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

For the reaction \( \mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3} \), the equilibrium constant \(K_c\) is 0.5 at a certain temperature. If initially, 1 mole of \(\mathrm{N_2}\) and 3 moles of \(\mathrm{H_2}\) are placed in a 1 L container, what is the equilibrium concentration of \(\mathrm{NH_3}\)?

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

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

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).