Buffer Solutions - Free Educational videos for Students in K-12

Buffer Solutions - By The Organic Chemistry Tutor

Transcript


00:00	in this video we're going to talk about buffer solutions
00:03	. A buffer solution is composed of a weak acid
00:07	and the conjugate base . So let me give you
00:11	some examples of that . HF , Hydrofluoric acid ,
00:18	that's a weak acid . The conjugate base of HF
00:22	is fluoride . All you gotta do is remove a
00:24	hydrogen . So we're gonna pair of fluoride with sodium
00:28	. So here we have the weak acid and the
00:30	conjugal weak base . Together these two components form a
00:34	buffer solution . Another example will be acidic acid which
00:39	is a weak acid . The conjugate base of acetic
00:43	acid is acetate C two H 302 minus and we'll
00:48	pick it up with the sodium ion . So sodium
00:50	acetate will be the week base these two combined .
00:56	They create a buffer solution . Or let's say if
00:59	we have hydra cynic acid , H . D .
01:01	N . With its conjugate base , sodium cyanide .
01:06	This too will create above solution . So a buffer
01:09	solution is composed of a weak acid and the congregate
01:12	with base . The purpose of a buffer solution is
01:16	to maintain a constant ph level throughout the solution .
01:20	It resists changes in its ph Now it has the
01:27	ability to do that because it contains both a weak
01:30	acid and a weak base . So let's see if
01:32	we have a solution of HF and sodium fluoride .
01:38	So we have a buffer solution . If we were
01:41	to add acid to this solution , let's say hydrochloric
01:44	acid Instead of HDL , let's use H 30 plus
01:51	. So if we were to add acid or anything
01:54	that would increase the HDL plus concentration , the acid
01:59	is going to react with the week based component of
02:01	the buffer solution . So it's gonna react with fluoride
02:08	. And so we're gonna get HF and water .
02:14	If fluoride wasn't there , the ph would decrease dramatically
02:21	when you have a high level of H 20 plus
02:23	or hydro knee um ions in a solution , the
02:26	pH will be very low and that's what assets do
02:31	whenever you add an asset to a solution , it
02:33	brings down the ph , but because fluoride reacts with
02:37	the incoming hydrogen ions , you're not going to have
02:40	a significant drop in ph Because they will quickly consume
02:44	these ions in the solution , create an HF and
02:47	one . Now let's look at the other situation .
02:52	Let's say if we add a strong base , like
02:53	sodium hydroxide , if we immediately increase the hydroxide ion
02:59	concentration , the ph of the solution , it's going
03:03	to go up . I mean , I take that
03:05	back the people which is going to go down rather
03:08	the ph is going to go up . These two
03:10	are related to each other . Remember the ph is
03:15	14 minus the ph of the solution . So when
03:19	the ph goes down , the ph goes up now
03:25	, once you add hydroxide , it's going to react
03:27	with the weak acid component of the buffers . So
03:30	it's going to react with HF HF is going to
03:33	quickly consume the hydroxide ions , creating fluoride and water
03:42	. And so because HF quickly reacts with hydroxide ,
03:46	we're not going to see a sharp decrease in appeal
03:49	wage of a solution nor will we see a sharp
03:51	increase in the ph , the ph will increase slightly
03:55	, but not by that much because these are being
03:58	reacted with HF . And so that's how a buffer
04:01	solution can maintain a relatively constant ph level in the
04:06	solution . It's by having the ability to react with
04:10	any incoming acid or any incoming base that is added
04:14	to the solution . Now , let's talk about how
04:18	we can calculate the ph of a buffer solution .
04:23	So let's say we have a solution with Hydrofluoric acid
04:27	and water and the solution also contains fluoride . So
04:41	in this reaction we have the weak acid and the
04:44	conjugate weak base . So we have a buffer solution
04:50	. Everything is in the acquis phase except water .
04:54	If we were to write the equation that is associated
04:57	with K . A . The acid dissociation constant ,
05:01	this will be equal to the concentration of the products
05:13	divided by the concentration of the reactant . So what
05:21	is not included because it's in the liquid phase .
05:24	Now what we're gonna do is we're gonna take the
05:26	negative log of both sides of the equation . So
05:31	we're gonna have negative log of K . A .
05:34	And that's going to be equal to negative log of
05:38	the right side of the equation . So we have
05:49	f minus Times H 30 plus divided by HF .
06:03	Now a property of logs allows us to split one
06:07	log into two separate logs . So for instance let's
06:11	say if we have log a B , we can
06:15	write this as log A plus , log B .
