Calculate the equilibrium concentration of Zn
2+ in a Zn(CN)
42-My Answer:-
Zn
2+(aq)+ 4CN (aq)
ZnCN
42-(aq) ; K
f =1×10
18Let x be the change in concentration as Zn
2+ dissociates. Because the initial Zn
2+ concentration is 0, the concentration at any times is x:
[tex]1.0\times10^{18}=\frac{[Zn(CN)_4]^{2-}}{[Zn^{2+}][CN^-]^4}=\frac{0.30-x}{x(4x)^4}[/tex]
[tex]1\times 10^{18}*(256x^5)=0.30- x[/tex]
Since x is very small in comparison to 0.30 M, drop x:
Solving this equation, we get x=6.51 × 10
-5 M
Is this answer correct?