds are: a) TlBr: \(K_{\mathrm{sp}} = 3.61 \times 10^{-6}\) b) Ni(OH)2: \(K_{\mathrm{sp}} = 5.63 \times 10^{-16}\) c) Cu3(AsO4)2: \(K_{\mathrm{sp}} = 8.59 \times 10^{-44}\)

Step-by-step solution

(Compound (a) TlBr Dissolution Reaction)

Writing the dissolution reaction for TlBr: TlBr (s) \(\rightleftharpoons\) Tl\(^{+}\) (aq) + Br\(^{-}\) (aq)

(Expression for \(K_{\mathrm{sp}}\) of TlBr)

The expression for the solubility product constant (\(K_{\mathrm{sp}}\)) of TlBr will be: \(K_{\mathrm{sp}} = [Tl^{+}] [Br^{-}]\) The molar solubility of TlBr is given as \(1.9 \times 10^{-3}\). Since the stoichiometry of the dissolution reaction is 1:1 for Tl\(^{+}\) and Br\(^{-}\), we can directly substitute the molar solubility into the expression for \(K_{\mathrm{sp}}\).

(Calculate \(K_{\mathrm{sp}}\) of TlBr)

Substituting the molar solubility into the expression: \(K_{\mathrm{sp}} = (1.9 \times 10^{-3})(1.9 \times 10^{-3})\) \(K_{\mathrm{sp}} = 3.61 \times 10^{-6}\)

(Compound (b) Ni(OH)2 Dissolution Reaction)

Writing the dissolution reaction for Ni(OH)2: Ni(OH)\(_2\) (s) \(\rightleftharpoons\) Ni\(^{2+}\) (aq) + 2OH\(^{-}\) (aq)

(Expression for \(K_{\mathrm{sp}}\) of Ni(OH)2)

The expression for the solubility product constant (\(K_{\mathrm{sp}}\)) of Ni(OH)\(_2\) will be: \(K_{\mathrm{sp}} = [Ni^{2+}] [OH^{-}]^2\) The molar solubility of Ni(OH)\(_2\) is given as \(5.2 \times 10^{-6}\). Since the stoichiometry of the dissolution reaction is 1:2 for Ni\(^{2+}\) and OH\(^{-}\), we need to take this into account when substituting the molar solubility into the expression for \(K_{\mathrm{sp}}\).

(Calculate \(K_{\mathrm{sp}}\) of Ni(OH)2)

Substituting the molar solubility into the expression, considering the stoichiometry: \(K_{\mathrm{sp}} = (5.2 \times 10^{-6})(2 \times 5.2 \times 10^{-6})^2\) \(K_{\mathrm{sp}} = 5.63 \times 10^{-16}\)

(Compound (c) Cu3(AsO4)2 Dissolution Reaction)

Writing the dissolution reaction for \(\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\): \(\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) (s) \(\rightleftharpoons\) 3Cu\(^{2+}\) (aq) + 2AsO\(_{4}^{3-}\) (aq)

(Expression for \(K_{\mathrm{sp}}\) of Cu3(AsO4)2)

The expression for the solubility product constant (\(K_{\mathrm{sp}}\)) of \(\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) will be: \(K_{\mathrm{sp}} = [Cu^{2+}]^3 [AsO_{4}^{3-}]^2\) The molar solubility of \(\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is given as \(3.7 \times 10^{-8}\). Since the stoichiometry of the dissolution reaction is 3:2 for Cu\(^{2+}\) and AsO\(_{4}^{3-}\), we need to take this into account when substituting the molar solubility into the expression for \(K_{\mathrm{sp}}\).

(Calculate \(K_{\mathrm{sp}}\) of Cu3(AsO4)2)

Substituting the molar solubility into the expression, considering the stoichiometry: \(K_{\mathrm{sp}} = (3 \times 3.7 \times 10^{-8})^3 (2 \times 3.7 \times 10^{-8})^2\) \(K_{\mathrm{sp}} = 8.59 \times 10^{-44}\)