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**Here is the problem:**

CH_{3}NH_{2}, methyl amine is a weak base with a K_{b}of 4.38 x 10¯^{4}. What would be the pH of a solution of 0.350 M methyl ammonium chloride, CH_{3}NH_{3}^{+}Cl¯?

**Some discussion before the solution:**

1) This problem type seems to be seldom asked. Much more popular is giving you the K_{a} of an acid and then asking about pH of a solution formed from the salt of the acid. The problem above give you the K_{b} 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 K_{w} = K_{a}K_{b}. We will use that equation to calculate the K_{a} of CH_{3}NH_{3}^{+} (the acid) from the K_{b} of CH_{3}NH_{2} (the base).

4) See this tutorial for how the equation is derived. A reminder: the equation above applies to conjugate acid-base pairs. CH_{3}NH_{2} (the base) and CH_{3}NH_{3}^{+} (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:

CH_{3}NH_{3}^{+}+ H_{2}O <===> CH_{3}NH_{2}+ H_{3}O^{+}

We must calculate the K_{a} for the above reaction. We will use the K_{b} of CH_{3}NH_{2} 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 K_{a} value. I will round off the final answer to the appropriate number of significant figures.

We now need to calculate the [H^{+}] using the K_{a} 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 NH_{4}NO_{3}? The ionization constant of the weak base NH_{3} is 1.77 x 10¯^{5}

**Solution:**

1) The chemical equation of interest is this:

NH_{4}^{+}+ H_{2}O ---> H_{3}O^{+}+ NH_{3}

2) We need the K_{a} of the ammonium ion:

K_{w}= K_{a}K_{b}1.00 x 10¯

^{14}= (K_{a}) (1.77 x 10¯^{5})K

_{a}= 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}MpH = - 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, CH_{3}NH_{3}Br? (K_{b} of CH_{3}NH_{2} = 4.4 x 10¯^{4}.)

I answered this question on Yahoo Answers.

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