Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed. Will it lead to the precipitation of copper iodate? (For cupric iodate K_sp = 7.4 × 10^–8).

Question

Predict expression for \( \alpha \) in terms of \( K_{eq} \) and concentration
C: \( A_2B_3 \, (aq) \rightleftharpoons 2A^{3+} (aq) + 3B^{2-} (aq) \)

Answer 1

Cu^{2+} = ce{IO3-} = {0.002}{2} = 0.001 M

Ion Product

Q = Cu^{2+} ce{IO3-}^2 = 10^{-3} * 10^{-3})^2 = 1.0 * 10^{-9} \)

Ksp = 7.4 * 10^{-8}

Q < Ksp → No precipitation

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed. Will it lead to the precipitation of copper iodate? (For cupric iodate K_sp = 7.4 × 10^–8).

Solution and Explanation

Ion Product

Top Questions on Law Of Chemical Equilibrium And Equilibrium Constant

Questions Asked in CBSE Class XI exam

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed. Will it lead to the precipitation of copper iodate? (For cupric iodate Ksp = 7.4 × 10–8).

Solution and Explanation

Ion Product

Top Questions on Law Of Chemical Equilibrium And Equilibrium Constant

Questions Asked in CBSE Class XI exam

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed. Will it lead to the precipitation of copper iodate? (For cupric iodate K_sp = 7.4 × 10^–8).