Answer:
8.96x10^-4
Explanation:
Step 1:
The equation for the reaction is given below:
PbCl2(s) <==> Pb^2(aq) + + 2Cl^-(aq)
Step 2:
Data obtained from the question. This includes:
Mass of PbCl2 = 0.2495 g
Volume of solution = 50.0 mL
Concentration of Pb^2+, [Pb^2] = 0.0159 M
Concentration of Cl^-, [Cl^-] = 0.0318 M
Equilibrium constant Kc =?
Step 3:
Determination of the mole of PbCl2 in 0.2495 g of PbCl2. This is illustrated below:
Mass of PbCl2 = 0.2495 g
Molar Mass of PbCl2 = 207 + (35.5x2) = 207 + 71 = 278g/mol
Number of mole = Mass/Molar Mass
Number of mole PbCl2 = 0.2495/278
Number of mole PbCl2 = 8.97x10^-4 mole
Step 4:
Determination of the molarity of PbCl2. This is shown below:
Mole of PbCl2 = 8.97x10^-4 mole
Volume = 50 mL = 50/1000 = 0.05 L
Molarity of PbCl2 =?
Molarity = mole /Volume
Molarity of PbCl2 = 8.97x10^-4/0.05
Molarity of PbCl2 = 0.01794 M
Step 5:
Determination of the equilibrium constant Kc. This is illustrated below:
PbCl2(s) <==> Pb^2(aq) + + 2Cl^-(aq)
The equilibrium constant for the reaction above is given by:
Kc = [Pb^2] [Cl^-]^2 / [PbCl2]
[Pb^2] = 0.0159 M
[Cl^-] = 0.0318 M
[PbCl2] = 0.01794 M
Kc =?
Kc = [Pb^2] [Cl^-]^2 / [PbCl2]
Kc = 0.0159 x (0.0318)^2 /0.01794
Kc = 8.96x10^-4
