A buffer is made up of 0.300 L each of 0.500 M KH2 PO4 and 0.317 M K2 HPO4. Assuming that volum

step 1 Okay, so in this problem, we're going to have a buffer. We're going to find its pH, and then we'

A buffer is made up of 0.300 L each of 0.500 M KH2 PO4 and...

A buffer is made up of $0.300 \mathrm{~L}$ each of $0.500 \mathrm{M} \mathrm{KH}_{2} \mathrm{PO}_{4}$ and $0.317 \mathrm{M}$ $\mathrm{K}_{2} \mathrm{HPO}_{4}$. Assuming that volumes are additive, calculate (a) the $\mathrm{pH}$ of the buffer. (b) the $\mathrm{pH}$ of the buffer after the addition of $0.0500 \mathrm{~mol}$ of $\mathrm{HCl}$ to $0.600 \mathrm{~L}$ of buffer. (c) the $\mathrm{pH}$ of the buffer after the addition of $0.0500 \mathrm{~mol}$ of $\mathrm{NaOH}$ to $0.600 \mathrm{~L}$ of buffer.

A buffer is made up of $0.300 \mathrm{~L}$ each of $0.500 \mathrm{M} \mathrm{KH}_{2} \mathrm{PO}_{4}$ and $0.317 \mathrm{M}$ $\mathrm{K}_{2} \mathrm{HPO}_{4}$. Assuming that volumes are additive, calculate
(a) the $\mathrm{pH}$ of the buffer.
(b) the $\mathrm{pH}$ of the buffer after the addition of $0.0500 \mathrm{~mol}$ of $\mathrm{HCl}$ to $0.600 \mathrm{~L}$ of buffer.
(c) the $\mathrm{pH}$ of the buffer after the addition of $0.0500 \mathrm{~mol}$ of $\mathrm{NaOH}$ to $0.600 \mathrm{~L}$ of buffer.

Key Concepts

Stoichiometry of Buffer Components

The stoichiometry in buffer systems involves understanding the amounts of the weak acid and conjugate base present and how they change when an external acid or base is added. Accurate calculations require accounting for the moles of each species before and after the reaction, as modifications in their ratios directly impact the resulting pH of the buffer solution.

Acid-Base Neutralization Reactions in Buffers

When strong acids or bases are added to a buffer solution, the acid or base is neutralized by the corresponding conjugate component of the buffer. This neutralization reaction prevents dramatic changes in pH. The stoichiometry of these reactions must be carefully accounted for, as changes in the concentrations of the buffer components will influence the overall pH according to the Henderson-Hasselbalch equation.

Buffer Solutions

A buffer solution is a system that contains a weak acid and its conjugate base (or a weak base and its conjugate acid) and is capable of resisting significant changes in pH upon the addition of small amounts of acid or base. This resistance occurs because the components in the buffer react with the added H+ or OH- ions, thereby minimizing the change in pH.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an essential formula in acid-base chemistry for calculating the pH of a buffer solution. It relates the pH of the solution to the pKa (or pKb) of the acid and the ratio of the concentrations of the conjugate base to the weak acid, making it a useful tool for predicting the behavior of buffer systems under various conditions.

Transcript

00:01 Okay, so in this problem, we're going to have a buffer.

00:04 We're going to find its ph, and then we're going to find its ph after we add either strong acid or strong base to it.

00:10 So first, let's go ahead and find the moles of each of the components.

00:15 So i have 0 .5 molar and 0 .30 liters to give me 0 .150 moles of our acid, our weak acid, which is h2 -p -o -4 negative.

00:30 And then 0 .317 molar, multiply it by its leaders, and that will give us the moles of our conjugate weak base.

00:49 And those are going to come in handy as we go forward here.

00:53 So in the first one, we're looking for the ph, just of the buffer, before we've added anything else to it.

01:00 So that's going to be h -plus is k -a times the concentration of the acid.

01:10 Over the concentration of the base.

01:15 So we've got our ka, which is going to be 6 .2 times 7 negative 8.

01:22 And then we'll just multiply by the concentration of our acid, which was 0 .5 molar, and divided it by the concentration of our base.

01:34 And then we'll get our concentration of hydrogen ion, which is 9 .8 times of the 8 -molar, h -plus.

01:43 And then if we go ahead and take minus 1 ,000, the log of that, we'll get the ph.

01:48 So for this, ph is 7 .01 for the buffer before we do anything to it.

01:55 Okay? in the next part, though, we're going to go ahead and add some strong acid.

02:02 Well, the strong acid, the h -plus, is going to react with the component that's the weak base.

02:08 So it's going to react with the h -p -o -4 -2 -minus to make some more weak acid.

02:19 H2, p .04 minus.

02:21 So some of the weak base is used up and some of the weak acid is produced.

02:25 I'm going to write a little table, okay, of initial change and final, just to keep track of what's going on here.

02:33 So i added 0 .05 moles of h plus to the 0 .095 moles of hydrogen phosphate that i have, and i started with 0 .150 moles of the dihydrogen.

02:50 Phosphate.

02:50 So you can see that h pluses are limiting reactant.

02:53 So this is going to be minus 0 .05.

02:56 So i must use up the same amount of this since it's 1 to 1 reaction.

03:00 And i'm going to add 0 .05 again because it's 1 to 1 ratio.

03:05 So i'll have none of this.

03:08 And after the reaction has taken place, i'll have 0 .045 moles.

03:14 And for our weak acid, we add those up and we're going to have 0 .2 .0 .0 .0.

03:20 Moles.

03:23 So if you look, you can see that we have a weak acid, a weak base and it's conjugate acid.

03:31 So we have a buffer...

Please give Ace some feedback