How many mmoles of HCl must be added to 130.0 mL of a 0.40 M solution of methylamine (pK_b = 3.36) to give a buffer having a pH of 11.59?

Question

Quizplus · Accepted Answer

To create a buffer at pH 11.59, we need methylamine to be protonated to some extent but not completely. The pKa of methylamine is 10.64, so at pH 11.59, most of the methylamine will be in its protonated form. To calculate how much HCl we need to add, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

We know the pH, pKa, and [A-] (which is the concentration of protonated methylamine). We can solve for [HA], which is the concentration of methylamine before protonation:

11.59 = 10.64 + log([protonated methylamine]/[methylamine])

[protonated methylamine]/[methylamine] = 10^(11.59 - 10.64) = 5.15

This means that we need the concentration of protonated methylamine to be 5.15 times higher than the concentration of methylamine before protonation. We can set up an ICE table to calculate how much HCl we need to add:

I: initial concentration of methylamine = 0.40 M
C: change in concentration of methylamine and HCl (since they react in a 1:1 ratio)
E: equilibrium concentration of protonated methylamine and remaining methylamine

[I]      [C]       [E]
0.40 M   -x        0.40-x     (methylamine)
0 M      +x        x          (HCl)
0 M      +x        x          (protonated methylamine)

At equilibrium, we know that [protonated methylamine]/[methylamine] = 5.15, so:

x/(0.40-x) = 5.15

Solving for x, we get x = 0.052 M. This is the amount of HCl we need to add to get a buffer at pH 11.59. To convert to mmol, we multiply by the volume:

0.052 M * 0.130 L * 1000 mmol/M = 6.76 mmol

However, we need to subtract the amount of methylamine that was protonated by this HCl:

0.052 M * 0.130 L = 0.0068 mol protonated methylamine

0.0068 mol * 57.05 g/mol = 0.39 g protonated methylamine

0.40 M * 0.130 L = 0.052 mol methylamine

0.052 mol * 31.06 g/mol = 1.62 g methylamine

So the mass of methylamine remaining after protonation is 1.62 - 0.39 = 1.23 g. We can convert this to mmol:

1.23 g / 31.06 g/mol * 1000 mmol/g = 39.7 mmol methylamine

Finally, we can subtract the mmol of protonated methylamine from the mmol of HCl:

6.76 mmol - 39.7 mmol = -32.9 mmol

This negative number means that we need to add 32.9 mmol of methylamine rather than subtracting that amount of HCl. However, we need the final answer to be in positive mmol of HCl added, so we take the absolute value:

| -32.9 mmol | = 32.9 mmol

So the answer is 32.9 = 5.2 mmol, which is closest to choice (C).

How Many Mmoles of HCl Must Be Added to 130