For which of the mixtures will Ag2SO4(s) precipitate

For which of the mixtures will Ag2SO4(s) precipitate? 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.20 M AgNO3(aq) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.30 M AgNO3(aq) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.40 M AgNO3(aq) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.50 M AgNO3(aq)

The Correct Answer and Explanation is:

To determine whether Ag2SO4(s) will precipitate, we need to calculate the ion concentrations in the solution after mixing, and then compare the resulting ion product (Q) to the solubility product constant (Ksp) for Ag2SO4. The solubility product constant (Ksp) for Ag2SO4 is Ksp=1.2×10−5K_{sp} = 1.2 \times 10^{-5}Ksp​=1.2×10−5.

Step-by-Step Process:

1. Write the dissociation equations:
   • For Na2SO4: Na2SO4(aq)→2Na+(aq)+SO42−(aq)Na_2SO_4 (aq) \rightarrow 2Na^+ (aq) + SO_4^{2-} (aq)Na2​SO4​(aq)→2Na+(aq)+SO42−​(aq)
     • For AgNO3:
     AgNO3(aq)→Ag+(aq)+NO3−(aq)AgNO_3 (aq) \rightarrow Ag^+ (aq) + NO_3^- (aq)AgNO3​(aq)→Ag+(aq)+NO3−​(aq)
2. Determine the concentrations of ions after mixing: We have two solutions being mixed. For each case, we can calculate the concentration of the relevant ions:
   • For Na2SO4:
     The number of moles of SO42−SO_4^{2-}SO42−​ in 150.0 mL of 0.10 M Na2SO4: moles of SO42−=0.10 M×0.150 L=0.015 moles\text{moles of } SO_4^{2-} = 0.10 \, M \times 0.150 \, L = 0.015 \, \text{moles}moles of SO42−​=0.10M×0.150L=0.015moles After mixing with 5.0 mL of AgNO3, the total volume becomes: Vtotal=150.0 mL+5.0 mL=155.0 mL=0.155 LV_{total} = 150.0 \, \text{mL} + 5.0 \, \text{mL} = 155.0 \, \text{mL} = 0.155 \, LVtotal​=150.0mL+5.0mL=155.0mL=0.155L The concentration of SO42−SO_4^{2-}SO42−​ is: [SO42−]=0.015 moles0.155 L=0.09677 M[SO_4^{2-}] = \frac{0.015 \, \text{moles}}{0.155 \, L} = 0.09677 \, M[SO42−​]=0.155L0.015moles​=0.09677M
   • For AgNO3:
     The number of moles of Ag+Ag^+Ag+ in 5.0 mL of 0.20 M AgNO3: moles of Ag+=0.20 M×0.005 L=0.001 moles\text{moles of } Ag^+ = 0.20 \, M \times 0.005 \, L = 0.001 \, \text{moles}moles of Ag+=0.20M×0.005L=0.001moles The concentration of Ag+Ag^+Ag+ is: [Ag+]=0.001 moles0.155 L=0.00645 M[Ag^+] = \frac{0.001 \, \text{moles}}{0.155 \, L} = 0.00645 \, M[Ag+]=0.155L0.001moles​=0.00645M
   Now calculate the ion product Q: Q=[Ag+]2×[SO42−]Q = [Ag^+]^2 \times [SO_4^{2-}]Q=[Ag+]2×[SO42−​] Q=(0.00645)2×0.09677=3.98×10−6Q = (0.00645)^2 \times 0.09677 = 3.98 \times 10^{-6}Q=(0.00645)2×0.09677=3.98×10−6 Since Q<KspQ < K_{sp}Q<Ksp​, Ag2SO4 will not precipitate in this case.

We repeat the same steps for the other concentrations of AgNO3:

Case 2: 5.0 mL of 0.30 M AgNO3

• Moles of Ag+Ag^+Ag+: 0.30 M×0.005 L=0.0015 moles0.30 \, M \times 0.005 \, L = 0.0015 \, \text{moles}0.30M×0.005L=0.0015moles
• Concentration of Ag+Ag^+Ag+: [Ag+]=0.0015 moles0.155 L=0.00968 M[Ag^+] = \frac{0.0015 \, \text{moles}}{0.155 \, L} = 0.00968 \, M[Ag+]=0.155L0.0015moles​=0.00968M
• Ion product QQQ: Q=(0.00968)2×0.09677=8.98×10−6Q = (0.00968)^2 \times 0.09677 = 8.98 \times 10^{-6}Q=(0.00968)2×0.09677=8.98×10−6 Since Q<KspQ < K_{sp}Q<Ksp​, Ag2SO4 will not precipitate.

Case 3: 5.0 mL of 0.40 M AgNO3

• Moles of Ag+Ag^+Ag+: 0.40 M×0.005 L=0.002 moles0.40 \, M \times 0.005 \, L = 0.002 \, \text{moles}0.40M×0.005L=0.002moles
• Concentration of Ag+Ag^+Ag+: [Ag+]=0.002 moles0.155 L=0.0129 M[Ag^+] = \frac{0.002 \, \text{moles}}{0.155 \, L} = 0.0129 \, M[Ag+]=0.155L0.002moles​=0.0129M
• Ion product QQQ: Q=(0.0129)2×0.09677=1.61×10−5Q = (0.0129)^2 \times 0.09677 = 1.61 \times 10^{-5}Q=(0.0129)2×0.09677=1.61×10−5 Since Q>KspQ > K_{sp}Q>Ksp​, Ag2SO4 will precipitate.

Case 4: 5.0 mL of 0.50 M AgNO3

• Moles of Ag+Ag^+Ag+: 0.50 M×0.005 L=0.0025 moles0.50 \, M \times 0.005 \, L = 0.0025 \, \text{moles}0.50M×0.005L=0.0025moles
• Concentration of Ag+Ag^+Ag+: [Ag+]=0.0025 moles0.155 L=0.0161 M[Ag^+] = \frac{0.0025 \, \text{moles}}{0.155 \, L} = 0.0161 \, M[Ag+]=0.155L0.0025moles​=0.0161M
• Ion product QQQ: Q=(0.0161)2×0.09677=2.49×10−5Q = (0.0161)^2 \times 0.09677 = 2.49 \times 10^{-5}Q=(0.0161)2×0.09677=2.49×10−5 Since Q>KspQ > K_{sp}Q>Ksp​, Ag2SO4 will precipitate.

Conclusion:

Ag2SO4 will precipitate in cases where the concentration of AgNO3 is 0.40 M or 0.50 M.