06:21	So let's say florida's a HBO plus is B .
06:25	And HFC . We could also do this . If
06:28	we have log A B , oversee , we can
06:35	split be from A . N . C . So
06:38	we can say this is log a oversea plus log
06:43	B . We don't have to separate it from C
06:47	. We can keep them together or we can separate
06:49	them Right now I want to take out H20 plus
06:54	out of this expression . So be would correspond to
06:58	H . 20 Plus . I'm going to keep A
07:00	N . C . The way it is , I'm
07:02	going to use this property of the log expression .
07:10	So this is going to be negative law K .
07:15	A . And that's going to equal . So we
07:20	need to distribute the negative sign . So keep that
07:21	in mind . Negative log . We're going to have
07:29	a oversea actually let's write log be first . So
07:35	this is gonna be negative log atrial plus . And
07:43	then we're gonna have log oversee with a negative sign
07:47	because this has to be distributed and then minus log
07:54	and then it's going to be the base over the
07:55	acid . So florida is the base . HF is
08:02	the acid . Now let's go ahead and delete some
08:13	of the stuff that we have here to make some
08:15	extra space . Now , some formulas that you need
08:19	to be familiar with . All right , these two
08:21	, you know that the ph is equal to the
08:23	negative log Of H . 30 . Plus . You've
08:28	seen that many times before at this point . Mhm
08:32	. Also , you know that the p K .
08:36	Is negative log K . A . So we can
08:40	replace this with the PKK . We can replace this
08:45	with ph so we have Pekka is equal to the
08:49	ph minus the log of the base . In this
08:57	case the bases f minus divided by the acid HF
09:11	. Now you can put this in parentheses to indicate
09:14	concentration . But since you have a ratio , it
09:16	could be in moles as well . Because most similarity
09:23	they have the same ratio given the same volume of
09:26	the solution . Now , what we're gonna do is
09:31	we're gonna isolate ph so I'm gonna take this term
09:35	and move it to the other side . It's negative
09:41	on the right side , but it's gonna be positive
09:43	on the left side . So the PKK plus log
09:50	of the base over the acid . And I'm going
09:53	to replace this with a more generic form of the
09:55	base , which is typically a minus . And the
10:00	generic form of an asset would be H . A
10:06	. On the right side . We still have ph
10:09	, I'm going to reverse these two equations . I'm
10:12	gonna put ph on the other side and I'm gonna
10:14	put this on the other side . So if we
10:16	simply reverse these two equations or just reverse the equation
10:22	, rather by switching both sides , we're going to
10:25	get the Henderson Hasselbach equation . So this equation helps
10:29	us to calculate the ph of a buffer solution .
10:33	So the ph is equal to the Peca of the
10:35	acid plus log base over acid . A minus is
10:44	the week based component of the acid , H .
10:47	A . Is the weak acid . So that's how
10:49	you can calculate the ph of a buffer solution .
10:53	Is by using the Henderson Hasselbach equation . So the
10:57	ph is dependent on the ratio of the contract base
11:04	to the conjugate acid . Now there are some things
11:13	that we can discern from this equation . What can
11:17	we say regarding the ph of a buffer solution when
11:21	the concentration of the weak acid is equal to the
11:25	concentration of the conjugate base ? If H a equals
11:30	a minus . What can we say about the ph
11:35	If this is 10 and that is 10 , 10/10
11:38	is one . If this is .5 , that's .5.5.5
11:43	is one . So whenever these are the same and
11:47	the 1-1 ratio , you get log of one ,
11:51	The log of one is 0 . So this disappears
11:55	. Thus we have , the ph is equal to
11:57	the P K . So whenever you have a buffer
12:01	with an equal amount of weak acid and weak base
12:05	, the ph will be equal to the PKK .
