In: Chemistry

# At 22 °C an excess amount of a generic metal hydroxide M(OH)2 is mixed with pure water. The resulting equilibrium solution has a pH of 10.26. What is the Ksp of the salt at 22 °C?

At 22 °C an excess amount of a generic metal hydroxide M(OH)2 is mixed with pure water. The resulting equilibrium solution has a pH of 10.26. What is the Ksp of the salt at 22 °C?

## Solutions

##### Expert Solution

$$\mathrm{M}(\mathrm{OH}) 2(\mathrm{~s})<->\mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})$$

$$\mathrm{Ksp}=\left[\mathrm{M}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}$$

1) Use the $$\mathrm{pH}$$ to get the $$\mathrm{pOH}$$

$$14-10.26=3.74$$

2) Use the $$\mathrm{pOH}$$ to get the $$\left[\mathrm{OH}^{-}\right]$$

$$\left[\mathrm{OH}^{-}\right]=10^{-} \mathrm{pOH}=10^{-3.74}=1.82 \times 10^{-4}$$

3) $$\left[\mathrm{M}^{2+}\right]$$ is half the value of $$\left[\mathrm{OH}^{-}\right]$$, therefore:

$$\left[\mathrm{M}^{2^{+}}\right]=1.82 \times 10^{-4}$$ divided by $$2=9.10 \times 10^{-5}$$

4) Put values into Ksp expression

$$\mathrm{Ksp}=3.01 \times 10^{-12}$$