contestada

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
?

Respuesta :

Eduard22sly

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

Otras preguntas