12:11	So let's say if you want to create a buffer
12:14	solution With or that centered at a ph of 4.5
12:22	. If you only create a buffer solution that's centered
12:24	at that ph well , you want to find an
12:26	acid With the PKK that is very close to 4.5
12:32	. Likewise , if you want to create a buffer
12:34	solution with a ph that's centered around six point out
12:39	, you want to look for an acid that has
12:41	a peek a of A value that is close to
12:45	six point close as possible . The closer the better
12:49	. So that's how you can create a buffer solution
12:52	that's centered at a given ph is by choosing the
12:55	appropriate acid with the PKK that is very close to
12:58	the ph . Now let's talk about the ratio of
13:04	base to acid and acid base log 10 is one
13:15	. Let's think about what that means . So let's
13:18	say if a minus is 10 times the value of
13:23	a change , Let's say this is one , M
13:28	and H a is 0.1 . One divided by .1
13:34	is 10 . Lot of 10 is one . And
13:39	let's say the peek A . Is for what's the
13:42	ph The ph is going to be one unit higher
13:49	whenever the ratio between acid debates or base to acid
13:53	. When it differs whenever you have a 1 10
13:55	ratio , the ph is going to be one unit
13:58	away from the PKK . If you have more of
14:01	the acid , it's gonna be one unit lower than
14:04	the peak . A if you have more of the
14:06	base , it's one unit higher Log of a 100
14:17	is to 100 is 10 squared . A property of
14:22	logs allows you to move the , export it to
14:24	the front . So a lot of 10 squared is
14:27	the same as too long . 10 and log 10
14:30	is one . So you get two times one ,
14:32	which is to So long of 100 is to .
14:36	Therefore , if the ratio between base to acid or
14:39	ask the base is 1-100 , The Ph will be
14:43	two units away from the PKK Log of 1000 ,
14:49	which is log 10 to the 3rd . That's equally
14:54	three . So if the ratio of acid base or
14:57	base to acid is 1000 The Ph will be three
15:01	units away from the PK . So let's say the
15:07	PKK of a certain asset . We're going to say
15:10	it's for actually , let's let's make it six .
15:19	And let's say the numbers on this line represents the
15:21	ph of the solution . When the Ph is six
15:25	, we're going to have equal amounts of basin acid
15:29	. So 50% will be based and 50% will be
15:38	the weak acid H . A . Under those conditions
15:42	, the ph will equal the PK . So we
15:49	can describe that as a percentage . Or we can
15:52	say , you know , we have one mauler of
15:56	the conjugate base And then one molar solution of the
16:00	conjugate acid At a Ph of seven . Let's keep
16:07	the amount of basis . Same . Let's say it's
16:09	one moller . The amount of acid will be 10
16:12	times less . It'll be .1 mm . Ahh .
16:19	So as you can see here , we have more
16:21	bass than acid . So the ph is going to
16:23	be higher than the PKK Will be the situation at
16:28	a page of eight . So if we have the
16:32	same amount of base , The amount of acid will
16:36	be 10 times less is going to be point 01
16:39	M . So any time the ph is well ,
16:45	anytime you have more based on acid , the ph
16:49	will be greater than the P . K . So
16:52	on the right side where a minus that is the
16:56	week based component of the buffer . If it's greater
16:59	than H . A . Or will be great in
17:01	the H . A . Under those conditions , the
17:04	ph will be greater than the PKK . Now ,
17:11	what if we had one M aging In .1 M
17:16	. A minus ? What's the ph ? So here
17:20	we have more of the acid than the base .
17:22	So the ph is going to be less than the
17:23	PK And the ratio is 1 - 10 . So
17:27	we're gonna be one unit away from the PKK .
17:29	Therefore the PhD will be , It's gonna be five
17:37	. Likewise , let's see if we have .01 M
17:41	a minus mhm and one M A chain . The
17:49	ratio between asset base Or based the asset is like
17:53	1 - 100 . So Lot 200 is too ,
17:58	we're gonna be two minutes away from the PKK Or
18:02	let's see if we have .001 of the base And
18:06	the one molar solution of the acid . The ratio
18:10	is 1-1000 one , divided by .001 is a 1000
18:14	. So we're gonna be three units away from the
18:16	PK . So that's how you can determine the ph
18:21	of a buffer solution conceptually by looking at the ratio
18:25	between based acid or acid base . And remember if
18:33	you have more of the asset , then the contract
18:37	base . The ph is going to be less than
18:44	the PK . And if you have equal amounts of
18:47	acid and base , the ph will be equal to
18:51	the PK whenever you have a buffer solution . Now
18:55	, let's go ahead and work on some practice problems
18:58	. Number one . What is the ph of a
19:01	solution Consistent of .75 molar acetic acid And .5 Molar
19:06	sodium acetate . And were given the acid association constant
19:11	of a city acid Is 1.8 times 10 to the
19:14	-5 . Feel free to pause the video if you
19:17	want to work on this problem . So what we're
19:22	gonna do is we're going to use the Henderson Hasselbach
19:25	equation because we have a weak acid and we have
19:28	the conjugal we base , which means we have a
19:30	buffer solution . The ph is going to equal the
19:37	Peca of the weak acid plus log of the base
19:43	. The base . We can write as a minus
19:45	. Or you can simply put B . For the
19:47	base and then A . For the acid . Well
19:50	, you can write a check if you want .
