### Hydrolysis calculations: salts of weak acids are bases

Note: in the first three problems, I give the Kb of the conjugate base (for example, the acetate ion in Example #1). Often, these problems are given with the Ka of the acid and you have to calculate the value of the Kb. You do so with this equation:

KaKb = Kw

You will see such a situation starting in the fourth example as well as scattered through the additional problems.

Example #1: What is the pH of a 0.100 M solution of sodium acetate? Kb = 5.65 x 10¯10. (I will use NaAc as shorthand)

Solution:

1) Here is the chemical reaction (net ionic) for the hydrolysis of NaAc:

Ac¯ + H2O ⇌ HAc + OH¯

2) Here is the Kb expression for Ac¯:

 [HAc] [OH¯] Kb = ---------------- [Ac¯]

3) We can then substitute values into the Kb expression in the normal manner:

 (x) (x) 5.65 x 10¯10 = ----------- 0.100 - x

4) Ignoring the minus x in the usual manner, we proceed to solve for the hydroxide ion concentration:

x = $\sqrt{\mathrm{\left(5.65 x 10¯10\right) \left(0.100\right)}}$

x = 7.52 x 10¯6 M = [OH¯]

5) We then calculate the pH. Since this is a base calculation, we need to do the pOH first:

pOH = - log 7.52 x 10¯6 = 5.124

pH = 14 - 5.124

pH = 8.876

Example #2: What is the pH of a 0.0500 M solution of KCN? Kb = 2.1 x 10¯5.

Solution:

1) Here is the chemical reaction (net ionic) for the hydrolysis of KCN:

CN¯ + H2O ⇌ HCN + OH¯

2) Here is the Kb expression for CN¯:

 [HCN] [OH¯] Kb = ---------------- [CN¯]

3) We can then substitute values into the Kb expression in the normal manner:

 (x) (x) 2.1 x 10¯5 = ----------- 0.050 - x

4) Ignoring the minus x in the usual manner, we proceed to sove for the hydroxide ion concentration:

x = $\sqrt{\mathrm{\left(2.1 x 10¯5\right) \left(0.050\right)}}$

x = 1.025 x 10¯3 M <--- this is the [OH¯]

5) We then calculate the pH. Since this is a base calculation, we need to do the pOH first:

pOH = - log 1.025 x 10¯3 = 2.99

pH = 14 - 2.99

pH = 11.01

Example #3: Find the pH of a 0.30 M solution of sodium benzoate, C6H5COONa. The Kb for C6H5COO¯ (benzoate ion) is 1.55 x 10¯10.

Solution:

C6H5COO¯ + H2O ⇌ C6H5COOH + OH¯
 (x) (x) 1.55 x 10¯10 = ----------- 0.30 - x

x = $\sqrt{\mathrm{\left(1.55 x 10¯10\right) \left(0.30\right)}}$

x = 6.8 x 10¯6 M <--- the [OH¯]

pOH = - log 6.8 x 10¯6 = 5.17

pH = 14 - 5.17

pH = 8.83

Example #4: Find the pH of a 0.20 M solution of sodium propionate (C2H5COONa), where the Ka of propionic acid = 1.34 x 10¯5. (Use the equation Kw = KaKb to go from the Ka of the acid to the Kb of its conjugate base.)

Solution:

1) We need to get the Kb of the propionate ion first:

 1.00 x 10¯14 Kb = ---------------- = 7.46 x 10¯10 1.34 x 10¯5

2) Now, the solution follows the pattern outlined in the tutorial:

C2H5COO¯ + H2O ⇌ C2H5COOH + OH¯
 (x) (x) 7.46 x 10¯10 = ---------- 0.20

x = $\sqrt{\mathrm{\left(7.46 x 10¯10\right) \left(0.20\right)}}$

x = 1.22 x 10¯5 M = [OH¯]

pOH = - log 1.22 x 10¯5 = 4.91

pH = 14 - 4.91

pH = 9.09

Example #5: Given that the Ka for HOCl is 3.5 x 10¯8, calculate the pH of a 0.102 M solution of Ca(OCl)2

Solution:

1) A 0.102 M solution of Ca(OCl)2 is 0.204 M in just OCl¯ ions (which are the conjugate base of the acid HOCl)

2) The chemical reaction of interest is:

OCl¯ + H2O ⇌ HOCl + OH¯

3) Conjugate pairs have the following relationship:

KaKb = Kw

(3.5 x 10¯8) (Kb) = 1.0 x 10¯14

Kb = 2.857 x 10¯7

4) For the chemical equilibrium given in (2), we have:

 (x) (x) 2.857 x 10¯7 = --------- 0.204

x = [OH¯] = 2.414 x 10¯4

pOH = -log 2.414 x 10¯4 = 3.62

pH = 14 - 3.62 = 10.38

Bonus Example: Determine the Ka of the weak acid HX knowing that a 0.10 M solution of LiX has pH = 8.90.

Solution technique: we will determine the Kb of X¯ (since that is what we have data for). Based on the fact that HX is the conjugate base, we will use KaKb = Kw to get the Ka.

Solution:

1) This is the reaction of interest:

X¯ + H2O ⇌ HX + OH¯

2) We need to fill in the right side of the Kb expression for X¯. The pH is where we start:

pH = 8.90

pOH = 14 - 8.90 = 5.10

3) We use the pOH to give us the two values in the numerator of the Kb expression for X¯

[OH¯] = 10¯pOH = 10¯5.10 = 7.9 x 10¯6 M

4) Now, we can write the Kb expression. Remember that [X¯] = [OH¯]:

 [7.9 x 10¯6] [7.9 x 10¯6] Kb = ---------------------------- 0.10

Kb = 6.2 x 10¯10

5) The Kb of X¯ is related to the Ka of HX (what we want to get) by way of Kw.

KaKb = Kw

(Ka) (6.2 x 10¯10) = 1.0 x 10¯14

Ka = 1.6 x 10¯5