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

AI Thread Summary
The discussion centers on calculating the equilibrium concentration of A2+ after mixing equal volumes of 3.60 M A2+ and 6.80 M B-. After establishing that [B-] is 0.40 M, the initial attempt at solving the problem led to an incorrect conclusion of 1.47 M for [A2+] at equilibrium. The correct approach, as clarified by participants, involves recognizing the correct stoichiometry of the reaction, which should be A2+ + 2B- ⇌ AB2. Ultimately, the consensus is that the equilibrium concentration of [A2+] is indeed 0.80 M, confirming the answer provided in the textbook. The importance of correctly interpreting the reaction and initial concentrations is emphasized in the discussion.
brbrett
Messages
35
Reaction score
0

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!
 
Physics news on Phys.org
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.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top