Hydrolysis calculations: Problems #1 - 10

Note: the problems are all mixed up. Some are salts of weak acids, some are salts of weak bases.

Problem #1: CH3NH2, methyl amine is a weak base with a Kb of 4.38 x 10¯4. What would be the pH of a solution of 0.350 M methyl ammonium chloride, CH3NH3+Cl¯?

Some discussion before the solution:

1) This problem type seems to be seldom asked. Much more popular is giving you the Ka of an acid and then asking about pH of a solution formed from the salt of the acid. The problem above give you the Kb of a base and asks you for the pH of a salt of that base.

2) A bit of a warning: watch out, because there could be a temptation to teach the OTHER problem type, then test on this problem type. You have been warned!!

3) The solution to this type of problem depends on knowing that Kw = KaKb. We will use that equation to calculate the Ka of CH3NH3+ (the acid) from the Kb of CH3NH2 (the base).

4) See this tutorial for how the equation is derived. A reminder: the equation above applies to conjugate acid-base pairs. CH3NH2 (the base) and CH3NH3+ (the acid) is the conjugate acid-base pair involved in the problem being discussed.

5) Another important point to know is how salts hydrolyze. In the problem being examined, we have the salt of a weak base hydrolyzing. See this tutorial for more explanation.

Solution:

1) Here is the relevant chemical equation for the above problem (chloride ion is a spectator ion, so I left it off):

CH3NH3+ + H2O ⇌ CH3NH2 + H3O+

2) We must calculate the Ka for the above reaction. We will use the Kb of CH3NH2 to do this:

Kw = KaKb

1.00 x 10¯14 = (x) (4.38 x 10¯4)

x = 1.00 x 10¯14 / 4.38 x 10¯4 = 2.2831 x 10¯11

3) Please note that I left some guard digits on the Ka value. I will round off the final answer to the appropriate number of significant figures.

4) We now calculate the [H3O+] using the Ka expression:

2.2831 x 10¯11 = [(x) (x)] / (0.350 − x)

neglect the minus x

x = 2.8268 x 10¯6 M

6) Since we have the [H3O+], our last step will be to calculate the pH:

pH = −log 2.8268 x 10¯6 = 5.549

Note that, since this is the salt of a weak base (which is itself an acid), the calculated pH is an acidic value. This is as it should be!

Problem #2: What is the pH of 0.410 M methylammonium bromide, CH3NH3Br? (Kb of CH3NH2 = 4.4 x 10¯4.)

Note the differences in wording between this problem and the one just prior. In the prior problem, chloride was the spectator ion. In this problem, it is bromide. In this problem, the Kb value has been rounded off as well as the concentration being different.

Solution:

1) The relevant chemical equation is the hydrolysis of methyl amine:

CH3NH3+ + H2O ⇌ CH3NH2 + H3O+

2) We must calculate the Ka for the above reaction. We will use the Kb of CH3NH2 to do this:

Kw = KaKb

1.00 x 10¯14 = (x) (4.4 x 10¯4)

x = 1.00 x 10¯14 / 4.4 x 10¯4 = 2.2727 x 10¯11

3) We now calculate the [H3O+] using the Ka expression:

2.2727 x 10¯11 = [(x) (x)] / (0.410)

x = 3.05257 x 10¯6 M

6) Since we have the [H3O+], our last step will be to calculate the pH:

pH = −log 3.05257 x 10¯6 = 5.515

Since the provided Kb value had two sig figs, the best answer to report for the pH is 5.52.

Problem #3: What is the pH of a solution prepared by adding 0.750 g of ammonium chloride to 125 mL of water? (Kb of NH3 is 1.77 x 10¯5)

Solution:

1) Molarity first:

MV = mass / molar mass

(x) (0.125 L) = 0.750 g / 53.4916 g/mol

x = 0.112167 M (we'll just keep a few guard digits and round off to a more appropriate value at the end)

2) The Ka for NH4+

Kw = KaKb

1.00 x 10¯14 = (Ka) (1.77 x 10¯5)

Ka = 5.65 x 10¯10

3) Use the Ka expression:

 (x) (x) 5.65 x 10¯10 = –––––––– 0.112167

x = 7.9608 x 10¯6 M (this is the H+ conc)

4) Calculate the pH:

pH = −log 7.9608 x 10¯6 = 5.099

Problem #4: Calculate the pH of a 0.970 M aqueous solution of hydrazine hydrochloride (H2NNH3Cl). (For hydrazine, H2NNH2, Kb = 1.26 x 10¯6.)

