What is the [A2+] at Equilibrium When Mixed with B-?

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brbrett
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Homework Statement


Equal Volumes of 3.60 M A2+ and 6.80 M B- are mixed. After the reaction, equilibrium is established with [B-] = 0.40 M.
What is the [A2+] at equilibrium?

Homework Equations


Ice table, algebra

The Attempt at a Solution


This is probably a really easy problem, but I seem to be getting the wrong answer. Probably just a small lapse, but anyways, here's my work.

Equilibrium constant not given.

Ice Table:
A2+ (aq) + 3B- ==><== AB3-
I 3.60 6.80 0
C +x -3x +x
E 3.60 - x 0.40 +x

By subtracting 0.40 from 6.80, I determined 3x to be 6.4.
Using the ratios, I divided 3x by 3 to get 2.13(3 repeating).
Next I subtracted x from 3.60, and got my answer or 1.47.

The answer key says I am wrong, and that [A2+] at equilibrium is 0.80 M.

So am I going about this wrong, or is the book wrong (it has been in the past)?
Thanks!
 
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brbrett said:
A2+ (aq) + 3B- ==><== AB3-

Shouldn't it be $$A^{2+} + 2B^{-} \rightleftharpoons AB_2$$

$$\begin{array}[c!c!c!c!]
\text{ }& A^{2+} & 2B{-}&AB_2\\
\text{Initial concentration}& 3.60 & 6.80 & 0\\
\hline
\text{Change}& -x&-2x&x\\
\hline
\text{Equilibrium concentration }& 3.6 -x & 6.8 -2x=(0.40) & x\\
\hline
\end{array}$$Anyhow answer in books seems incorrect.
 
Last edited:
As long as the reaction is not given in the problem statement it can be everything.
 
It says "equal volumes of...are mixed". So immediately after mixing the initial concentrations are 1.8M and 3.4M. But as Borek says, if they do not specify the reaction there is no unique answer.
 
The reaction was specified in this case, and it is the one I posted above. This text has a lot of errors, so I'm not too surprised it's wrong this time. Good to double check though.
 
brbrett said:
The reaction was specified in this case, and it is the one I posted above.

Then the answer given is the correct one. See mjc123 post.