Hydrolysis calculations: salts of weak bases are acids

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Problem #1: What is the pH of a 0.0500 M solution of ammonium chloride, NH4Cl. Ka = 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 = square root of [(5.65 x 10¯10) (0.0500)]

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

Problem #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 = square root of [(2.70 x 10¯11) (0.100)]

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

Problem #3:

Given the pKa for ammonium ion is 9.26, 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¯9.26

Ka = 5.4954 x 10¯10

3) Write the dissociation equation and Ka expression:

NH4+ ⇔ H+ + NH3

Ka = ([H+] [NH3]) / [NH4+]

4) Insert values into Ka expression and solve:

5.4954 x 10¯10 = [(x) (x)] / 0.100

x = 7.4131 x 10¯6 M

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

-log 7.4131 x 10¯6 = 5.13

There is a comment at the end of the calculation tutorial for salts of weak acid, just above the practice problems. Please read it if you skipped it before. Also, with specific reference to salts of weak bases

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Practice Problems

Problems 1 & 2 are salt of weak acid problems. Please do them first.

3) 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.

4) 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.)

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