Answer:
The concentration of [tex]FeSCN^{2+}[/tex] is 0.0574 M.
Explanation:
[tex]Fe^{3+}+SCN^-\rightleftharpoons FeSCN^{2+}[/tex]
Concentration of ferric ions ,[tex][Fe^{3+}] =0.048M[/tex]
Concentration of thiocyanate ions ,[tex][SCN^-]=0.014M[/tex]
Equilibrium constant for the reaction = [tex]K_c=85.49[/tex]
An expression of an equilibrium constant is given by :
[tex]K_c=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}[/tex]
[tex]85.49=\frac{x}{0.048 M\times 0.014M}[/tex]
[tex]x=85.49\times 0.048 M\times 0.014M=0.05745 M\approx 0.0574 M[/tex]
The concentration of [tex]FeSCN^{2+}[/tex] is 0.0574 M.
The concentration of FeM+ is 0.057 M.
The equilibrium constant is a value that shows the extent to which a reactant is being converted to products under the given conditions. Let the thiocyanide be M.
Now, the reaction equation is; Fe3+(aq) + M-(aq) ⇄ FeM2+. Now the Kc of the reaction is 85.49.
Hence;
85.49 = [FeM2+]/ [Fe3+] [M-]
85.49 = x/[0.048 M] [0.014 M]
x = 85.49 [0.048 M] [0.014 M]
x = 0.057 M
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