12/1 in-class activity

The pK_a of CH₃NH₃⁺ is 10.62.  A solution is originally 1.072 M in CH₃NH₂ (methylamine) and 0.836 M CH₃NH₃⁺.

1.) What is the initial pH of the buffer?  Hint: use the Henderson-Hasselbalch equation.

pH = pK_a + log([base]/[acid])                     (Henderson-Hasselbalch equation)

[base] = [CH₃NH₂] = 1.072 M

[acid] = [CH₃NH₃⁺] = 0.836 M

pH = 10.73

                         

2.) 3 mL of 1.5 M HBr is added to a 15 mL solution of the buffer above.  Taking the volume change into consideration, what is the pH of the new solution?  What is the pH of a 15 mL solution of the buffer after 3 mL of pure water is added?

Use M₁V₁=M₂V₂ to find the new concentrations of acid and base in solution.

[base] = [CH₃NH₂] = 0.8933 M

[acid] = [CH₃NH₃⁺] = 0.697 M

[HBr] = 0.25 M

Here, we assume that the HBr reacts completely with the base, methylammine.  We therefore subtract 0.25 M from the concentration of the base.  We now get the concentrations:

[base] = [CH₃NH₂] = 0.6433 M

[acid] = [CH₃NH₃⁺] = 0.947 M

We plug these values into the Henderson-Hasselbalch equation.

pH = 10.47

pH decreases.  This makes sense.

For the second question, we use M₁V₁=M₂V₂ to determine the new concentrations.  We plug these new values into the Henderson-Hasselbalch equation and determine the new pH.  However, since the concentrations of acid and base decrease by the same proportion, the ratio of the two stays the same, and the pH is the same as in problem 1.  pH = 10.73

3.) 5 mL of 1.7 M NaOH is added to 12 mL of the buffer solution.  Taking the volume change into consideration, what is the pOH of the new solution?  What is the pOH of a 12 mL solution of the buffer after 5 mL of pure water is added?

Use M₁V₁=M₂V₂ to find the new concentrations of acid and base in solution

[base] = [CH₃NH₂] = 0.7567 M

[acid] = [CH₃NH₃⁺] = 0.590 M

[NaOH] = 0.50 M

Here, we assume that the NaOH reacts completely with the acid, methylammonia.  We therefore subtract 0.50 M from the concentration of the base.  We now get the concentrations:

[base] = [CH₃NH₂] = 1.257 M

[acid] = [CH₃NH₃⁺] = 0.090 M

We plug these values into the Henderson-Hasselbalch equation.

pH = 11.76, pOH = 2.24

pH increases.  This makes sense.

For the second question, we use M₁V₁=M₂V₂ to determine the new concentrations.  We plug these new values into the Henderson-Hasselbalch equation and determine the new pH.  However, since the concentrations of acid and base decrease by the same proportion, the ratio of the two stays the same, and the pH is the same as in problem 1.  pH = 10.73, pOH = 3.27