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21 October, 02:50

The Kp of calcium iodate is 7.1 x 107 What is the concentration of Ca2 + in equilibrium with Ca 3.3 x 10-2 M? 2.3.3 * 10M b. 8.1 x 10 SM c. 2.2 * 10-SM d. 6.5 x 10 M e. 7.1 x 10-3 M

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  1. 21 October, 03:12
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    Hey there!:

    Given Ksp of Ca (IO₃) ₂ = 7.1 x 10⁻⁷

    At the equilibrium cation will be:

    Ca (IO₃) ₂ (s) ⇌ Ca²⁺ (aq) + 2 IO₃⁻ (aq):

    Ksp = [ Ca²⁺ ] [ IO₃⁻ ]² given [ IO₃⁻ ]² = 3.3 x 10⁻² M

    therefore:

    Ksp = [ Ca²⁺ ] [ IO₃⁻ ]²

    7.1 x 10⁻⁷ = [ Ca²⁺ ] [ 3.3 x 10⁻² ] ²

    [ Ca²⁺ ] = [ 7.1 x 10⁻⁷ ] / [ 3.3 x 10⁻² ] ²

    [ Ca²⁺ ] = [ 7.1 x 10⁻⁷ ] / [ 10.89 x 10⁻⁴ ]

    [ Ca²⁺ ] = 6.5 x 10⁻⁴ M
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