pH of Weak Acids and Bases - Percent Ionization - Ka & Kb - Free Educational videos for Students in K-12

pH of Weak Acids and Bases - Percent Ionization - Ka & Kb - By The Organic Chemistry Tutor

Transcript


00:00	in this video , we're going to talk about how
00:02	to calculate the ph of a weak acid solution as
00:06	well as the ph of a week based solution .
00:09	We're also going to talk about percent organization and some
00:12	of the problems associated with weak assess on the basis
00:15	. But let's review some basics . Now , if
00:18	you recall when a strong acid like hydrochloric acid reacts
00:23	with water because it's a strong acid , it will
00:27	dissociate completely . So we could use a single arrow
00:30	to represent this reaction . So if the initial concentration
00:37	of H . D . I would say it's zero
00:39	five as it dissociates completely , the hydrogen ion concentration
00:44	will be the same . So you can just calculate
00:48	the ph of the solution . Using this formula ,
00:51	the ph is equal to negative log of HBO plus
00:56	. And if you need to calculate the P .
00:57	O . H , let's save a strong based solution
01:00	. You could use this negative log oh h minus
01:04	. I remember the ph plus the P O .
01:07	H . As up to 14 at 25°C. . So
01:13	if you know the P . O . H ,
01:15	the ph is simply 14 minus the P . O
01:17	. H . Now I'm dealing with a weak acid
01:22	solution . The situation is different . A weak acid
01:28	doesn't associate completely . So we're gonna use the double
01:32	arrow to indicate a situation of equilibrium . So this
01:37	reaction is reversible . Yeah . So when dealing with
01:46	a weak acid or a weak base or let's say
01:50	a weak acid , you need to use something called
01:52	a K . A . The acid dissociation constant ,
01:55	it's an equilibrium constant and it's equal to the concentration
01:59	of the products divided by the concentration of the reactant
02:07	. What is the liquid ? So remember , liquids
02:09	and solids are not included in the equilibrium expression .
02:12	Everything else is any acquis face , It's dissolved in
02:16	water . So you do include them in the equilibrium
02:19	expression . Now , once you calculate H 30 plus
02:26	using ice table , so you can calculate the ph
02:29	using the equation that we just considered now , I'm
02:34	dealing with basis in this case any streets a weak
02:43	base . You need to use the base dissociation constant
02:48	, which is K . B . KB . Like
02:52	K . A . Is equal to the concentration of
02:54	the products divided by the concentration of the reactant .
03:04	Now use nice tables . You want to calculate the
03:06	concentration of hydroxide because once you have that , you
03:10	can calculate the P . O . H . Using
03:12	this formula and then once you have the P .
03:16	O . H . You could find the ph .
03:26	Now sometimes you may need to calculate KB from K
03:29	. A . Or vice versa . Remember K .
03:32	A . Times KB Is equal to K . W.
03:36	, which is one times 10 to the -14 .
03:40	And this is temperature dependent . So this equation holds
03:43	true at 25°C. . Now sometimes you may need to
03:52	calculate the atrial plus concentration from ph and this formula
03:57	will help you to do that . It's equal to
03:59	10 raised to the negative ph . Now if you
04:04	need to calculate the hydroxide ion concentration it's 10 ,
04:09	raise the negative P . O . H . And
04:13	also the hydrogen ion concentration times the hydroxide iron concentration
04:22	at 25 degrees Celsius . This is equal to K
04:24	. W . The auto ionization constant for water and
04:28	that's one Times 10 to the -14 . So those
04:32	are some other formulas that you want to be familiar
04:35	with . Now let's talk about acid strength . I've
04:42	covered this in some other videos but nevertheless it's a
04:46	good concept to review acid strength increases with increasing K
04:52	values . So if you have to acids with two
04:56	different K values , the stronger acid is the one
04:59	with the higher K value . Now , acid strength
05:04	is inversely related to peek A . So the stronger
05:07	acid is going to have the lower PK value ,
05:11	peek A . Is equal to negative log of K
05:16	. And PKB is equal to negative log KB .
