Practice Problem Answers

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1) 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.

Answer

  (x) (x)
1.55 x 10¯10 = ----------------
  0.30 - x

x = square root of [(1.55 x 10¯10) (0.30)]

x = 6.8 x 10¯6 M = [OH¯]

pOH = - log 6.8 x 10¯6 = 5.17

pH = 14 - 5.17 = 8.83

2) Find the pH of a 0.20 M solution of sodium propionate (C2H5COONa), where the Ka of propionic acid = 1.34 x 10¯5. (Hint: Kw = KaKb)

Answer

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  

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

  (x) (x)
7.46 x 10¯10 = ----------------
  0.20 - x

x = square root of [(7.46 x 10¯10) (0.20)]

x = 1.22 x 10¯5 M = [OH¯]

pOH = - log 1.22 x 10¯5 = 4.91

pH = 14 - 4.91 = 9.09

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.

Answer

The chemical reaction is: C6H5NH3+ + H2O ---> C6H5NH2 + H3O+

The calculations for the pH are:

  (x) (x)
2.3 x 10¯5 = ----------------
  0.0400 - x

x = square root of [(2.3 x 10¯5) (0.0400)]

x = 9.6 x 10¯4 M = [H3O+]

pH = - log 9.6 x 10¯6 = 3.02

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.)

Answer

First, we calculate the Ka:

  1.00 x 10¯14  
Ka = ---------------- = 6.25 x 10¯6
  1.6 x 10¯9  

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

x = square root of [(6.25 x 10¯6) (0.015)]

x = 3.06 x 10¯4 M = [H3O+]

pH = - log 3.06 x 10¯6 = 3.51

Here is a very nice bonus problem, as a reward for working through the above 4 problems:

Determine the Ka of the weak acid HX knowing that a 0.10 M solution of LiX has pH = 8.90. (Hint: you have to work backwards from the example calculations and practice problems.)

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