Enough water is added to the buffer in Question 30 to make...

Enough water is added to the buffer in Question 30 to make the total volume 10.0 L. Calculate (a) the $\mathrm{pH}$ of the buffer. (b) the $\mathrm{pH}$ of the buffer after the addition of $0.0300 \mathrm{~mol}$ of $\mathrm{HCl}$ to $0.710 \mathrm{~L}$ of diluted buffer. (c) the $\mathrm{pH}$ of the buffer after the addition of $0.0300 \mathrm{~mol}$ of $\mathrm{NaOH}$ to $0.710 \mathrm{~L}$ of diluted buffer. (d) Compare your answers to Question $30(\mathrm{a})-(\mathrm{c})$ 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 30 to make the total volume 10.0 L. Calculate
(a) the $\mathrm{pH}$ of the buffer.
(b) the $\mathrm{pH}$ of the buffer after the addition of $0.0300 \mathrm{~mol}$ of $\mathrm{HCl}$ to $0.710 \mathrm{~L}$ of diluted buffer.
(c) the $\mathrm{pH}$ of the buffer after the addition of $0.0300 \mathrm{~mol}$ of $\mathrm{NaOH}$ to $0.710 \mathrm{~L}$ of diluted buffer.
(d) Compare your answers to Question $30(\mathrm{a})-(\mathrm{c})$ 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 Capacity

Buffer capacity refers to the amount of strong acid or base that can be added to a buffer solution before a significant change in pH occurs. It depends on the absolute concentrations of the buffering agents; a more concentrated buffer has a higher capacity and can better resist changes in pH.

Buffer Solution

A buffer solution is composed of a weak acid and its conjugate base, or a weak base and its conjugate acid, and is designed to maintain a stable pH when small amounts of strong acid or base are added. The key characteristic is the ability to moderate pH changes through a reversible equilibrium between the acid and base components.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an algebraic expression that relates the pH of a buffer to the pKa of the weak acid and the ratio of the concentrations of the conjugate base to the acid. This equation is central to calculating the pH of a buffer and analyzing system equilibria under various conditions.

Effect of Dilution on Buffer pH

Dilution of a buffer solution generally does not change the pH significantly because the ratio of the concentrations of the acidic and basic components remains constant. However, the absolute concentration of buffering species decreases, which can affect the buffer's ability to neutralize added acids or bases.

Transcript

00:01 So let's start by calculating the moles of our weak acid and weak base.

00:05 So we've got 0 .2 molar and 0 .355 liters.

00:13 So that gives us 0 .0710 moles of our bicarbonate.

00:21 And then for the carbonate, 0 .134 molar times the same volume.

00:29 So this is going to be 0 .0476 moles of carbonate.

00:35 So we have a buffer.

00:36 We have a weak acid in its conjugate base.

00:41 So the concentration of h plus in a buffer is ka times a concentration of the acid over the concentration of the base.

00:55 And since that's moles per liter over moles per liter, the volumes cancel out.

01:03 And we can also write this equation as ka times moles of acid over moles of base.

01:13 So i'm going to use it that way.

01:14 So our ka is 4 .7 times 10 to negative 11.

01:21 Our moles of acid is 0 .0710, 0476.

01:29 So this is going to give us 7 times 10 to the negative 11 molar h plus.

01:35 So if we take minus the log of that, we'll get the ph, which is 10 .15.

01:41 Okay.

01:44 Diluting our solution doesn't actually change our ph.

01:51 Okay, and this next part, we've got 10 liters of our buffer, but we're only going to use 0 .710 of it.

01:58 So let's do some ratios to figure out how many moles.

02:01 If i have 0 .710 moles of my bicarbonate in 10 liters, when i only use 0 .710 liters, how many moles of bicarbonate in 10 liters? bicarbonate will i end up with using.

02:17 So i have 0 .0504 moles of my bicarbonate to react here.

02:25 And let's do the same thing here.

02:28 0 .0476 moles in 10 liters of my buffer.

02:36 How much would be if i only use 0 .710 liters of it? okay, so that's going to give us 0 .00338 moles.

02:49 Of our carbonate.

02:53 Okay, and we're going to take some h plus and add it to our buffer.

02:58 So that h plus is going to react with our carbonate.

03:05 It reacts with the weak base to make more weak acid.

03:11 So we're starting with 0 .3 -0 -0 -0 -0 -0 -0 -8 moles there, and 0 .003 -38 moles, and 0 .00504 moles to start with.

03:24 Well, you can see that your limiting reactant is your carbonate, so minus .00338.

03:32 We'll subtract the same thing here, and we'll add the same thing over here because it's all one to one...

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