Answer:
[tex][OH^-]= 1.12 \times 10^{-2} \ M[/tex]
Explanation:
[tex]\text{From the information given:}[/tex]
[tex]K_{sp} \ of \ Zn(OH)_2 \ is \ 3.0 \times 10^{-16}[/tex]
[tex]K_f \ of \ Zn(OH)_2 = 4.6 \times 10^{17}[/tex]
[tex]\mathtt{K_{sp} \ \ \ \ \ \ \ \ \mathtt{Zn(OH)_{2(s)}} \to Zn^{2+}_{(aq)} + 2OH^{-}_{(aq)} }[/tex]
[tex]\mathtt{K_{f} \ \ \ \ \ \ \ \ \mathtt{Zn^{2+}_{(aq)} } + 4OH^-_{(aq)} \to Zn(OH)^{2-}_{4}_{(aq)}}[/tex]
[tex]\mathtt{K \ \ \ \ \ \ \ \ \mathtt{{Zn(OH)_{2}_{(s)}}} + 2OH^{-}_{(aq)} \to Zn(OH)^{2-} _{4(aq) }}[/tex]
[tex]\text{To calculate the minimum concentration of OH}^{-}[/tex]
[tex]K = \dfrac{Zn(OH)^-_4}{[OH^-]^2} \\ \\[/tex]
[tex][OH^-]^2= \dfrac {0.017}{1.380 \times 10^2}[/tex]
[tex][OH^-]^2=1.232 \times 10^{-4}[/tex]
[tex][OH^-]= 1.12 \times 10^{-2} \ M[/tex]
The minimum concentration of OH- required to dissolve [tex]1.7\times10^-2\; mol[/tex] [tex]Zn(OH)_2[/tex] in a liter of solution is [tex]1.12 \times 10^-^2 \; M[/tex].
Concentration refers to the accumulation of a substance in a specific place.
The solubility equilibrium
[tex]ksp\;of\; Zn(OH)_2 = 3.0 \times 10^-^1^6\\kf \;of\; Zn(OH)_2 = 4.6 \times10^1^7[/tex]
Calculating the minimum concentration
[tex]k =\dfrac{Zn(OH)^-^4}{[OH^-]^2} \\\\[/tex]
[tex][OH^-]^2 = \dfrac{0.017}{1.380 \times 10^2} \\\\[OH^-]^2 = 1.232 \times 10^-^4\\[OH^-]^2 = 1.12 \times 10^-^2[/tex]
Thus, the minimum concentration of OH- required is [tex]1.12 \times 10^-^2 \; M[/tex]
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