Ksp calculation of Cr(OH)2

Question

At $25~\mathrm{^\circ C}$, $10.24~\mathrm{mg}$ of $\ce{Cr(OH)2}$ are dissolved in enough water to make $125~\mathrm{mL}$ of solution. When equilibrium is established, the solution has a $\mathrm{pH}$ of $8.49$. Estimate $K_\text{sp}$ for $\ce{Cr(OH)2}$ (Ans : $1.47\times 10^{-17}$)

I have calculated it but not have the same answer as this. I tried it this way:

1. $\mathrm{pH}$ is $8.49$, then $\mathrm{pOH}$ is $5.51$. I use this formula: $\mathrm{pOH} = -\log[\ce{OH}]$
2. The $[\ce{OH}]$ is $10^{-5.51}$
3. Find molarity $\ce{Cr}$: $(10.24\cdot(1/1000)\cdot(1/86))/(0.125)=9.525\times 10^{-4}$
4. Substitute into $K_\text{sp} = [\ce{Cr}][\ce{OH}]^2$

But after I substitute value of $[\ce{Cr}] = 9.525\times 10^{-4}$ and $[\ce{OH}]$ of $(2\times 10^{-5.51})^2$, I get answer of $3.6\times 10^{-14}$.

Where have I gone wrong? Any Suggestions?

Answer 1

Your calculation of $[\ce{OH-}] = 10^{-5.51}\ \mathrm{mol\ l^{-1}}$ is correct.

The pH is well above 7. Therefore, the present $[\ce{OH-}]$ is mainly a result of the dissociation of $\ce{Cr(OH)2}$:

$$\ce{Cr(OH)2 <=> Cr^2+ + 2 OH- }$$

Thus, you know $[\ce{Cr^2+}] \approx \tfrac{1}{2} [\ce{OH-}] = \tfrac{1}{2}\cdot10^{-5.51}\ \mathrm{mol\ l^{-1}} $.

$K_\text{sp} = [\ce{Cr^2+}][\ce{OH-}]^2 = \tfrac{1}{2}\cdot10^{-5.51} \cdot \left(10^{-5.51}\right)^2 = 1.48\cdot 10^{-17}$