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12 August, 17:27

onsider the reversible dissolution of lead (II) chloride. P b C l 2 (s) - ⇀ ↽ - P b 2 + (a q) + 2 C l - (a q) PbClX2 (s) ↽--⇀PbX2 + (aq) + 2ClX - (aq) Suppose you add 0.2393 g of P b C l 2 (s) PbClX2 (s) to 50.0 mL of water. When the solution reaches equilibrium, you find that the concentration of P b 2 + (a q) PbX2 + (aq) is 0.0159 M and the concentration of C l - (a q) ClX - (aq) is 0.0318 M. What is the value of the equilibrium constant, Kc, for the dissolution of P b C l 2 PbClX2?

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  1. 12 August, 19:48
    0
    9.34x10^-4

    Explanation:

    Step 1:

    The balanced equation for the reaction.

    PbCl2 (s) Pb^2 + (aq) + 2Cl^ - (aq)

    Step 2:

    Data obtained from the question:

    Mass of PbCl2 = 0.2393 g

    Volume = 50mL

    concentration of Pb^2+, [Pb^2+] = 0.0159 M

    Concentration of Cl^-, [Cl^-] = 0.0318 M

    Equilibrium constant, Kc = ?

    Step 3:

    Determination of the number of mole PbCl2.

    The number of mole of PbCl2 can be obtained as follow:

    Molar Mass of PbCl2 = 207 + (35.5x2) = 278g/mol

    Mass of PbCl2 = 0.2393 g

    Number of mole = Mass / Molar Mass

    Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

    Step 4:

    Determination of Molarity of PbCl2.

    At this stage we shall obtain the molarity of PbCl2. This is shown below:

    Mole of PbCl2 = 8.61x10^-4 mole

    Volume = 50mL = 50/1000 = 0.05L

    Molarity of PbCl2 = ?

    Molarity = mole / Volume

    Molarity of PbCl2 = 8.61x10^-4/0.05

    Molarity of PbCl2 = 0.01722 M

    Step 5:

    Determination of the equilibrium constant Kc.

    PbCl2 (s) Pb^2 + (aq) + 2Cl^ - (aq)

    The equilibrium constant Kc for the equation above is given by:

    Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

    [Pb^2+] = 0.0159 M

    [Cl^-] = 0.0318 M

    [PbCl2] = 0.01722 M

    Kc = ?

    Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

    Kc = 0.0159 x (0.0318) ^2 / 0.01722

    Kc = 9.34x10^-4
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