Question

The addition of 0.242 L of 1.92 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions as AgCl and PbCl2. The total mass of the resulting precipitate is 65.08 g. Find the mass of PbCl2 and AgCl in the precipitate. Calculate the mass of PbCl2 and AgCl in grams.

Answer 1

Answer:

Mass PbCl₂ = 50.24g

Mass AgCl = 14.84g

Explanation:

The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:

Ag⁺ + Cl⁻ → AgCl(s)

Pb²⁺ + 2Cl⁻ → PbCl₂(s)

If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:

Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = 0.00719X moles of Cl⁻ from PbCl₂

And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:

(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = 0.45409 - 0.00698X moles of Cl⁻ from AgCl

Moles of Cl⁻ that were added in the KCl solution are:

0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.

Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)

0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles

0.45409 + 0.00021X = 0.46464

0.00021X = 0.01055

X = 0.01055 / 0.00021

X = 50.24g

As X = Mass PbCl₂

Mass PbCl₂ = 50.24g

And mass of AgCl = 65.08 - 50.24

Mass AgCl = 14.84g

Answer 2

The masses of the compounds in the precipitate can be found my knowing

the number of moles of chloride ion contributed by each compound.

• The mass of PbCl₂ in the precipitate is approximately 49.24 g

• The mass of AgCl in the precipitate is approximately 15.84 g

Reasons:

The given parameter are;

Volume of KCl solution added = 0.242 L

Concentration of KCl solution = 1.92 M KCl

The ions in the solution to which KCl is added = Ag⁺ and Pb²⁺ ions

Precipitates formed = AgCl and PbCl₂

The mass of the precipitate = 65.08 g

Required:

The mass of PbCl₂ and AgCl in the precipitate

Solution;

Number of moles of chloride ions in a mole of PbCl₂ = 2 moles

Number of moles of chloride ions in a mole of AgCl = 1 mole

Let X represent the mass of PbCl₂ in the precipitate, we have;

The mass of AgCl in the precipitate = 65.08 g - X

$\mathrm{Number \ of \ moles \ of \ PbCl_2} = \dfrac{X \, g}{278.1 \, g} =\mathbf{ \dfrac{X }{278.1}}$

Number of moles of chloride ions from PbCl₂ is therefore;

$\mathrm{Number \ of \ moles \ of \ Cl^- from \ PbCl_2} =\mathbf{ 2 \times \dfrac{X }{278.1} \ moles \ of \ Cl^-}$

$\mathrm{Number \ of \ moles \ of \ AgCl \ in \ the \ precipitate} = \dfrac{65.08 -X }{143.32}$

$\mathrm{Number \ of \ moles \ of \ Cl^- from \ AgCl} = \mathbf{ \dfrac{65.08 -X }{143.32}} \ moles \ of \ Cl^-$

The number of moles of chloride ions from one mole of KCl = 1 mole

Number of moles of chloride ions from 0.242 L of 1.92 M KCl is therefore;

0.242 L × 1.92 moles/L = 0.46464 moles

Number of moles of chloride ions from KCl = 0.46464 moles

$0.46464 \ moles \ from \ KCl = \overbrace{ \dfrac{ 2 \times X }{278.1} + \dfrac{65.08 -X }{143.32}} \ moles \ in \ PbCl_2 \ and \ AgCl$

Which gives;

$\displaystyle \frac{192}{896089} \cdot X + \frac{1627}{3583} = \frac{1452}{3125}$

Therefore;

$\displaystyle X = \frac{\frac{1452}{3125} - \frac{1627}{3583} }{ \frac{192}{896089} } = \frac{105864850549}{2149800000} \approx \mathbf{ 49.24}$

The mass of PbCl₂ in the precipitate, X ≈ 49.24 g

The mass of AgCl in the precipitate = 65.08 g - 49.24 g ≈ 15.84 g

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