## Chemistry and Chemical Reactivity (9th Edition)

1. Calculate the molar mass $(PbSO_4)$: 207.2* 1 + 32.07* 1 + 16* 4 ) = 303.27g/mol 2. Calculate the number of moles $(PbSO_4)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.055}{ 303.27}$ $n(moles) = 1.8\times 10^{- 4}$ 3. Find the concentration in mol/L $(PbSO_4)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 1.8\times 10^{- 4}}{ 0.25}$ $C(mol/L) = 7.3\times 10^{- 4}$ ------ Now, find the solubility: 4. Write the $K_{sp}$ expression: $PbSO_4(s) \lt -- \gt 1Pb^{2+}(aq) + 1S{O_4}^{2-}(aq)$ $2.5 \times 10^{-8} = [Pb^{2+}]^ 1[S{O_4}^{2-}]^ 1$ 5. Considering a pure solution: $[Pb^{2+}] = 1S$ and $[S{O_4}^{2-}] = 1S$ $2.5 \times 10^{-8}= ( 1S)^ 1 \times ( 1S)^ 1$ $2.5 \times 10^{-8} = 1S^ 2$ $2.5 \times 10^{-8} = S^ 2$ $\sqrt [ 2] {2.5 \times 10^{-8}} = S$ $1.6 \times 10^{-4} = S$ - This is the molar solubility value for this salt. ** Since the concentration added is greater than the solubility value, there will be precipitation, and only $1.6 \times 10^{-4}M$ will dissolve. 6. Find the number of moles: $Concentration(M) = \frac{n(mol)}{V(L)}$ $1.6 \times 10^{-4} = \frac{n(mol)}{0.25}$ $1.6 \times 10^{-4} * 0.25 = n(mol)$ $4.0 \times 10^{-5} moles = n(mol)$ 8. Calculate the mass $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $mm(g/mol) * n(mol) = mass(g)$ $303.27 * 4 \times 10^{-5} = mass(g)$ $0.012 = mass(g)$