Respuesta :
Answer:
A. 0.000128 M is the solubility of M(OH)2 in pure water.
B. [tex]3.23\times 10^{-6} M[/tex] is the solubility of [tex]M(OH)_2[/tex] in a 0.202 M solution of [tex]M(NO_3)_2[/tex].
Explanation:
A
Solubility product of generic metal hydroxide = [tex]K_{sp}=8.45\times 10^{-12}[/tex]
[tex]M(OH)_2\rightleftharpoons M^{2+}+2OH^-[/tex]
S 2S
The expression of a solubility product is given by :
[tex]K_{sp}=[M^{2+}][OH^-]^2[/tex]
[tex]K_{sp}=S\times (2S)^2=4S^3[/tex]
[tex]8.45\times 10^{-12}=4S^3[/tex]
Solving for S:
[tex]S=0.000128 M[/tex]
0.000128 M is the solubility of M(OH)2 in pure water
B
Concentration of [tex]M(NO_3)_2[/tex] = 0.202 M
Solubility product of generic metal hydroxide = [tex]K_{sp}=8.45\times 10^{-12}[/tex]
[tex]M(OH)_2\rightleftharpoons M^{2+}+2OH^-[/tex]
S 2S
So, [tex][M^{2+}]=0.202 M+S[/tex]
The expression of a solubility product is given by :
[tex]K_{sp}=[M^{2+}][OH^-]^2[/tex]
[tex]8.45\times 10^{-12}=(0.202 M+S)(2S)^2[/tex]
Solving for S:
[tex]S=3.23\times 10^{-6} M[/tex]
[tex]3.23\times 10^{-6} M[/tex] is the solubility of [tex]M(OH)_2[/tex] in a 0.202 M solution of [tex]M(NO_3)_2[/tex].