Question

Ignoring activities, determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 4.94. The Ksp for Zn(CN)2 is 3.0 × 10-16. The Ka for HCN is 6.2 × 10-10.

Answer #1

pH = 4.94

-log [H^{+}] = 4.94

[H^{+}] = 10^{-4.94} = 1.15 x
10^{-5}

Zn(CN)_{2} (s)
Zn^{2+}(aq) + 2CN^{-}(aq)

Ksp = [Zn^{2+}] [CN^{-}]^{2} = 3 x
10^{-16}

CN^{-}(aq) + H^{+}(aq)
HCN(aq)

Ka = [CN^{-}] [H^{+}] / [HCN] = 6.2 x
10^{-10}

[CN^{-}]_{Total} = [CN^{-}] + [HCN] =
2[Zn^{2+}]

Zn(CN)_{2}(s) + 2H^{+} ==⇒ Zn^{+2} +
2HCN

Need Keq for the above reaction:

[Zn^{+2}] [HCN]^{2} /
[H^{+}]^{2}

Now,

Ka = [CN^{-}] [H^{+}] / [HCN] = 6.2 x
10^{-10}

Square the above:

[CN^{-}]^{2} [H^{+}]^{2} /
[HCN]^{2} = (6.2 x 10^{-10})^{2} = 3.84 x
10^{-19}

[HCN]^{2} / [CN^{-}]^{2}[
H^{+}]^{2} = 1 / (3.84 x 10^{-19}) = 2.6 x
10^{18}

Multiply the above equation by the Ksp

([Zn^{+2}] [CN^{-}]^{2}) x (
[HCN]^{2}/ [CN^{-}]^{2}
[H^{+}]^{2} ) = (2.6 x 10^{18}) x (3.0 x
10^{-16})

[Zn^{+2}] [HCN]^{2} /
[H^{+}]^{2} = 780

Let x = molar solubility of Zn(CN)2

Then [Zn^{+2}] = x ; [HCN] = 2x and [H^{+}] = 1.15
x 10^{-5}

x(2x)^{2} / (1.15 x 10^{-5})^{2} = 780

4x^{3} = 780 x (1.15 x 10^{-5})^{2}

4x^{3} = 1.03 x 10^{-7}

x^{3} = 2.58 x 10^{-8}

x = 3.0 x 10^{-3}

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Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of
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Determine the [ZN] [CN] and [HCN] in a saturated solution of
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