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Question: The volume (in mL) of 0.125 M $AgNO_3$ required to quantitatively precipitate chloride ions in 0.3g ...

The volume (in mL) of 0.125 M AgNO3AgNO_3 required to quantitatively precipitate chloride ions in 0.3g of [Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3 is _______.

M[Co(NH3)6]Cl3M_{[Co(NH_3)_6]Cl_3} = 267.46g/mol

MAgNO3M_{AgNO_3} = 169.87g/mol

Report the nearest integer as the answer.

Answer

27

Explanation

Solution

To quantitatively precipitate chloride ions from [Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3 using AgNO3AgNO_3, we first need to determine the number of chloride ions that are outside the coordination sphere and thus precipitable.

  1. Dissociation of the complex:
    The complex [Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3 dissociates in water as follows:
    [Co(NH3)6]Cl3(aq)→[Co(NH3)6]3+(aq)+3Cl−(aq)[Co(NH_3)_6]Cl_3 (aq) \rightarrow [Co(NH_3)_6]^{3+} (aq) + 3Cl^- (aq)
    This indicates that 1 mole of [Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3 yields 3 moles of precipitable chloride ions (Cl−Cl^-).

  2. Moles of [Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3:
    Given mass of [Co(NH3)6]Cl3=0.3 g[Co(NH_3)_6]Cl_3 = 0.3 \text{ g}
    Given molar mass of [Co(NH3)6]Cl3=267.46 g/mol[Co(NH_3)_6]Cl_3 = 267.46 \text{ g/mol}
    Moles of [Co(NH3)6]Cl3=MassMolar Mass=0.3 g267.46 g/mol≈0.00112106 mol[Co(NH_3)_6]Cl_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.3 \text{ g}}{267.46 \text{ g/mol}} \approx 0.00112106 \text{ mol}

  3. Moles of precipitable chloride ions (Cl−Cl^-):
    Since 1 mole of the complex yields 3 moles of Cl−Cl^-,
    Moles of Cl−=3×Moles of [Co(NH3)6]Cl3Cl^- = 3 \times \text{Moles of } [Co(NH_3)_6]Cl_3
    Moles of Cl−=3×0.00112106 mol≈0.00336318 molCl^- = 3 \times 0.00112106 \text{ mol} \approx 0.00336318 \text{ mol}

  4. Reaction with AgNO3AgNO_3:
    The reaction between AgNO3AgNO_3 and Cl−Cl^- ions is:
    AgNO3(aq)+Cl−(aq)→AgCl(s)+NO3−(aq)AgNO_3 (aq) + Cl^- (aq) \rightarrow AgCl (s) + NO_3^- (aq)
    From the stoichiometry, 1 mole of AgNO3AgNO_3 reacts with 1 mole of Cl−Cl^-.
    Therefore, moles of AgNO3AgNO_3 required = Moles of Cl−Cl^-
    Moles of AgNO3AgNO_3 required ≈0.00336318 mol\approx 0.00336318 \text{ mol}

  5. Volume of 0.125 M AgNO30.125 \text{ M } AgNO_3 required:
    Given concentration of AgNO3=0.125 MAgNO_3 = 0.125 \text{ M} (or 0.125 mol/L0.125 \text{ mol/L})
    Volume (in Liters) = Moles of AgNO3Concentration of AgNO3\frac{\text{Moles of } AgNO_3}{\text{Concentration of } AgNO_3}
    Volume (in Liters) = 0.00336318 mol0.125 mol/L≈0.02690544 L\frac{0.00336318 \text{ mol}}{0.125 \text{ mol/L}} \approx 0.02690544 \text{ L}

  6. Convert volume to milliliters:
    Volume (in mL) = Volume (in Liters) ×1000\times 1000
    Volume (in mL) = 0.02690544×1000≈26.90544 mL0.02690544 \times 1000 \approx 26.90544 \text{ mL}

  7. Report the nearest integer:
    Rounding 26.90544 mL to the nearest integer gives 27 mL.