Consider the reversible dissolution of lead(II) chloride.

P
b
C
l
2
(
s
)
−
⇀
↽
−
P
b
2
+
(
a
q
)
+
2
C
l
−
(
a
q
)

Suppose you add 0.2495 g of
P
b
C
l
2
(
s
)
to 50.0 mL of water. When the solution reaches equilibrium, you find that the concentration of
P
b
2
+
(
a
q
)
is 0.0159 M and the concentration of
C
l
−
(
a
q
)
is 0.0318 M.

What is the value of the equilibrium constant, Kc, for the dissolution of
P
b
C
l
2
?

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

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be