When 200 ml 5.15 x 10^-4 M Ba(NO3)2 is added to 150 ml of 8.25 x 10^-3 M Na2SO4. ksp for BaSO4 is 1.8 x 10^- Will preciptation occur

To determine whether precipitation will occur when mixing these solutions, we need to calculate the ion product (Q) of the potential precipitate, barium sulfate (BaSO₄), and compare it to the solubility product constant (Ksp) for BaSO₄. If Q exceeds Ksp, precipitation will occur.

1. Break Down the Problem

Calculate the concentrations of Ba²⁺ and SO₄²⁻ ions in the mixed solution and then determine the ion product Q.

2. Relevant Concepts

• Dilution Formula: $C_1V_1 = C_2V_2$
• Ion Product (Q): $Q = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$
• Ksp for BaSO₄: $1.8 \times 10^{-10}$

3. Analysis and Detail

1. Calculate the concentration of Ba²⁺ ions after mixing:

   • Initial concentration of Ba(NO₃)₂ = $5.15 \times 10^{-4} \, \text{M}$
   • Volume of Ba(NO₃)₂ = 200 ml = 0.2 L
   • Total volume after mixing = 200 ml + 150 ml = 350 ml = 0.35 L
   • Using dilution formula: $[\text{Ba}^{2+}] = \frac{5.15 \times 10^{-4} \, \text{M} \times 0.2 \, \text{L}}{0.35 \, \text{L}} = 2.943 \times 10^{-4} \, \text{M}$

2. Calculate the concentration of SO₄²⁻ ions after mixing:

   • Initial concentration of Na₂SO₄ = $8.25 \times 10^{-3} \, \text{M}$
   • Volume of Na₂SO₄ = 150 ml = 0.15 L
   • Using dilution formula: $[\text{SO}_4^{2-}] = \frac{8.25 \times 10^{-3} \, \text{M} \times 0.15 \, \text{L}}{0.35 \, \text{L}} = 3.536 \times 10^{-3} \, \text{M}$

3. Calculate the ion product Q: $Q = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = (2.943 \times 10^{-4} \, \text{M})(3.536 \times 10^{-3} \, \text{M}) = 1.04 \times 10^{-6}$

4. Verify and Summarize

• Compare Q with Ksp:
  • $Q = 1.04 \times 10^{-6}$
  • $Ksp = 1.8 \times 10^{-10}$
• Since $Q > Ksp$, precipitation of BaSO₄ will occur.

Final Answer

Precipitation of BaSO₄ will occur because the ion product $Q = 1.04 \times 10^{-6}$ is greater than the solubility product constant $Ksp = 1.8 \times 10^{-10}$.