Answer:
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