Solution:

1) Calculate the Ka of H2NNH3+, the hydrazinium ion:

KaKb = Kw

(Ka) (1.26 x 10¯6) = 1.00 x 10¯14

Ka = 7.9365 x 10¯9

2) Hydrazinium hydrolyzes as follows:

H2NNH3+ + H2O ⇌ H3O+ + H2NNH2

3) Solve the Ka expression for [H3O+]:

 (x) (x) 7.9365 x 10¯9 = ––––– 0.970

x = $\sqrt{\mathrm{\left(7.9365 x 10¯9\right) \left(0.970\right)}}$

x = 0.00008774 M

4) The pH:

pH = −log 0.00008774 = 4.057

Problem #5: What is the pH of a 0.502 M solution of NH4NO3? The ionization constant of the weak base NH3 is 1.77 x 10¯5

Solution:

1) The chemical equation of interest is this:

NH4+ + H2O ⇌ H3O+ + NH3

2) We need the Ka of the ammonium ion:

Kw = KaKb

1.00 x 10¯14 = (Ka) (1.77 x 10¯5)

Ka = 5.65 x 10¯10

3) Calculate the hydrogen ion concentration, then pH:

5.65 x 10¯10 = [(x) (x)] / 0.502

x = 1.684 x 10¯5 M

pH = −log 1.684 x 10¯5 = 4.774

Problem #6: A 0.50 M solution of NaNO2 is prepared. Calculate the pH of this solution.

Solution:

1) The conjugate acid of NO2¯ is HNO2. Look up its Ka:

4.0 x 10¯4

2) The nitrous anion behaves as a base in solution. Write the chemical reaction:

NO2¯ + H2O ⇌ HNO2 + OH¯

3) Write the Kb expression for the nitrous anion and solve it for the hydroxide ion concentration:

 (x) (x) 2.5 x 10¯11 = ––––– 0.50

x = 3.5355 x 10¯6 M <--- this is the [OH¯]

4) pOH, then pH

pOH = −log 3.5355 x 10¯6 = 5.45

pH = 14 − 5.45 = 8.55

Problem #7: Calculate the pH of a 0.360 M aqueous solution of ethylamine hydrochloride (C2H5NH3Cl). (For ethylamine, C2H5NH2, Kb = 5.60 x 10¯4.)

Solution:

1) The chemical reaction for the hydrolysis of ethylamine hydrochloride is:

C2H5NH3+ + H2O ⇌ H3O+ + C2H5NH2

Note that the chloride ion (a spectator ion) has been eliminated.

2) A Ka expression for C2H5NH3+ can be written:

 [H3O+] [C2H5NH2] Ka = –––––––––––––––– [C2H5NH3+]

3) Let us solve the Ka expression:

 (x) (x) 1.7857 x 10¯11 = –––––– 0.360

x = 2.53545 x 10¯6 M <--- this is the [H3O+]

Note: the value for the Ka came from this relationship:

KaKb = Kw

4) Get the pH:

pH = −log 2.53545 x 10¯6 = 5.596

Problem #8: What is the pH of an aqueous solution of 0.620 M KNO2?

Comment: to solve this problem, we need to know something more. What we need is the Ka for nitrous acid, HNO2. When looking around the Internet, there are several values used in various problems. I decided to use 4.3 x 10¯4.

1) Write the ionization equation for KNO2:

NO2¯ + H2O ⇌ HNO2 + OH¯

2) Use Ka of nitrous acid to calculate the Kb for nitrous ion:

(4.3 x 10¯4) (x) = 1.0 x 10¯14

x = 2.3256 x 10¯11 (I'll round off at the end.)

3) Calculate [OH¯]:

2.3256 x 10¯11 = [(x) (x)] / 0.620

x = 3.797 x 10¯6 M

4) Calculate pH

pOH = −log 3.797 x 10¯6 = 5.42

pH = 14.00 − 5.42 = 8.58

Problem #9: Calculate the pH of a 0.200 M solution of ethanolammonium chloride, HOCH2CH2NH3Cl. The weak base ethanolamine, HOCH2CH2NH2, has a base ionization constant (Kb) of 3.1 x 10¯5.

Solution:

1) Ethanolammonium ion is the conjugate acid of the weak base ethanolamine, HOCH2CH2NH2

2) Conjugate pairs have this relationship for their K's:

KaKb = Kw

3) Calculate the Ka of the ethanolammonium ion:

(1 x 10¯14) / (3.1 x 10¯5) = 3.22581 x 10¯10

4) Ethanolammonium ion hydrolyzes in water:

HOCH2CH2NH3+ + H2O ⇌ H3O+ + HOCH2CH2NH2

5) Solve the Ka expression for HOCH2CH2NH3+:

 (x) (x) 3.22581 x 10¯10 = –––––– 0.200

x = 8.0322 x 10¯6 M <--- this is the [H3O+]

6) Calculation of the pH:

pH = −log 8.0322 x 10¯6

pH = 5.10 (rounded to two sig figs)

Problem #10: Calculate the pH of a 0.100 M NaCN solution. The Ka for HCN, the conjugate acid of CN¯, equals 6.17 x 10¯10

Solution:

1) Write the hydrolysis equation for the cyanide anion:

CN¯ + H2O ⇌ OH¯ + HCN

2) Determine the Kb for CN¯:

KaKb = Kw

(6.17 x 10¯10) (Kb) = 1.00 x 10¯14

Kb = 1.62074 x 10¯5

3) Solve the Kb expression for the hydroxide ion concentration:

 (x) (x) 1.62074 x 10¯5 = –––––– 0.100

x = 0.001273 M

4) First the pOH, then the pH:

pOH = −log 0.001273 = 2.895

pH = 14 − 2.895 = 11.105