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**Problem #1:** A 0.150 M solution of acetic acid (shorthand formula = HAc) is found to be 1.086% dissociated. What is the K_{a}?

**Solution:**

The equation for the dissociation of HAc is:

HAc <===> H^{+}+ Ac¯

The K_{a} expression is:

K_{a}= ([H^{+}] [Ac¯]) / [HAc]

We need to know the three values on the right.

1) We find [H^{+}] using the concentration and the percent dissociation:

(0.01086) (0.15) = 1.629 x 10¯^{3}M

2) For the [Ac¯], we use the 1:1 stoichiometry from the above equation to find:

[Ac¯] = [H^{+}] = 1.629 x 10¯^{3}M

3) For the [HAc], we use the concentration given. We could subtract 1.629 x 10¯^{3} from it, but this is usually not done. Warning: you might have a teacher who does ask you to do it. You might want to check to make sure.

4) Now, we insert into the K_{a} expression and solve:

x = [(1.629 x 10¯^{3}) (1.629 x 10¯^{3})] / 0.15x = 1.77 x 10¯

^{5}

Note: in April 2009, when the ChemTeam solved the above example on paper, he wrote 0.0186 rather than 0.01086. The ChemTeam is embarrassed!

**Problem #2:** A generic weak acid, formula = HA, has a concentration of 0.200 M and is 1.235% dissociated. Determine the K_{a}.

**Solution:**

Please note the use of a generic weak acid in the question. The solution technique for this type of problem works for almost all weak acids. In an introductiory course, the technique works for all acids, since the more unusual cases are reserved for a more advanced course. Hence, the use of a generic acid.

1) [H^{+}] = (0.200) (0.01235) = 2.47 x 10¯^{3} M

2) [A¯] = [H^{+}] = (0.200) (0.01235) = 2.47 x 10¯^{3} M

3) [HA] = 0.200

4) Insert into the K_{a} expression:

x = [(2.47 x 10¯^{3}) (2.47 x 10¯^{3})] / 0.200x = 3.05 x 10¯

^{5}

**Problem #3:** A 0.300 M solution of formic acid, HCOOH, is 2.45% ionized. Calculate the ionization constant for formic acid.

**Solution:**

(0.0245) (0.300) will give [H^{+}] and [HCOO¯]

The answer is 1.80 x 10¯^{4}

**Problem #4:** What is the K_{a} of a 9.5 x 10¯^{2} M solution of a monoprotic weak acid that has a percent dissociation of 0.62%?

The solution is left to the reader.

**Problem #5:** A 1.3 M solution of the weak base methylamine (CH_{3}NH_{2}) has a percent ionization of 0.72%. What is its K_{b}?

**Solution:**

1) Write the ionization equation for methylamine and its K_{b} expression:

CH_{3}NH_{2}+ H_{2}O <==> CH_{3}NH_{3}^{+}+ OH¯Use B in place of CH

_{3}NH_{2}:K

_{b}= ([HB^{+}] [OH¯]) / [B]

2) Use percent ionization to determine [OH¯]:

(1.3) (0.0072) = 0.00936 MThis is also [HB

^{+}].

3) Calculate K_{b}:

K_{b}= [(0.00936) (0.00936)] / 1.3K

_{b}= 6.7 x 10^{-5}

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