Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Win,odd Dhamnekar on January 11, 2021, 09:21:11 AM
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Calculate the equilibrium concentration of Zn2+ in a Zn(CN)42-
My Answer:-
Zn2+(aq)+ 4CN (aq) :rarrow: ZnCN42-(aq) ; Kf =1×1018
Let x be the change in concentration as Zn2+ dissociates. Because the initial Zn2+ 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?
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Yes, but take care with some matters of presentation. Write the concentration of the complex as [Zn(CN)42-]. (I know we sometimes use square brackets in writing complexes, but this is a fairly simple one, and it's important to distinguish charges from exponents.) Also, there is no such species as "4CN". You should write [CN-] (don't forget the minus charge). [CN-] = 4x. We have discussed this before. Do you understand what the numbers mean in an equation?
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As per your reply, i made the necessary changes in my answer. I think it now comply with the rules for the presentation in chemistry.
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There's another point that occurs to me. You say "Since x is very small in comparison to 0.30, drop x". It's a good idea to check such an assumption at the end of your calculation. You could simply add a line:
"6.51 x 105 << 0.30, therefore the assumption was valid."
It does no harm to write this explicitly in your answer; even if you don't, it's good to get into the habit of making that check in your own mind whenever you use such an assumption.