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

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

KaKb = Kw

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

Example #1: What is the pH of a 0.0500 M solution of ammonium chloride, NH4Cl. Ka of NH4+ = 5.65 x 10¯10.

Solution:

1) Here is the chemical reaction (net ionic) for the hydrolysis of NH4Cl:

NH4+ + H2O ⇌ NH3 + H3O+

2) Here is the Ka expression for NH4+:

 [NH3] [H3O+] Ka = ––––––––––– [NH4+]

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

 (x) (x) 5.65 x 10¯10 = ––––––––– 0.0500 − x

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

x = $\sqrt{\mathrm{\left(5.65 x 10¯10\right) \left(0.0500\right)}}$

x = 5.32 x 10¯6 M = [H3O+]

5) We then calculate the pH directly from the [H3O+] value:

pH = −log 5.32 x 10¯6 = 5.274

Example #2: What is the pH of a 0.100 M solution of methyl ammonium chloride (CH3NH3Cl). Ka of the methyl ammonium ion (CH3NH3+ = 2.70 x 10¯11)

Solution:

1) Here is the chemical reaction (net ionic) for the hydrolysis of CH3NH3+:

CH3NH3+ + H2O ⇌ CH3NH2 + H3O+

2) Here is the Ka expression for CH3NH3+:

 [CH3NH2] [H3O+] Ka = –––––––––––––– [CH3NH3+]

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

 (x) (x) 2.70 x 10¯11 = –––––––– 0.100 − x

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

x = $\sqrt{\mathrm{\left(2.70 x 10¯11\right) \left(0.100\right)}}$

x = 1.64 x 10¯6 M = [H3O+]

5) We then calculate the pH directly from the [H3O+] value:

pH = −log 1.64 x 10¯6 = 5.784

Example #3: Given the pKa for ammonium ion is 9.248, what is the pH of 1.00 L of solution which contains 5.45 g of NH4Cl (the molar mass of NH4Cl = 54.5 g mol¯1.)

Solution:

1) Determine molarity of the solution:

5.45 g / 54.5 g mol¯1 = 0.100 mol

0.100 mol / 1.00 L = 0.100 M

2) Determine Ka for NH4Cl:

Ka = 10¯pKa = 10¯9.248

Ka = 5.64937 x 10¯10

3) Write the equation for the hydrolysis of the ammonium ion and Ka expression:

NH4+ + H2O ⇌ H3O+ + NH3

 [H3O+] [NH3] Ka = –––––––––––– [NH4+]

4) Insert values into Ka expression and solve:

 (x) (x) 5.64937 x 10¯10 = –––––– 0.100

x = 7.51623 x 10¯6 M

5) Take negative log of this value (this value being the [H3O+]) for the pH:

pH = −log 7.51623 x 10¯6 = 5.124

Example #4: Aniline is a weak organic base with the formula C6H5NH2. The anilinium ion has the formula C6H5NH3+ and its Ka = 2.3 x 10¯5.

(a) write the chemical reaction showing the hydrolysis of anilinium ion.
(b) calculate the pH of a 0.0400 M solution of anilinium ion.

Solution:

1) The chemical reaction for the hydrolysis is:

C6H5NH3+ + H2O ⇌ C6H5NH2 + H3O+

2) The calculations for the pH are:

 (x) (x) 2.3 x 10¯5 = –––––––– 0.0400 − x

x = $\sqrt{\mathrm{\left(2.3 x 10¯5\right) \left(0.0400\right)}}$

x = 9.6 x 10¯4 M = [H3O+]

pH = −log 9.6 x 10¯6 = 3.02

Example #5: Pyridine (C5H5N) is a weak base and reacts with HCl as follows:

C5H5N + HCl ---> C5H5NH+ + Cl¯

What is the pH of a 0.015 M solution of the pyridinium ion (C5H5NH+)? The Kb for pyridine is 1.6 x 10¯9. (Hint: calculate the Ka for the pyridinium ion and use it in the calculation.)

Solution:

1) First, we calculate the Ka:

KaKb = Kw

 1.00 x 10¯14 Ka = –––––––––– 1.6 x 10¯9

Ka = 6.25 x 10¯6

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

C5H5NH+ + H2O ⇌ H3O+ + C5H5N

 (x) (x) 6.25 x 10¯6 = –––––– 0.015

x = $\sqrt{\mathrm{\left(6.25 x 10¯6\right) \left(0.015\right)}}$

x = 3.06 x 10¯4 M <--- this is the [H3O+]

pH = −log 3.06 x 10¯6 = 3.51

Example #6: Calculate the [H+] and the pH of a 0.580 M solution of NH4Cl (Kb of NH3 = 1.77 x 10¯5)

Solution:

1) The key ion is the ammonium ion. It is a weak acid and ionizes as follows:

NH4+ + H2O ⇌ H3O+ + NH3

 [H3O+] [NH3] Ka = –––––––––––– [NH4+]

2) We must determine the value for the Ka of NH4+:

KaKb = Kw

(Ka) (1.77 x 10¯5) = 1.00 x 10¯14

Ka = 5.65 x 10¯10

3) Determine the [H+]:

 (x) (x) 5.65 x 10¯10 = –––––– 0.580

x = 1.81 x 10¯5 M

4) Determine the pH:

pH = −log 1.81 x 10¯5 = 4.742

Bonus Example: The pH of a 0.160 M CH3NH3Cl solution is 5.500. What is the value of Kb for CH3NH2?

Solution technique: we will determine the Ka of CH3NH3Cl (since that is what we have data for). Based on the fact that CH3NH2 is the conjugate base, we will use KaKb = Kw to get the Kb.

Solution:

1) Here's the Ka expression we are interested in:

 [H3O+] [CH3NH2] Ka = –––––––––––––––– [CH3NH3+]

2) It comes from this reaction, the hydrolysis of CH3NH3+:

CH3NH3+ + H2O ⇌ H3O+ + CH3NH2

3) The concentrations of two components of the Ka expression come from the pH:

H3O+ = 10¯5.500 = 3.16228 x 10¯6 M

This is also the concentration of the CH3NH2.

4) Solve for the Ka:

 (3.16228 x 10¯6) (3.16228 x 10¯6) Ka = –––––––––––––––––––––––––––– 0.160

Ka = 6.25 x 10¯11

5) Solve for the Kb:

KaKb = Kw

(6.25 x 10¯11) (Kb) = 1.00 x 10¯14

Kb = 1.6 x 10¯4