19:52	I like to write be over a base over acid
19:56	. Now the first thing we need to do is
19:58	calculate the P . T . A . Of the
20:00	weak acid and that's negative log of the K .
20:03	A . So we have the K . Value Which
20:08	is 1.8 times 10 To the -5 . So the
20:20	P . K . of the city acid is 4.7447
20:27	. So now let's plug it into this formula so
20:29	that we have the PKK . It's this number and
20:34	then plus log the base . This is the based
20:39	component of the buffer . So the concentration is .5
20:43	Mueller And the concentration of the weak acid component is
20:48	.75 Mueller . The unit similarity will cancel because we're
20:53	dealing with a ratio here . So let's go ahead
20:58	and plug this in 4.7447 plus log .5 divided by
21:04	.75 . This is you can run into 4.57 .
21:13	So that's the answer for this problem . So that's
21:16	how we can calculate the ph of a buffer solution
21:20	. Now , there's one more thing I want to
21:21	mention regarding this problem notice that we have for more
21:26	of the acid than the base because we have more
21:31	of the acid than the base . The ph is
21:35	going to be less than the peak . A as
21:39	we discussed earlier in this video Here , we can
21:41	see the ph is 4.57 . The PKA is 4.74
21:46	. So the ph is less than the PKK because
21:49	we have more of the acid component than the base
21:52	component of the buffer . Number two , What is
21:56	the ph of a solution containing .15 moles of ammonium
22:01	chloride And 1.5 moles of Ammonia . And were given
22:05	the KB , the based association constant of NH dream
22:13	. So what should we do here ? What ?
22:16	Once again we have a buffer solution , We have
22:18	the weak acid and we have the weak base ,
22:22	the consequent weak base . Now in this case notice
22:25	that we have more of the base than the acid
22:30	. We have more of ammonia then the morning chloride
22:34	. So what can we say about the P .
22:35	H . And P K . The ph is going
22:40	to be greater . And then the PKK because we
22:46	have more of the base than the acid . Let's
22:48	go ahead and calculate the P K . Were given
22:50	the KB . So we need to calculate the PKB
22:52	. 1st PKB is negative log of KB . So
23:02	this is gonna be negative log of 1.8 Times 10
23:07	to the -5 . And we've seen that value .
23:17	So this is 4.7 44 73 Now the PKK plus
23:28	the PKB , These two , they add up to
23:30	14 . So to calculate the P . K .
23:34	It's gonna be 14 minus the PKB . So 14
23:41	-4 0.7 4473 . That gives us a peek a
23:54	value of 9.2 5 5 - seven . So since
24:05	we have more bass than acid , we know that
24:08	the answer the ph of the solution Has to be
24:11	greater than the P . K . It has to
24:13	be greater than 9.25 5-7 . Let's run that to
24:18	nine 255 So let's go ahead and calculate the ph
24:24	of the solution . But before we do that ,
24:28	notice that the ratio of base to acid is 10-1
24:33	, 1.5 divided by .15 is 10 . Whenever the
24:38	ratio is 10 to 1 or 1 to 10 ,
24:41	we know that ph will differ From the PKK by
24:44	one unit . And since we have more bass than
24:47	acid , one unit higher than this number , It's
24:50	gonna be 10.255 . Therefore conceptually , we know that
24:55	the ph of this solution should be this value .
25:00	And we're going to confirm it by calculation So we
25:02	should get 10.255 . So let's write the buffer equation
25:08	, ph is equal to P K plus log based
25:15	over acid . The PKK is nine .255 And then
25:23	we can add the 27 as well to get a
25:25	more accurate result and then plus log Now , as
25:29	was mentioned before , this could be in eunice similarity
25:33	or it could be in moles . Both hands will
25:36	cancel . So we're going to plug in the most
25:39	. We have 1.5 moles over point 15 moles .
25:46	So we can cross out these two units . Now
25:51	let's plug this into our calculator , 9.255-7 plus log
25:59	1.5 over .15 . This will give you 10.2 5
26:06	5 - seven . So that's the ph of the
26:10	solution . So we could see why it's one unit
26:14	higher Is because the ratio of based asset is 10-1
26:19	. Number three , Calculate the ph of a solution
26:23	containing 15g of Hydrofluoric acid and 21 g of sodium
26:28	fluoride In 750 ml of solution . So once again
26:32	we have a buffer solution but this time well given
26:36	the grams of the acid and a base before we
26:39	had the polarity . And during the second problem we
26:41	had the moles for this one . All we need
26:44	to do is convert grams to moles . And then
26:47	we could use the buffer equation . So let's do
26:49	that first . So we have 15 g of HF
26:55	. To convert it to moles . We need to
26:56	know the molar mass . The molar mass of hydrogen
27:02	is approximately one For flooring is 19 . So this
27:07	is approximately 20 g per remote . So one mole
27:11	of HF Has a mass of approximately 20 gramps .
