### pH calculations for the salt of a weak acid and a weak baseSome Discussion and Examples

The first example is a very general treatment, without determining the pH. The second example has a much more detailed discussion before deriving a formula used to calculate the pH of the solution of a salt of a weak acid and a weak base.

Example #1: An aqueous solution of NH4CN is prepared and its pH measured? Which of these do you expect?

 (a) pH < 7 (b) pH = 7 (c) pH > 7 (d) unknown (molarity needed)

Discussion:

An aqueous solution of NH4CN contains NH4+ ions and CN¯ ions (ignoring water and the other stuff). By itself in solution, NH4+ ions are acidic (conjugate acid of the weak base NH3). By itself in solution, CN¯ ions are basic (conjugate base of the weak acid HCN). In this first example, I will, ascertain if the solution is overall acidic or neutral by comparing the Ka of NH4+ and the Kb of CN¯.

Solution:

1) Consider the behavior of NH4+ (an acid) in solution:

NH4+ + H2O ⇌ H3O+ + NH3

The Kb for NH3 is 1.77 x 10¯5

 Kw 1.00 x 10¯14 Ka = ––– = –––––––––– = 5.65 x 10¯10 (the Ka for the ammonium ion) Kb 1.77 x 10¯5

2) Consider the behavior of CN¯ (a base) in solution:

CN¯ + H2O ⇌ HCN + OH¯

The Ka for HCN is 6.17 x 10¯10

 Kw 1.00 x 10¯14 Kb = ––– = –––––––––– = 1.62 x 10¯5 (the Kb for the cyanide ion) Ka 6.17 x 10¯10

This is the key comparison:

1.62 x 10¯5 >> 5.65 x 10¯10

The Kb of CN¯ is much greater than the Ka of NH4+.

The Kb (of CN¯) is almost 300,000 times the size of the Ka (of NH4+).

Therefore:

⟹ CN¯ is a MUCH stronger base than NH4+ is an acid.

⟹ The solution will be basic.

⟹ The pH will be greater than 7.

Example #2: What is the pH of a solution of NH4CN? (Kb of ammonia is 1.77 x 10¯5 and the Ka of HCN is 6.17 x 10¯10)

Introduction:

When considering the pH of a solution of the salt of a weak acid and a weak base, it turns out that the concentration of the salt does not play a role in the calculation.

Below, instead of using the actual formula of a salt, I will use HA and BOH for the weak acid and the weak base. For the salt, I will use the formula AB, rather than BA.

Even though the final answer contains no concentration values, we will still have to use concentrations as steps along the way to the final answer. For the salt, I will use c for its concentration. c will also be the concentration of the acid anion, A¯.

A colleague authored an explanation of Example #2 using a concept called 'extent of reaction.' I used a concept called 'degree of hydrolysis' in my explanation. His contribution is in the form of a PDF and you may find it here.

Thank you, LRD.

Lengthy Discussion:

1) The dissociation of HA and its Ka expression:

HA ⇌ H+ + A¯

 [H+] [A¯] Ka = –––––––– [1] [HA]

2) The dissociation of BOH and its Kb expression:

BOH ⇌ B+ + OH¯

 [B+] [OH¯] Kb = ––––––––– [2] [BOH]

3) Assume AB to be a strong electrolyte, ionizing 100% in solution:

AB ---> A¯ + B+

4) Let A+ and B¯ hydrolyze in solution and assign concentrations:

 B+ + A¯ + H2O ⇌ HA + BOH c (1 − h) c (1 − h) ch ch

Where c = concentration and h = the degree of hydrolysis.

5) Write a hydrolysis constant expression and substitute in the concentrations

 [HA] [BOH] Kh = ––––––––––– [3] [A¯] [B+]

 (ch) (ch) Kh = ––––––––––––––– c (1 − h) * c (1 − h)

6) Assume h to be small relative to 1, giving:

 (ch) (ch) Kh = ––––––––––– c (1) * c (1)

Kh = h2

h = $\sqrt{\mathrm{Kh}}$

7) From a consideration of [1], [2], and [3], we find this:

 Kw Kh = –––––– KaKb

The consideration of [1], [2], and [3] to give the above equation is detailed in a separate file.

8) Combining the results of step 6 and step 7:

h = [Kw / (KaKb)]12       [4]

By the way, this equation does not resolve to 1. The Ka and Kb are not of a conjugate pair, so their product does not equal Kw.

 HA ⇌ H+ + A¯ ch [H+] [c (1 − h)]

Where c = concentration and h = degree of hydrolysis.

10) Write the Ka expression and substitute in the concentrations:

 [H+] [A¯] Ka = –––––––– [HA]

 [H+] * [c (1 − h)] Ka = –––––––––––––– ch

11) Assume h to be small relative to 1, giving:

 [H+] * c Ka = ––––––– ch

Ka = [H+] / h

[H+] = Ka * h

12) Utilizing equation [4]:

[H+] = Ka * [Kw / (KaKb)]12

[H+] = Ka12 * [Kw / Kb]12

[H+] = [(KwKa) / Kb]12

13) Change to pH:

−log [H+] = − 12 log[(KwKa) / Kb]

pH = 12(−log Kw + −log Ka − (−log Kb))

pH = 12(pKw + pKa − pKb)

14) Since pKw = 14:

pH = 12(14 + pKa − pKb)

pH = 7 + 12(pKa − pKb) <--- this formula is what we will use

Solution:

pH = 7 + 12(pKa − pKb)

pH = 7 + 12(9.210 − 4.752)

pH = 7 + 12(4.458)

pH = 7 + 2.229 = 9.229