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Question: Assuming complete precipitation of AgCl, calculate the sum of the molar concentration of all the ion...

Assuming complete precipitation of AgCl, calculate the sum of the molar concentration of all the ions if 2 lit of 2M Ag2SO4Ag_2SO_4 is mixed with 4 lit of 1 M NaCl solution is:

A

4M

B

2M

C

3 M

D

2.5 M

Answer

2 M

Explanation

Solution

Here's how to calculate the final molar concentration:

  1. Calculate initial moles of reactants and their ions:

    • For Ag2SO4Ag_2SO_4:

      • Volume = 2 L, Concentration = 2 M
      • Moles of Ag2SO4=Volume×Concentration=2 L×2 M=4 molesAg_2SO_4 = \text{Volume} \times \text{Concentration} = 2 \text{ L} \times 2 \text{ M} = 4 \text{ moles}
      • Upon dissociation: Ag2SO4(aq)→2Ag+(aq)+SO42−(aq)Ag_2SO_4(aq) \rightarrow 2Ag^+(aq) + SO_4^{2-}(aq)
      • Moles of Ag+Ag^+ ions = 2×4=8 moles2 \times 4 = 8 \text{ moles}
      • Moles of SO42−SO_4^{2-} ions = 1×4=4 moles1 \times 4 = 4 \text{ moles}

    • For NaCl:

      • Volume = 4 L, Concentration = 1 M
      • Moles of NaCl = Volume ×\times Concentration = 4 L×1 M=4 moles4 \text{ L} \times 1 \text{ M} = 4 \text{ moles}
      • Upon dissociation: NaCl(aq)→Na+(aq)+Cl−(aq)NaCl(aq) \rightarrow Na^+(aq) + Cl^-(aq)
      • Moles of Na+Na^+ ions = 1×4=4 moles1 \times 4 = 4 \text{ moles}
      • Moles of Cl−Cl^- ions = 1×4=4 moles1 \times 4 = 4 \text{ moles}

  2. Identify the precipitation reaction and limiting reactant:

    The reaction is between Ag+Ag^+ and Cl−Cl^- to form AgCl precipitate:

    Ag+(aq)+Cl−(aq)→AgCl(s)Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)

    Initial moles:

    • Ag+Ag^+: 8 moles
    • Cl−Cl^-: 4 moles

    Since the stoichiometric ratio is 1:1, 4 moles of Cl−Cl^- will react with 4 moles of Ag+Ag^+. Therefore, Cl−Cl^- is the limiting reactant.

  3. Calculate moles of ions remaining in solution after precipitation:

    • Moles of Ag+Ag^+ remaining = Initial Ag+Ag^+ - Consumed Ag+Ag^+ = 8 moles−4 moles=4 moles8 \text{ moles} - 4 \text{ moles} = 4 \text{ moles}
    • Moles of Cl−Cl^- remaining = Initial Cl−Cl^- - Consumed Cl−Cl^- = 4 moles−4 moles=0 moles4 \text{ moles} - 4 \text{ moles} = 0 \text{ moles} (all precipitated)
    • Moles of Na+Na^+ remaining = 4 moles4 \text{ moles} (spectator ion, does not react)
    • Moles of SO42−SO_4^{2-} remaining = 4 moles4 \text{ moles} (spectator ion, does not react)

    Total moles of ions in solution = Moles of Ag+Ag^+ + Moles of Na+Na^+ + Moles of SO42−SO_4^{2-}

    Total moles of ions = 4 moles+4 moles+4 moles=12 moles4 \text{ moles} + 4 \text{ moles} + 4 \text{ moles} = 12 \text{ moles}

  4. Calculate the total volume of the solution:

    Total volume = Volume of Ag2SO4Ag_2SO_4 solution + Volume of NaCl solution

    Total volume = 2 L+4 L=6 L2 \text{ L} + 4 \text{ L} = 6 \text{ L}

  5. Calculate the sum of the molar concentration of all the ions:

    Sum of molar concentration = Total moles of ionsTotal volume\frac{\text{Total moles of ions}}{\text{Total volume}}

    Sum of molar concentration = 12 moles6 L=2 M\frac{12 \text{ moles}}{6 \text{ L}} = 2 \text{ M}