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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 = KaKb1.00 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)] / 0.502x = 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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