05:26	So you want to add these formulas to your notes
05:29	if you haven't done so already . Now the PKK
05:33	plus the PkB . These two , they add up
05:36	to 14 at 25°C. . Now the next thing we
05:42	need to talk about is how to calculate percent association
05:47	also referred to as percent organization . Beset organization is
05:53	equal to X divided by the acid concentration . That
05:59	is the original . Ask the concentration that was dissolved
06:02	in a solution times 100 X represents the amount of
06:08	acid that is associated . And when dealing with assets
06:13	for the most part X is usually the HBO plus
06:18	concentration . So you can find it from doing an
06:24	ice table . Or if you know the ph you
06:25	can find it that way as well when dealing with
06:29	weak basis X . Is usually the hydroxide concentration .
06:36	So those are the formulas that we're gonna be using
06:38	throughout this video to calculate different things . So let's
06:42	go ahead and work on some practice problems . Let's
06:45	put these equations to good use . So what is
06:48	the ph of a .75 molar acetic acid solution and
06:52	were given to K . So first let's write the
06:55	dissociation reaction of acidic acid when it's combined with water
07:01	. So because it's a weak acid , it's going
07:04	to be a reversible reaction . So we need to
07:06	put two arrows Whenever you add a weak acid in
07:10	water is going to generate the hydro Nehemiah on H30
07:12	plus and also the conjugate base , which in this
07:16	case is a state . So what we're gonna do
07:19	is we're gonna make an ice table . The initial
07:24	amount of acetic acid is .75 and this is gonna
07:29	be units of polarity . And for the products we're
07:32	going to start with zero . Now the reaction has
07:35	no choice but to shift to the right , he
07:37	can't shift the left . So the products will increase
07:41	by X . And the reaction will decrease by X
07:45	. We don't have to worry about water because what
07:48	is a pure liquid ? Mhm . Everything else is
07:51	an acquis face . So as you recall , you
07:55	can't have any liquids or solids in the equal in
07:58	expression when you're dealing with K . Or K .
08:01	B . Or any type of K . So if
08:06	we are the first two rows , it's gonna be
08:08	X and 0.75 minus X . Now , in order
08:14	to calculate the ph of the solution , we need
08:17	to calculate the concentration of HBO plus , which we
08:20	could see that it's equal to X . So we've
08:23	got to find the value of X . So we
08:25	can write the expression for K . The acid dissociation
08:29	constant . It's equal to the concentration of the products
08:33	which can be found on the right side . That's
08:34	H . 30 Plus and the way I am divided
08:39	by the reactant and we can't use water since it's
08:43	a liquid . Now , K . A . Is
08:47	equal to 1.8 Times 10 to the -5 A trio
08:53	plus and asking they're both equal to X acidic acid
08:58	At Equilibrium . It's .75 -X . Now because tends
09:03	to -5 is pretty small . We could ignore this
09:07	X . So therefore we can say that K .
09:10	A . Or this number is equal to X squared
09:15	Divided by .75 . Now let's get rid of a
09:22	few things . We really don't need this anymore .
09:25	But keep in mind that HBO plus is equal to
09:28	X . Now in order to find the value of
09:31	X , we need to cross multiply . So one
09:33	times x squared is X squared . Yeah . And
09:39	1.8 Times 10 to -5 Multiply by .75 . That's
09:47	equal to 1.35 Times 10 : -5 . Now let's
09:53	take the square root of both sides . So the
09:55	square root of that number Is going to be 3.674
10:01	Times 10 : -3 . That's the concentration of a
10:04	trio plus . Yeah , now that we have the
10:11	concentration of HBO plus we can calculate the ph of
10:14	the solution . The ph is equal to negative log
10:19	of the hydro me . Um I I concentration .
10:22	So it's a negative log of 3.674 Times 10 to
10:26	the ministry . So you should get about 2.43 .
10:39	So that's the answer for this problem . That's the
10:41	ph of the solution . So if you want to
10:46	quickly get the answer whenever you have a weak acid
10:49	and if you're given the concentration and the K .
10:51	Value and you want to find the ph let's simply
10:55	review the steps first . Use this equation . It's
10:58	gonna be K . A . Which is equal to
11:01	X squared divided by the acid concentration minus X .
