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?

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