# Calculate [Cr^3+] in equilibrium with Cr(OH4)- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0

I tried setting up a RICE chart with the initial conditions being:

$$\begin{array}{c|c} \hline \text{Species} & \text{Concentration / M} \\ \hline \ce{Cr^3+} & 0.010 \\ \ce{OH-} & 10^{-4.0} \\ \ce{Cr(OH)4-} & 0 \\ \hline \end{array}$$

I'm assuming I should use the given $K_\mathrm{f}$ value for $\ce{Cr(OH)4-}$ ($8 \times 10^{29}$).

However, when I solve for the change I end up with a negative value for the concentration of $\ce{Cr^{3+}}$? The correct answer is supposed to be $10^{-16}$.

• You probably mean $\ce{Cr^{3+}}$ and not $\ce{Cr3+}$. Here is a link on how to properly typeset chemical equations and other formulae using MathJax. Also, please write down the balanced reaction and we can proceed from there. – Todd Minehardt Jan 31 '17 at 2:07
• I forgot to mention that you do not have nitrate in your query. You'll need to account for that in the overall reaction. – Todd Minehardt Jan 31 '17 at 2:13
• You don't have to worry about the nitrate. The answer is the same as if you threw some $\ce{Cr(OH)3}$ into a solution buffered at pH 10. Until you know what you are doing don't ski[p steps by doing some of this in your head. Write out reaction, then the equilibrium equation, solve the equilibrium equation for $\ce{[Cr^{3+}]}$, and then do the RICE chart. – MaxW Jan 31 '17 at 4:25
• I've written out the full RICE chart and tried setting it up multiple ways, with no luck. – khajiit Feb 9 '17 at 22:30
• Could you please add the source of the question? (Book title, page number or question number) – orthocresol Mar 26 '17 at 22:57

I know this problem and the answer the book gives, and unfortunately the book is in error. Let me explain. Adding 0.010 mol of Cr(NO3)3 will also give you 0.010 mol of $\ce{Cr^3+}$, as we can assume the $\ce{Cr^3+}$ will completely dissociate from the $\ce{NO^3-}$. The $\ce{OH^-}$ concentration can be found from the pH, and it is 0.0001 $\frac{moles}{L}$.

The "before reaction starts" gets set up accordingly:

$\ce{Cr^3+ + 4OH- <=> Cr(OH)4-}$

0.010 M .0001 M 0 M

Notice, however, that there is a 4:1 ratio of $\ce{OH^-}$ to $\ce{Cr^3+}$. It takes 4 $\ce{OH^-}$ for every 1 $\ce{Cr^3+}$, and $\ce{OH^-}$ is quite obviously the limiting reactant. So after the reaction, what is the concentration of $\ce{Cr^3+}$? Let's make it 1:1, then.

$\frac{.0001 M}{4}$ = $\ce{2.5 x 10^{-5}}$. And this is essentially the [$\ce{Cr^3+}$] and [$\ce{Cr(OH)4-}$]. However, this is not what the book says. They meant to put the concentration of $\ce{Cr(OH)4-}$ as .010 M, not $\ce{Cr^3+}$. Let's try it that way, and see what we get!

Cr(OH)4- <=> Cr3+ + 4OH-

Here, in the equation, the concentrations are:

$\ce{Cr(OH)4-}$ = .010 M

$\ce{Cr^3+}$ = 0

4$\ce{OH^-}$ = .0001 M

-x +x +4x

.010 - x +x .0001 + 4x

Because .010 - x and .0001 + 4x are so small, let .010 - x = .010 and .0001 + 4x = .0001.

The $\ce{K_f}$ = 8 x $10^{29}$, but that is for the equation when it looks like:

$$\ce{Cr(OH)4- <=> Cr3+ + 4OH-}$$

We switched it around to find out how much $\ce{Cr(OH)4-}$ would dissolve to make $\ce{Cr^3+}$, so we have to use $\frac{1}{Kf}$, which = 1.25 x $10^{-30}$.

$\ce{K_f}$ = $\frac{[{Cr^3+}][{OH-}]^4}{[\ce{Cr(OH)4-}]}$

$\ce{K_f}$ = $\frac{(x)(.0001)^4}{(.010)}$

1.25 x $10^{-30}$ = $\frac{1 x 10^{-16}x}{.010}$

You can take it from there. The answer will be 1.25 x $10^{-16}$ M.