Calculate the pH of a solution of the salt, given the Kb of the base


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Here is the problem:

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 asking you for the pH of a salt of that base.

2) A bit of a warning: for some teachers, there would 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.

The solution.

Here is the relevant chemical equation for the above problem:

CH3NH3+ + H2O <===> CH3NH2 + H3O+

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

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

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.

We now need to calculate the [H+] 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

Since this is the [H+], 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 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 #3: Codeine is a narcotic pain reliever that forms a salt with HCl. What is the pH of 0.064 codeine hydrochloride (pKb = 5.80)?

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

I answered this question on Yahoo Answers.


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