Respuesta :
Explanation:
We have to determine the molar solubility of [tex]Zn(CN)_{2}[/tex] in the given solution.
So, let molar solubility is x.
[tex]Zn(CN)_{2} \rightleftharpoons Zn^{2+} + 2CN^{-}[/tex]
0 x 2x
Also, HCN upon dissociation gives [tex]HCN \rightleftharpoons H^{+} + CN^{-}[/tex]
So, concentration of hydrogen ion equals the concentration of cynaide ion. Therefore, calculate the concentration of hydrogen ion using the pH value as follows.
pH = -log[tex][H^{+}][/tex]
1.33 = -log[tex][H^{+}][/tex]
antilog (-1.33) = [tex][H^{+}][/tex]
[tex][H^{+}][/tex] = [tex]4.68 \times 10^{-2}[/tex]
Let molar solubility of hydrogen and cyanide ions is "a".
[tex]HCN \rightleftharpoons H^{+} + CN^{-}[/tex]
0 a a
Also, [tex]K_{sp}[/tex] = [tex][Zn^{2+}][CN^{-}]^{2}[/tex]
= x [tex](x + a)^{2}[/tex]
= [tex]x \times a^{2}[/tex] ( x + a = a because a >> x)
x = [tex]\frac{K_{sp}}{a^{2}}[/tex]
= [tex]\frac{3.0 \times 10^{-16}}{(4.68 \times 10^{-2})^{2}}[/tex]
= [tex]1.369 \times 10^{-13} M[/tex]
Thus, we can conclude that molar solubility of [tex]Zn(CN)_{2}[/tex] in the given solution is [tex]1.369 \times 10^{-13} M[/tex].