11:06	Now , if K is very small , If it's
11:09	like 10 to negative five or less , typically you
11:12	can ignore this X value . If K is not
11:15	very small , then you may have to use the
11:17	quadratic former . So once you get the equation in
11:22	this form A X squared plus bx plus C .
11:26	If you can't ignore this , X . Here's the
11:28	quadratic form that you need , it's negative B plus
11:31	or minus the square root of B squared minus four
11:34	ac divided by uh two way . So keep this
11:41	in mind . If you're going to use the quadratic
11:42	formula , you need to rearrange the equation until it's
11:46	in this form . So on one side of the
11:48	equation you have to have a zero . Now when
11:51	you're dealing with X , I mean when you're dealing
11:53	with K A X is equal to H 20 plus
11:59	. So once you find the value of X ,
12:00	you can calculate the ph using this equation number two
12:09	. What is the ph of a .25 moller ammonia
12:13	solution ? And were given the KB , the based
12:17	association constant . So since we have K B .
12:23	Basically we have a weak base in the solution and
12:26	a weak base will generate hydrogen ions . So we
12:30	need to use KB to find X . Which X
12:33	. Is going to represent the hydroxide ion concentration .
12:35	Once we have that we could find the ph of
12:38	the solution and then the ph so first let's try
12:42	to reaction . When you put a weak base with
12:47	water , it's going to be a reversible reaction and
12:51	you're going to get the conjugate acid in this case
12:53	an H four plus and hydroxide . So let's make
12:57	an ice table just as we did before the initial
13:02	concentration for ammonia Is .25 . The products will begin
13:07	at zero and the reaction will initially shift towards the
13:11	right . So adding the first two rows , we're
13:15	gonna have 20.25 -1 . And X . And X
13:21	. So KB like any other equilibrium constant . K
13:25	. It's going to be the concentration of the products
13:29	divided by the reactant and don't include water . Now
13:36	KB We know it's 1.8 Times 10 to -5 and
13:42	H four plus and o h minus . They're both
13:44	equal to X . X times X is X squared
13:47	And any street is point to five for -X .
13:51	But since KB is small , We could ignore the
13:55	X . That's next to the .25 . So we're
13:58	gonna get this equation . So let's cross multiply .
14:02	So we're going to have X squared is equal to
14:07	.25 times 1.8 times 10 to -5 And that's 4.5
14:14	Times 10 : -6 . Now , let's take the
14:19	square root of both sides . The Square Root of
14:25	4.5 times 10 : -6 . That's 2.1 Times 10
14:32	. Race to the monastery now that we have the
14:36	value of X , We have the concentration of hydroxide
14:42	, which means we could find the ph of the
14:44	solution . Mhm . Mhm . So the P O
14:52	H is negative log Of the hydroxide concentration or of
14:58	2.12 Times 10 to the monastery . So this is
15:09	equal to 2.67 . And whenever you have the ph
15:13	of a solution , You can easily find the ph
15:18	it's going to be 14 minus the peel age .
15:23	Yeah , So the ph is about 11.33 . And
15:33	so that's the answer for this problem . So this
15:37	is what you need to do if we need to
15:39	find a ph of a solution that contains a weak
15:42	base number three , what is the ph of a
15:49	point for Mueller ammonium chloride solution ? So in this
15:55	problem , we want to find the ph of a
15:57	salt solution . Ammonium chloride is an ionic compound and
16:02	were given the KB of any stream . So what
16:05	should we do in this problem ? We need to
16:07	realize that NH four plus is a weak acid because
16:11	NH treat the contract base is a weak base .
