Question

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?

Answer 1

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

Answer 2

$K_c=9.34*10^{-4}$

Let's solve this question step by step:

Step 1:

The balanced equation for the reaction.

$PbCl_2(s)$ ⇌ $Pb^{2+} (aq)+2Cl^-(aq)$

Given that:

Mass of PbCl₂ = 0.2393 g

Volume = 50mL

Concentration of $[Pb^{2+}]$  = 0.0159 M

Concentration of $[Cl^-]$= 0.0318 M

To find:

Equilibrium constant, Kc =?

Step 2:

Determination of the number of mole PbCl₂.

The number of moles of PbCl₂ can be calculated as:

Molar Mass of PbCl₂ = 207 + (35.5x2) = 278g/mol

Mass of PbCl₂ = 0.2393 g

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of PbCl_2}= \frac{0.2393}{278} =8.61*10^{-4} moles$ Number of moles of PbCl₂ $=\frac{0.2393}{278} =8.61*10^{-4}moles$  

Step 3:

Determination of Molarity of PbCl₂.

 Mole of PbCl₂ = $8.61*10^{-4} moles$

Volume = 50mL = 50/1000 = 0.05L

Molarity of PbCl₂ =?

Molarity = mole /Volume

Molarity of PbCl₂ =$\frac{8.61*10^{-4}moles}{0.05 L} =0.0172M$  

Molarity of PbCl₂ = 0.01722 M

Step 4:

Determination of the equilibrium constant Kc.

$PbCl_2(s)$ ⇌ $Pb^{2+} (aq)+2Cl^-(aq)$

The equilibrium constant Kc for the equation above is given by:

$K_c=\frac{[Pb^{2+}][Cl^-]}{[PbCl_2]}$

$[Pb^{2+}]$  = 0.0159 M

$[Cl^-]$= 0.0318 M

[PbCl₂] = 0.01722 M

 

$K_c=\frac{0.0159*(0.0318)^2}{0.0172} \\\\K_c=9.34*10^{-4}$

The equilibrium constant Kc =$9.34*10^{-4}$

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