Question

question 3 of 24
determine the value of ksp for cu(io₃)₂ by constructing an ice table, writing the solubility constant expression, and solving the expression. the molar solubility of cu(io₃)₂ is 2.7 × 10⁻³ m at this temperature. complete parts 1 - 2 before submitting your answer.
fill in the ice table with the appropriate value for each involved species to determine concentrations of all reactants and products at this temperature.
cu(io₃)₂(s) ⇌ cu²⁺(aq) + 2 io₃⁻(aq)
initial (m)
change (m)
equilibrium (m)
reset
0 2.7 × 10⁻³ - 2.7 × 10⁻³ 5.4 × 10⁻³ - 5.4 × 10⁻³ 1.00 2.00
4.00 - 2.00 - 1.00 - 4.00 1.4 × 10⁻⁸ 7.3 × 10⁻⁸ 2.0 × 10⁻⁵ 9.0 × 10⁻⁹
8.1 × 10⁻⁸ 1.5 × 10⁻⁸ - 1.4 × 10⁻⁸ - 7.3 × 10⁻⁸ - 2.0 × 10⁻⁵

Explanation:

Step 1: Initial concentrations

• For \(Cu(IO_{3})_{2}(s)\), since it is a solid, its concentration is not included in the solubility - product expression (we can represent it as "—" in the ICE table).
• Initially, before dissolution, the concentration of \(Cu^{2 +}(aq)\) is \(0\) (assuming no \(Cu^{2+}\) ions are present from other sources).
• Initially, before dissolution, the concentration of \(IO_{3}^{-}(aq)\) is \(0\) (assuming no \(IO_{3}^{-}\) ions are present from other sources).

Step 2: Change in concentrations

• Let \(s\) be the molar solubility of \(Cu(IO_{3})_{2}\). The balanced dissolution equation is \(Cu(IO_{3})_{2}(s)

ightleftharpoons Cu^{2 +}(aq)+2IO_{3}^{-}(aq)\).

• For \(Cu(IO_{3})_{2}(s)\), as it dissolves, the change in its "concentration" (again, not in the traditional molar - concentration sense for solids, but in terms of the reaction progress) is \(-s\). Given \(s = 2.7\times10^{-3}M\), we can represent it as \(- 2.7\times10^{-3}\) (but this is more of a formality for the ICE - table structure as solids don't have a molar - concentration in the solution - phase expression).
• For \(Cu^{2 +}(aq)\), using the stoichiometry of the reaction (\(1\) mole of \(Cu(IO_{3})_{2}\) gives \(1\) mole of \(Cu^{2+}\)), the change in concentration is \(+s=+2.7\times 10^{-3}\).
• For \(IO_{3}^{-}(aq)\), using the stoichiometry (\(1\) mole of \(Cu(IO_{3})_{2}\) gives \(2\) moles of \(IO_{3}^{-}\)), the change in concentration is \(+2s\). Substituting \(s = 2.7\times10^{-3}M\), we get \(+2\times(2.7\times10^{-3})=+5.4\times10^{-3}\).

Step 3: Equilibrium concentrations

• For \(Cu(IO_{3})_{2}(s)\), after dissolution, its "concentration" (again, non - traditional for solids) is still represented as "—".
• For \(Cu^{2 +}(aq)\), using the formula \(C = C_{initial}+ \Delta C\), \(C_{Cu^{2+}}=0 + 2.7\times10^{-3}=2.7\times10^{-3}\).
• For \(IO_{3}^{-}(aq)\), using the formula \(C = C_{initial}+\Delta C\), \(C_{IO_{3}^{-}}=0 + 5.4\times10^{-3}=5.4\times10^{-3}\).

The solubility - product expression \(K_{sp}=[Cu^{2 +}][IO_{3}^{-}]^{2}\).

Substitute \( [Cu^{2 +}]=2.7\times10^{-3}M\) and \( [IO_{3}^{-}]=5.4\times10^{-3}M\) into the \(K_{sp}\) expression:

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The ICE - table filled as follows:

	\(Cu(IO_{3})_{2}(s)\)	\(Cu^{2 +}(aq)\)	\(2IO_{3}^{-}(aq)\)
Change (M)	\(-2.7\times10^{-3}\)	\(+2.7\times10^{-3}\)	\(+5.4\times10^{-3}\)
Equilibrium (M)	—	\(2.7\times10^{-3}\)	\(5.4\times10^{-3}\)

Answer:

The \(K_{sp}\) value for \(Cu(IO_{3})_{2}\) is approximately \(7.9\times 10^{-8}\).