Respuesta :
Answer:
The molar solubility of the metal thiocyanate is [tex]1.127\times 10^{-4} M[/tex].
Explanation:
Concentration of potassium thiocyanate = 0.421 M
Concentration of thiocyanate ion =[tex][SCN^-]= 0.421 M[/tex]
Concentration of metal ion =[tex][M^{2+}]= ?[/tex]
The solubility product of metal thiocyanate = [tex]K_{sp}=2.00\times 10^{-5}[/tex]
[tex]M(SCN)_2\rightleftharpoons M^{2+}+2SCN^-[/tex]
S 2S
At equilbrium
S (2S+0.421)
The expression of solubility product is given by :
[tex]K_{sp}=[M^{2+}]\times [SCN^-]^2[/tex]
[tex]2.00\times 10^{-5}=S\times (2S+0.421)^2[/tex]
Solving for S:
[tex]S = 1.128\times 10^{-4} M[/tex]
[tex][M^{2+}]=\frac{2.00\times 10^{-5}}{(0.421 M)^2}=1.127\times 10^{-4} M[/tex]
The molar solubility of the metal thiocyanate is [tex]1.127\times 10^{-4} M[/tex].