27:19	So we can cross out the unit grams of HF
27:23	. It's gonna be 15 divided by 20 , Which
27:25	is .75 . So that's how much moles of acid
27:30	we have in this solution . Now let's start with
27:34	21 g of the base . Any f . Let's
27:37	convert that to mel's sodium has an atomic mass of
27:44	approximately 23 And for Florida's 19 23 Plus 19 is
27:49	42 . So one more . Love any F .
27:55	Has a mass of approximately 42 g . 21 divided
28:04	by 42 is .5 . So we get .5 moles
28:10	of sodium fluoride . So now that we have the
28:13	moles of the acid and base , we can now
28:17	calculate the ph but first , let's calculate the P
28:21	K . The PKK , which is negative log of
28:24	K . A and K . Is that number ?
28:27	So we have the negative log of 7.2 times 10
28:31	to the -4 . And so that's going to be
28:41	3.14 27 So now that we have that , let's
28:48	use the Henderson Hasselbach equation to get the ph of
28:53	the solution . So the peka is 3.14-7 plus log
29:03	of the base . The bases , by the way
29:07	, will the ph b greater than or less than
29:10	a PK ? So notice that we have more of
29:14	the acid than the base . So because we have
29:18	more of the acid than the base , the ph
29:22	is going to be less than a PK . So
29:26	the ph should be less than 3.14 - seven .
29:32	The solution is acidic . So now let's go ahead
29:38	and continue with this equation . So the bases .5
29:43	moles . The acid is point 75 moles . So
29:56	let's plug this in 3.14-7 plus log .5 over .75
30:03	. This is going to be approximately 2.97 , Which
30:07	is less than 3.14 . So that's the ph of
30:11	the solution . So any time you have more of
30:15	the acid component than the basic component , the ph
30:17	will be less than the P . K . Number
30:20	four . What is the Peca of an unknown weak
30:23	acid ? If the ph of the solution was measured
30:27	to be 5.62 When H is .45 And a minuses
30:33	.85 . So we want to calculate the P .
30:37	K . A . If we know the ph and
30:39	the concentration of the weak acid and the conjugal weak
30:42	base . So let's start with the buffer equation ,
30:46	ph is equal to P . K . Plus log
30:53	base over acid . Now , what we need to
30:56	do is we need to isolate PK . So I'm
30:59	going to take this term , move it to the
31:01	other side . So it's positive on the right side
31:05	. It's going to be negative on the left side
31:13	. Now I'm going to switch the left side of
31:14	the equation with the right side of the equation .
31:17	So we could say that the PKK of the unknown
31:20	weak acid is going to be the ph of the
31:23	solution minus log of the base over the acid .
31:30	So this is how we can calculate the peak a
31:33	of an unknown weak acid . All we need to
31:35	know is the ph of the solution and how much
31:38	of the acid and base that we've dissolved in a
31:40	solution . So the ph is 5.62 . The base
31:51	is the concentration of a minus . So that's .85
31:58	. The acid is the concentration of H . J
32:02	. So that's .45 . So this is going to
32:16	be 5.34 . So this is the peak a of
32:21	the unknown acid . Now , let's see if our
32:25	answer makes sense . So whenever the base is greater
32:31	in concentration on quantity , then the acid we know
32:35	that the ph is going to be greater than the
32:37	PKK . Is that the situation here , while the
32:42	ph The ph is 5.62 , the PKK It's 5.34
32:51	. So we can clearly see that the ph is
32:52	greater than the peka . So this answer makes sense
00:0-1	.

Summarizer

DESCRIPTION:

This chemistry video tutorial explains how to calculate the pH of a buffer solution using the henderson hasselbalch equation. It explains the concept, components, and function of a buffer solution. A buffer solution consist of a weak acid and its conjugate weak base counterpart. It's purpose is to maintain a relatively constant pH value. This video discusses the relationship between the pH and pKa values with the relative amounts of weak acid and weak base components in the buffer solution. This video contains plenty of examples and practice problems.

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