Predicting Precipitation Using Ksp and Solution Mixing

Q1. Does a precipitate form when 0.10 L of $8.0 \times 10^{-3}$ M Pb(NO$_3$)$_2$ is added to 0.40 L of $5.0 \times 10^{-3}$ M Na$_2$SO$_4$?

Background

Topic: Solubility Product (Ksp) and Precipitation

This question tests your ability to predict whether a precipitate will form when two solutions are mixed, based on the concentrations of ions and the solubility product constant (Ksp) for the possible precipitate.

Key Terms and Formulas

• Ksp (Solubility Product Constant): The equilibrium constant for the dissolution of a sparingly soluble salt.

• Q (Reaction Quotient): Calculated using the concentrations of ions after mixing. Compare Q to Ksp to predict precipitation.

• Key formula:

$Q = [\text{Pb}^{2+}][\text{SO}_4^{2-}]$

If $Q > K_{sp}$, a precipitate forms. If $Q < K_{sp}$, no precipitate forms.

Step-by-Step Guidance

1. Calculate the total volume after mixing: $V_{\text{total}} = 0.10\,\text{L} + 0.40\,\text{L} = 0.50\,\text{L}$.

2. Find the moles of Pb$^{2+}$ and SO$_4^{2-}$ from each solution:

   Moles Pb$^{2+}$: $0.10\,\text{L} \times 8.0 \times 10^{-3}\,\text{mol/L}$

   Moles SO$_4^{2-}$: $0.40\,\text{L} \times 5.0 \times 10^{-3}\,\text{mol/L}$

3. Calculate the concentrations of Pb$^{2+}$ and SO$_4^{2-}$ after mixing, using the total volume:

   $[\text{Pb}^{2+}] = \frac{\text{moles Pb}^{2+}}{0.50\,\text{L}}$

   $[\text{SO}_4^{2-}] = \frac{\text{moles SO}_4^{2-}}{0.50\,\text{L}}$

4. Set up the expression for $Q$ using the calculated concentrations:

   $Q = [\text{Pb}^{2+}][\text{SO}_4^{2-}]$

Try solving on your own before revealing the answer!