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Consider the insoluble compound zinc carbonate , ZnCO3 . The zinc ion also forms a complex with hydroxide ions . Write a balanced net ionic equation to show why the solubility of ZnCO3 (s) increases in the presence of hydroxide ions and calculate the equilibrium constant for this reaction. For Zn(OH)42- , Kf = 2.9×1015 . Use the pull-down boxes to specify states such as (aq) or (s).

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Answer:

The net ionic equation is  [tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]

The equilibrium constant is  [tex]K = 4.06 *10^{4}[/tex]

Explanation:

From the question we are that

      The  [tex]K_f = 2.9 *10^{15 }[/tex]

The ionic equation is chemical represented as

    Step 1

         [tex]ZnCO_3 _{(s)}[/tex]  ⇔   [tex]Zn^{2+} _{aq} + CO_3^{2-} _{aq}[/tex]   The  solubility product constant for stage is     [tex]K_{sp} = 1.4*10^{-11}[/tex]

 Step 2

        [tex]Zn^{2+} _{(aq)} + 4 0H^{-} _{(aq)}[/tex]    ⇔  [tex][Zn(OH_4)]^{2-} _{(aq)}[/tex]  The formation constant for this step is given as [tex]K_f = 2.9 *10^{15 }[/tex]

 The net reaction is  

           [tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]

The equilibrium constant is mathematically evaluated as

         [tex]K = K_{sp} * K_f[/tex]

substituting values

         [tex]K = 1.4*10^{-11} * 2.9 *10^{15}[/tex]

        [tex]K = 4.06 *10^{4}[/tex]

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