To solve the problem, we need to determine whether a precipitate will form when 100 mL of 0.06 M magnesium nitrite (Mg(NO2)2) is mixed with 50 mL of 0.06 M sodium oxalate (Na2C2O4), and if so, which ion will remain in excess after the precipitation.
### Step-by-Step Solution:
1. **Calculate the moles of Mg(NO2)2 and Na2C2O4:**
- For Mg(NO2)2:
\[
\text{Moles of Mg(NO2)2} = \text{Volume (L)} \times \text{Molarity (M)} = 0.1 \, \text{L} \times 0.06 \, \text{mol/L} = 0.006 \, \text{mol} = 6 \, \text{mmol}
\]
- For Na2C2O4:
\[
\text{Moles of Na2C2O4} = \text{Volume (L)} \times \text{Molarity (M)} = 0.05 \, \text{L} \times 0.06 \, \text{mol/L} = 0.003 \, \text{mol} = 3 \, \text{mmol}
\]
2. **Determine the concentration of ions after mixing:**
- Total volume after mixing = 100 mL + 50 mL = 150 mL = 0.150 L
- Concentration of Mg²⁺ ions:
\[
[\text{Mg}^{2+}] = \frac{6 \, \text{mmol}}{150 \, \text{mL}} = \frac{6 \, \text{mmol}}{0.150 \, \text{L}} = 0.04 \, \text{M}
\]
- Concentration of C2O4²⁻ ions:
\[
[\text{C2O4}^{2-}] = \frac{3 \, \text{mmol}}{150 \, \text{mL}} = \frac{3 \, \text{mmol}}{0.150 \, \text{L}} = 0.02 \, \text{M}
\]
3. **Calculate the ionic product (IP):**
- The ionic product (IP) is given by:
\[
\text{IP} = [\text{Mg}^{2+}] \times [\text{C2O4}^{2-}] = 0.04 \times 0.02 = 0.0008 = 8.0 \times 10^{-4} \, \text{M}^2
\]
4. **Compare IP with Ksp:**
- Given Ksp of MgC2O4 = \(8.6 \times 10^{-5} \, \text{M}^2\)
- Since \(IP = 8.0 \times 10^{-4} \, \text{M}^2\) is greater than \(Ksp = 8.6 \times 10^{-5} \, \text{M}^2\), a precipitate will form.
5. **Determine which ion remains in excess:**
- The equilibrium concentration of ions at saturation can be calculated from Ksp:
\[
[\text{Mg}^{2+}] = [\text{C2O4}^{2-}] = \sqrt{Ksp} = \sqrt{8.6 \times 10^{-5}} \approx 0.00927 \, \text{M}
\]
- Compare this with the initial concentrations:
- Initial [Mg²⁺] = 0.04 M
- Initial [C2O4²⁻] = 0.02 M
- After precipitation, the concentration of Mg²⁺ will be reduced to 0.00927 M, while C2O4²⁻ will be reduced to the same value. Therefore, the excess concentration of Mg²⁺ will be:
\[
0.04 - 0.00927 \approx 0.03073 \, \text{M}
\]
- The C2O4²⁻ will be reduced from 0.02 M to 0.00927 M, leaving it in lesser concentration than Mg²⁺.
### Conclusion:
A precipitate of magnesium oxalate (MgC2O4) will form, and excess magnesium ions will remain in the solution.
### Final Answer:
A precipitate will form, and excess magnesium ions will remain in the solution.
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