Solid lead nitrate is added to a solution that is \(0.020 \mathrm{M}\) in \(\mathrm{OH}^{-}\) and \(\mathrm{SO}_{4}^{2-}\). Addition of the lead nitrate does not change the volume of the solution. (a) Which compound, \(\mathrm{PbSO}_{4}\) or \(\mathrm{Pb}(\mathrm{OH})_{2}\left(K_{s p}=2.8 \times 10^{-16}\right)\), will precipitate first? (b) What is the \(\mathrm{pH}\) of the solution when \(\mathrm{PbSO}_{4}\) first starts to precipitate?

First, let's find the Ksp values for both lead sulfate (PbSO₄) and lead hydroxide (Pb(OH)₂). For PbSO₄, we can find its Ksp in a solubility table (it's not given), and for Pb(OH)₂, it's given as \(2.8 \times 10^{-16}\).

Next, we will determine the concentrations of lead ions required for PbSO₄ and Pb(OH)₂ to begin precipitating. We can use the Ksp expressions for this calculation: For PbSO₄: \[K_{sp} = [Pb^{2+}] [SO_{4}^{2-}]\] For Pb(OH)₂: \[K_{sp} = [Pb^{2+}] [OH^-]^2\] Let's plug in the known values and solve for the lead ion concentrations for each compound: For PbSO₄: Given that \([SO_{4}^{2-}] = 0.020\text{M}\), let's denote \([Pb^{2+}]\) as \(x\) and solve the Ksp expression: \[K_{sp} = x (0.020) \Rightarrow x = \frac{K_{sp}}{0.020}\] For Pb(OH)₂: Given that \([OH^-] = 0.020\text{M}\) and \(K_{sp} = 2.8 \times 10^{-16}\), \[2.8 \times 10^{-16} = x (0.020)^2 \Rightarrow x = \frac{2.8 \times 10^{-16}}{(0.020)^2}\] Now let's calculate the values of \(x\) for both compounds:

Now that we have calculated the lead ion concentrations required for the precipitation of PbSO₄ and Pb(OH)₂, we need to compare these values: For PbSO₄: \[x = \frac{K_{sp}}{0.020}\] For Pb(OH)₂: \[x = \frac{2.8 \times 10^{-16}}{(0.020)^2}\] The compound with the lowest x-value will precipitate first. Let's calculate these values and compare:

Since we determined that PbSO₄ precipitates first, now we need to find the pH of the solution when that happens. We know the concentration of hydroxide ions (\([OH^-]\)), so we can calculate the pOH value and then the pH value: \[pOH = -log[OH^-]\] \[pH = 14 - pOH\] By substituting the given \([OH^-]\) value, we can find the pH of the solution when PbSO₄ starts to precipitate.