Enough water is added to the buffer in Question 29 to make...

Enough water is added to the buffer in Question 29 to make the total volume $10.0 \mathrm{~L}$. 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 diluted 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 diluted buffer. (d) Compare your answers to Question $29(\mathrm{a})-(\mathrm{c})$ with your answers to (a)-(c) in this problem. (e) Comment on the effect of dilution on the pH of a buffer and on its buffer capacity.

Enough water is added to the buffer in Question 29 to make the total volume $10.0 \mathrm{~L}$. 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 diluted 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 diluted buffer.
(d) Compare your answers to Question $29(\mathrm{a})-(\mathrm{c})$ with your answers to
(a)-(c) in this problem.
(e) Comment on the effect of dilution on the pH of a buffer and on its buffer capacity.

Key Concepts

Buffer Solutions

Buffer solutions are mixtures that resist changes in pH when small amounts of acid or base are added. They are typically composed of a weak acid and its conjugate base or a weak base and its conjugate acid. The key characteristic of a buffer is that it maintains the pH relatively constant by neutralizing added acids or bases, which is essential for many biochemical and chemical processes.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a mathematical formula used to calculate the pH of a buffer solution. It relates the pH to the pKa (or pKb) and the ratio of the concentration of the conjugate base to the acid. This equation is a fundamental tool for understanding and predicting the behavior of buffer systems during acid-base reactions.

Acid-Base Stoichiometry in Buffers

Acid-base stoichiometry in the context of buffers involves calculating how the addition or removal of small quantities of acid or base alters the amounts of the buffer components. Through careful stoichiometric calculations, one can determine the new concentrations of the buffer's conjugate acid and base after the addition of a strong acid or base, which then informs the pH calculation using the Henderson-Hasselbalch equation.

Effect of Dilution on pH and Buffer Capacity

Dilution of a buffer solution typically does not significantly change its pH because the ratio of the concentrations of the conjugate acid-base pair remains constant. However, dilution does affect the buffer capacity, which is the ability of the buffer to resist pH changes upon addition of acid or base. A more diluted buffer has a lower buffer capacity, meaning it can neutralize less acid or base before a significant pH change occurs.

Transcript

00:01 Okay, so we have a buffer here.

00:03 So let's start by finding the moles of each component.

00:06 Okay, so 0 .5 molar, 0 .3 liters will give us 0 .150 moles of the weak acid, which is h2 -p -o -4 minus, so hydrogen, dihydrogen phosphate.

00:24 And then we'll go ahead and find the moles of its conjugate base.

00:39 Okay, so we have a weak acid and its conjugate base, so we have a buffer.

00:43 So we can go ahead and find the hydrogen ion concentration of that buffer.

00:52 So let's see, we've got h -plus is k -a times the concentration of our acid over the concentration of our base.

01:08 But since concentrations are moles over liters, we can probably see that those two volumes will be in the same container, so they're going to cancel out.

01:21 So we can also write this as k .a.

01:26 Times the moles of our acid, h2, b04 negative, over the moles of our base.

01:37 So we've got our ka, so we'll go ahead and plug that in.

01:41 So we've got 6 .2 times 10 to negative 8.

01:45 That's our ka.

01:47 And then moles of acid, 5 moles, over 0 .095 moles.

01:54 So it gives us 9 .8 times 10 and negative 8 molar h plus.

02:05 So the ph is 7 .01.

02:11 Diluting the solution doesn't actually change the ph because of what i showed you here with these volumes.

02:18 So you can use concentration or moles.

02:20 So it doesn't matter actually that that solution was diluted.

02:23 It won't change the ph at all.

02:27 And the second part, though, we're going to dilute it, and then we're only going to use 0 .6 liters of it.

02:35 So i'm going to have to figure out how many moles of each substance i had.

02:39 So if i had 0 .15 moles in the 10 liters, if i only use 0 .6 liters, how many moles will i have in my 0 .6 liters? so i'm going to see that i get 9 times 10, the negative 3 moles of our dihydrogen phosphate.

03:06 And i'm going to do the same thing there for the other one, the conjugate base.

03:12 So 0 .095 moles of it were present in 10 liters.

03:18 So how many moles will be present in only 0 .6 liters? so now i'll get the moles of its conjugate base, so 5 .7.

03:30 Times 10 to negative 3 moles of the hydrogen phosphate.

03:39 And now what's going to happen is we're going to add some strong acid and it's going to react with the hydrogen phosphate.

03:49 So i'm going to write that the hydrogen ion reacts with our hydrogen phosphate, our weak base, to make some weak acid.

04:05 And we'll keep track of what's happening here, initial change and final.

04:08 I'm adding 0 .05 moles of my strong acid to 0 .0057 moles of my weak base.

04:20 And i have some weak acid to start.

04:24 So in this case, our weak base is going to be our limiting reactive.

04:31 So all of this is going to react here.

04:34 So i'll use that much of the h -plus, and i'll make this much of our weak acid.

04:43 Because those are all one -to -one ratios...

Please give Ace some feedback