Humboldt State University ® Department of Chemistry

Richard A. Paselk
 

General Chemistry

Spring 2009
Exercise: Chemical Equilibrium

© R. Paselk 2008
 
 

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Chemical Dissociation Equilibrium Problems

1. What is the concentration of hydrogen ion in a 0.20 M solution of acetic acid (CH3COOH) if Ka = 1.8 x 10-5 @ 25°C? ( Ka is the equilibrium constant for an acid dissociation.)

First need to write the equation for the dissociation of the acid:

  CH3COOH equilibrium arrows CH3COO- + H+
Before reaction 0.20 M   0 0
Let [H+] = x
@ Equilibrium 0.20M - x   x   x

At equilibrium then, Ka = [CH3COO-][H+] / [CH3COOH]

Substituting, Ka = (x)(x) / (0.20 - x) = 1.8 x 10-5

Since Ka is small, assume x<<0.20. Can then simplify:

Ka = x2/ (0.20) = 1.8 x 10-5

x2 = 3.6 x 10-6

x = 1.9 x 10-3

and x = 1.9 x 10-3 is smaller than the error of significant figures, Assumption OK!

At equilibrium:

CH3COOH = 0.20 M

CH3COO- = 1.9 x 10-3 M

H+= 1.9 x 10-3 M

 

2. Calculate the solubility of calcium carbonate. CaCO3 is used by organisms to make shells etc. Vast deposits of these shells make up limestone and (after heat and compression) marble deposits. An important consideration in determining whether shells can be formed, limestone deposited etc. is the solubility of calcium carbonate. For the reaction of solid calcium carbonate in equilibrium with its ions, K = 8.7 x 10-9.

First we need to ask, what is meant by solubility? Evidently, we are asking how much of a solid dissolves to give the solvated ions, or in water the hydrated ions. Dissolving calcium carbonate in water can then be described chemically with the following equation:

CaCO3(s) equilibrium arrows Ca2+(aq) + CO32-(aq)

*In general, we will not use the subscript (aq) for hydrated ions since all ions in water solution are hydrated. It is necessary to show the subscripts for solid and gas however.

The equilibrium expression for this reaction can then be written as:

K = [Ca2+] [CO32-] / [CaCO3(s)]

But recall that the activity (the effective or apparent concentration of a substance in a chemical system) is 1 for any solid (remember, a solid is a crystalline substance, and therefore a pure phase). We can thus write the simpler expression for the equilibrium:

K = [Ca2+] [CO32-] = 8.7 x 10-9

We can now determine the solubility of calcium carbonate in water.

  CaCO3(s) equilibrium arrows Ca2+ + CO32-
Before reaction 1   0 0
[Ca2+] = [CO32-]; Let [Ca2+] = x
@ Equilibrium 1   x   x

x2 = 8.7 x 10-9

Thus [Ca2+] = [CO32-] = (8.7 x 10-9)1/2 = 9.3 x 10-5

So the solubility of calcium carbonate = 9.3 x 10-5 M.

 

3. When excess solid lead(II) chloride is dissolved in water at 25°C the equilibrium concentration of lead(II) ion is found to be 1.62 x 10-2 M. Calculate the equilibrium constant for this reaction giving a saturated solution.

The chemical equation for this reaction is:

PbCl2(s) equilibrium arrows Pb2+ + 2 Cl-

From the chemical equation we can see that twice as much chloride ion as lead ion results when lead chloride is dissolved,

  PbCl2(s) equilibrium arrows Pb2+ + 2 Cl-
Before reaction 1   0 0
Let [Pb2+] = x
@ Equilibrium 1   x   2x

thus a saturated solution of lead chloride will have:

Pb2+ = 1.62 x 10-2 M

Cl- = (1.62 x 10-2 M) = 3.24 x 10-2 M

Writing the equilibrium expression then gives:

K = [Pb2+] [Cl-]2/[PbCl2(s)]

But since the effective concentration of a substance is 1 in a pure phase, such as a crystalline solid,

K = [Pb2+] [Cl-]2

Substituting these values into the solubility product expression then gives:

K = (1.62 x 10-2)(3.24 x 10-2)2

K = 1.7 x 10-5

4. What is the solubility of ferric hydroxide ("rust") in a neutral aqueous solution where [OH-] = 1 x 10-7M. For the reaction of solid ferric hydroxide in equilibrium with its ions, K = 4 x 10-38.

First we need to write out the chemical reaction for the disolution of solid ferric hydroxide = iron(III) hydroxide:

Fe(OH)3(s) equilibrium arrows Fe3+ + 3 OH-

K = [Fe3+] [OH-]3 = 4 x 10-38

For solubility we are asking how many moles of iron(III) hydroxide dissolve in a liter of water. Here we see that for each Fe(OH)3(s) dissolved one [Fe3+] goes into solution, so [Fe3+] = solubility!

[Fe3+] = (4 x 10-38) / [OH-]3

but [OH-] = 1 x 10-7M

so [Fe3+] = (4 x 10-38) / (10-7)3

[Fe3+] = 4 x 10-17 M

In addition to these exercises you should work on the text materials.

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© R A Paselk

Last modified 23 November 2009