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: The solubility product constant, which represents the maximum product of ion concentrations that can exist in solution without forming a precipitate.

• Q: The reaction quotient, calculated from the actual concentrations of ions after mixing. If $Q > K_{sp}$, a precipitate forms.

• Key formula:

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

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 new concentration of Pb$^{2+}$ after mixing, using dilution:

   $[\text{Pb}^{2+}] = \frac{(8.0 \times 10^{-3}\,\text{mol/L}) \times 0.10\,\text{L}}{0.50\,\text{L}}$

3. Find the new concentration of SO$_4^{2-}$ after mixing, using dilution:

   $[\text{SO}_4^{2-}] = \frac{(5.0 \times 10^{-3}\,\text{mol/L}) \times 0.40\,\text{L}}{0.50\,\text{L}}$

4. Calculate $Q$ using the formula $Q = [\text{Pb}^{2+}][\text{SO}_4^{2-}]$ with the concentrations you just found.

Try solving on your own before revealing the answer!

Final Answer: A precipitate will form if $Q > K_{sp}$ for PbSO$_4$.

After calculating the concentrations and $Q$, compare $Q$ to the known $K_{sp}$ for PbSO$_4$ (typically $K_{sp} = 1.6 \times 10^{-8}$). If $Q$ exceeds this value, a precipitate forms.

This approach uses dilution and the solubility product to predict precipitation.