V. Advanced Weak Acid and Weak Base Equilibria
V-1. Introduction to Acidic
and Basic Salts
V-2. Sample Problem V-3. List of Problems V-4. Practice Problems

Introduction

The following problem is an example of determining the pH, and therefore the equilibrium [H+], when an acidic salt is added to water. For a reminder of how to begin, see the document on the general solution of equilibria problems.


Sample Problem

What is the pH of a 0.500 M solution of NH4Cl in water?

1. In any equilibrium problem, the first step is to identify the nature of the specie dissolving in water. NH4Cl (ammonium chloride) contains the spectator ion Cl-, so NH4Cl is a salt and completely dissociates:
NH4Cl(s) ---> NH4+(aq) + Cl-(aq)

Cl-(aq) is a spectator ion and will not be involved in any equilibria. NH4+(aq) can dissociate to produce NH3 (aq) and H+(aq), so the solution will be acidic.

NH4+(aq) <--> NH3 (aq) + H+(aq)

Kb for ammonia is 1.8x10-5. Using Ka * Kb = 1.00x10-14, then Ka for NH4+(aq) is 5.6x10-10.

The pre-equilibrium concentration of NH4+(aq) is given, and is 0.500 M. The pre-equilibrium concentration of H+(aq) is 1.0x10-7 due to the dissociation of pure water. We can consider the pre-equilibrium [H+] to be 0.00 M since it is much lower than 0.500 M. The pre-equilibrium concentration of NH3 (aq) is 0.00 M.

2. First balance the reaction: NH4+(aq) <--> NH3 (aq) + H+(aq)

Then write the equilibrium constant expression for this reaction:

     [H+][NH3]
Ka = ----------
       [NH4+]

3.

    [H+][NH3]
Q = ----------
      [NH4+]

Since the pre-equilibrium concentrations of the products are zero, Q is zero. Thus, Q < Keq and the reaction will proceed in the forward direction.

4. For each mol of NH4+ that dissociates, 1 mole each of H+ and NH3 forms:

NH4+H+NH3
[ ]o0.500 M~0.00 M0.00 M
Delta[ ]-x M+x M+x M
[ ]eq(0.500-x) M(0.00+x) M(0.00+x) M

5. Once you have the expressions for the equilibrium concentrations, you can use the equilibrium constant expression to solve for x:

     [H+][NH3]
Ka = ----------
       [NH4+]

               x2
5.6x10-10 = --------
           (0.500-x)
There are two ways to approach this problem from here. You can rearrange the equation above and solve for x using the quadratic equation, or you can make the approximation that x will be much less than 0.500 M and drop the x in the denominator.

Making this approximation gives:

              x2
5.6x10-10 = ------
            0.500

x2 = 2.8x10-10

x = 1.66x10-5

Before continuing, check that the approximation was valid:

                    x2
5.6x10-10 = -------------------
           (0.500 - 1.66x10-5)

x2 = 2.8x10-10

x = 1.66x10-5

We get the same answer for x, 1.66x10-5, so the approximation was valid.

Now find what the problem requested:

[H+] = x = 1.66x10-5 M

pH = -log[H+]

pH = -log(1.66x10-5)

pH = 4.78

Note that Ka was known only to 2 significant figures, so the answer after taking the log has 2 digits past the decimal point.

Check results:

    (1.66x10-5)(1.66x10-5)
Q = ----------------------
       (0.500-1.66x10-5)
Q = 5.5x10-10. Q = Ka, so the system is at equilibrium.


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