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 =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 =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 5.35 g of NH4Cl (the molar mass of NH4Cl = 53.5 g mol¯1.)
Solution:
1) Determine molarity of the solution:
5.35 g / 53.5 g mol¯1 = 0.100 mol0.100 mol / 1.00 L = 0.100 M
2) Determine Ka for NH4Cl:
Ka = 10¯pKa = 10¯9.248Ka = 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 =
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 =
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 MThis 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 − 3.16228 x 10¯6 Use the 5% approximation.
(3.16228 x 10¯6) (3.16228 x 10¯6) Ka ≈ –––––––––––––––––––––––––––– 0.160 Ka = 6.25 x 10¯11
Determining the validity of the approximation is left to the student.
5) Solve for the Kb:
KaKb = Kw(6.25 x 10¯11) (Kb) = 1.00 x 10¯14
Kb = 1.6 x 10¯4