Some Discussion and Examples

The hydrolysis of salts in water

Intro to hydrolysis calculations

Hydrolysis of the salt of a weak acid

Hydrolysis of the salt of a weak base

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 NH_{4}CN 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 NH_{4}CN contains NH_{4}^{+} ions and CN¯ ions (ignoring water and the other stuff). By itself in solution, NH_{4}^{+} ions are acidic (conjugate acid of the weak base NH_{3}). 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 K_{a} of NH_{4}^{+} and the K_{b} of CN¯.

**Solution:**

1) Consider the behavior of NH_{4}^{+} (an acid) in solution:

NH_{4}^{+}+ H_{2}O ⇌ H_{3}O^{+}+ NH_{3}The K

_{b}for NH_{3}is 1.77 x 10¯^{5}

K _{w}1.00 x 10¯ ^{14}K _{a}=––– = –––––––––– = 5.65 x 10¯ ^{10}(the K_{a}for the ammonium ion)K _{b}1.77 x 10¯ ^{5}

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

CN¯ + H_{2}O ⇌ HCN + OH¯The K

_{a}for HCN is 6.17 x 10¯^{10}

K _{w}1.00 x 10¯ ^{14}K _{b}=––– = –––––––––– = 1.62 x 10¯ ^{5}(the K_{b}for the cyanide ion)K _{a}6.17 x 10¯ ^{10}

This is the key comparison:

1.62 x 10¯^{5}>> 5.65 x 10¯^{10}The K

_{b}of CN¯ is much greater than the K_{a}of NH_{4}^{+}.The K

_{b}(of CN¯) is almost 300,000 times the size of the K_{a}(of NH_{4}^{+}).Therefore:

⟹ CN¯ is a MUCH stronger base than NH

_{4}^{+}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 NH_{4}CN? (K_{b} of ammonia is 1.77 x 10¯^{5} and the K_{a} 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 K_{a} expression:

HA ⇌ H^{+}+ A¯

[H ^{+}] [A¯]K _{a}=–––––––– [1] [HA]

2) The dissociation of BOH and its K_{b} expression:

BOH ⇌ B^{+}+ OH¯

[B ^{+}] [OH¯]K _{b}=––––––––– [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¯ + H _{2}O ⇌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] K _{h}=––––––––––– [3] [A¯] [B ^{+}]

(ch) (ch) K _{h}=––––––––––––––– c (1 − h) * c (1 − h)

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

(ch) (ch) K _{h}=––––––––––– c (1) * c (1) K

_{h}= h^{2}h = $\sqrt{\mathrm{Kh}}$

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

K _{w}K _{h}=–––––– K _{a}K_{b}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 = [K_{w}/ (K_{a}K_{b})]^{1⁄2}[4]By the way, this equation does not resolve to 1. The K

_{a}and K_{b}are not of a conjugate pair, so their product does not equal K_{w}.

9) We return to the acid dissociation equation and assign concentrations:

HA ⇌ H ^{+}+ A¯ ch [H ^{+}][c (1 − h)] Where c = concentration and h = degree of hydrolysis.

10) Write the K_{a} expression and substitute in the concentrations:

[H ^{+}] [A¯]K _{a}=–––––––– [HA]

[H ^{+}] * [c (1 − h)]K _{a}=–––––––––––––– ch

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

[H ^{+}] * cK _{a}=––––––– ch K

_{a}= [H^{+}] / h[H

^{+}] = K_{a}* h

12) Utilizing equation [4]:

[H^{+}] = K_{a}* [K_{w}/ (K_{a}K_{b})]^{1⁄2}[H

^{+}] = K_{a}^{1⁄2}* [K_{w}/ K_{b}]^{1⁄2}[H

^{+}] = [(K_{w}K_{a}) / K_{b}]^{1⁄2}

13) Change to pH:

−log [H^{+}] = −^{1}⁄_{2}log[(K_{w}K_{a}) / K_{b}]pH =

^{1}⁄_{2}(−log K_{w}+ −log K_{a}− (−log K_{b}))pH =

^{1}⁄_{2}(pK_{w}+ pK_{a}− pK_{b})

14) Since pK_{w} = 14:

pH =^{1}⁄_{2}(14 + pK_{a}− pK_{b})pH = 7 +

^{1}⁄_{2}(pK_{a}− pK_{b}) <--- this formula is what we will use

**Solution:**

pH = 7 +^{1}⁄_{2}(pK_{a}− pK_{b})pH = 7 +

^{1}⁄_{2}(9.210 − 4.752)pH = 7 +

^{1}⁄_{2}(4.458)pH = 7 + 2.229 = 9.229

The hydrolysis of salts in water

Intro to hydrolysis calculations

Hydrolysis of the salt of a weak acid