# Finding the concentration of each ion in saturated solution of lead(II) hydroxide

### Question

If the solubility of $$\ce{Pb(OH)2}$$ is $$\pu{0.155 g/L},$$ then the concentration of each ion in a saturated solution of $$\ce{Pb(OH)2}$$ is

A. $$[\ce{Pb^2+}] = \pu{0.155 g/L}$$ and $$[\ce{OH-}] = \pu{0.155 g/L}$$
B. $$[\ce{Pb^2+}] = \pu{0.052 g/L}$$ and $$[\ce{OH-}] = \pu{0.103 g/L}$$
C. $$[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$$ and $$[\ce{OH-}] = \pu{1.29E-3 mol/L}$$
D. $$[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$$ and $$[\ce{OH-}] = \pu{6.43E-4 mol/L}$$

C. $$[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$$ and $$[\ce{OH-}] = \pu{1.29E-3 mol/L}$$

### My Approach

What I did is that I wrote the dissociation equation:

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

Then I used to mole ratio to discover that:

\begin{align} γ(\ce{Pb^2+}) &= \pu{0.155 g/L}\\ γ(\ce{OH-}) &= \pu{0.31 g/L} \end{align}

I then converted the mass concentration $$γ$$ into molar concentration $$c$$ since the $$\pu{mol/L}$$ is commonly used to describe solubility:

$$\pu{0.155 g/L} × \frac{\pu{1 mol}}{\pu{207.2 g}} = \pu{7.48E-4 mol/L}$$

However, that's not the answer. What did I do wrong?

The mistake you made is that you assumed mass concentrations of the ions from the stoichiometric relation, which is only applicable to the amount of substance, and, as a consequence, amount concentration.

You have to convert mass concentration $$γ$$ to the molar concentration c first:

$$c(\ce{Pb(OH)2}) = \frac{γ(\ce{Pb(OH)2})}{M(\ce{Pb(OH)2})} = \frac{\pu{0.155 g L-1}}{\pu{241.21 g mol-1}} = \pu{6.43E-4 mol L-1}$$

And, since it's a saturated solution, then a nearly complete dissociation can be assumed:

$$[\ce{Pb^2+}] = c(\ce{Pb(OH)2}) = \pu{6.43E-4 mol L-1}$$

$$[\ce{OH-}] = 2 × c(\ce{Pb(OH)2}) = 2 × \pu{6.43E-4 mol L-1} = \pu{1.29E-3 mol L-1}$$

• RE: "And, since it's a saturated solution, then a nearly complete dissociation can be assumed:..." Another way of saying this is that the concentration of $\ce{[OH-]}$ is high enough that the autodissociation of water can be ignored. In yet other words, essentially all the $\ce{OH-}$ is from the dissolution of the $\ce{Pb(OH)_2}$.
– MaxW
Dec 16 '19 at 9:05