16:15	So let's write the reaction when it's added to water
16:22	. So because it's a weak acid , it's going
16:24	to donate a proton to water and so we're going
16:28	to get the H 30 plus ion and also an
16:31	industry . So , NH four plus is the acid
16:37	. It's the proton , donor water is the basis
16:41	. The proton , except er a straight A pluses
16:43	the conjugate acid . NH three is the conjugate base
16:49	. Now let's make a nice table . So initially
16:56	This is gonna be .4 And as usual this is
16:59	00 . The reaction is going to shift to the
17:02	right , increasing the products by x , decreasing reactant
17:07	by X as well . Now in this problem ,
17:16	should we use K A or K B . What
17:21	do we have in the solution ? Do we have
17:24	a weak acid or a weak base ? We don't
17:28	have any street in the solution . Initially We have
17:32	NH four plus and that's a weak acid , which
17:35	means we need to use K A . K is
17:39	always associated with HBO Plus . If you have something
17:43	that produces hydroxide ads , that's going to be associated
17:47	with KB , which is a base Or which is
17:50	four basis . So we have a weak acid in
17:53	a solution . We need to use K . A
17:54	. But were given KB . So how can we
17:57	find K A from KB . K A times KB
18:05	is equal to K . W , which is the
18:08	auto ionization constant of water And that's equal to one
18:12	times 10 to -14 . So therefore K A .
18:17	Is basically equal to that number divided by K .
18:20	B . And we have KB . It's 1.8 Times
18:29	10 to the -5 . So let's go ahead and
18:33	divide these two numbers . So this is about 5.56
18:42	Times 10 to -10 . So now that we have
18:48	K , we can write the equilibrium expression which is
18:52	going to be products of reactions . So we know
18:54	it's going to be X squared Divided by .4 -X
19:12	. Now K is very , very small . So
19:16	therefore we can neglect this X value . It's going
19:21	to be insignificant . Now let's cross multiply . So
19:28	we're gonna have X squared which is equal to 0.4
19:35	Times 5.56 Times 10 : -10 . As usual ,
19:45	we're going to take the square view of both sides
19:48	and calculate the value of X . So x is
19:54	1.49 Times 10 : -5 . Now , because we're
20:00	dealing with K A , X represents the H .
20:03	30 . Plus concentration . Which means we can now
20:07	find ph directly . So the ph is gonna be
20:11	negative log of the hydrogen ion concentration which is equal
20:15	to X . So therefore the ph is about 4.83
20:28	. That's the answer , number four . What is
20:32	the ph of a 1.5 molar sodium fluoride solution ?
20:37	So we're given the kaing of Hydrofluoric acid . Let's
20:40	go ahead and find a ph So we don't have
20:43	to worry about sodium . It's a spectator ion so
20:45	let's ignore it by the way . Feel free to
20:48	pause this video and work on this problem yourself .
20:51	See if you can get the answer . So fluoride
20:55	is the ion of entrance because it's associated with HF
21:00	. If HF is a weak acid fluoride is basic
21:04	enough to change the ph of the solution . So
21:07	florida is a weak base is going to react with
21:09	water and it does so reversible . E generating the
21:13	conjugate acid HF . And because florida is a weak
21:17	base is going to generate a small amount of hydroxide
21:20	ions . Does changing the ph of the solution .
21:25	So we're gonna make our ice table as usual And
21:28	the initial amount of Florida is 1.5 . The products
21:32	are zero and they're going to increase by X .
21:35	Fluoride will decrease by X . And as you can
21:39	see , the ice table for the most part is
21:40	pretty much the same . Now , we need to
21:45	find KB . Because fluoride is a weak base ,
21:49	we have sodium fluoride in the solution , not HF
21:52	, we're going to have a small amount of HF
21:53	. But for the most part we have a large
21:56	amount of Sudan florida . So we need to use
21:58	KB . Yeah , K B is going to equal
22:03	the products divided by the reaction . So let's go
22:12	ahead and find KB . K B is going to
22:14	be equal to K . W which is one times
22:18	10 to the negative 14 divided by K . A
22:22	. Which is this number ? Yeah , So KB
22:36	is 1.39 Times 10 to the -11 . HF and
22:42	hydroxide . They're both equal to X . So X
22:45	times X is X squared and f minus is going
22:51	to be 1.5 minus X . But we could ignore
22:53	that X . Since KB is very small , let's
23:01	cross multiplying . So this is going to be X
23:04	squared , which is equal to let's get rid of
23:09	this X . Yes , 1.39 times 10 to -11
23:15	Times 1.5 And that's 2.0 85 Times 10 : -11
23:23	. Now let's take the square root of both sides
23:27	. So X is equal to 4.56 , 6 Times
23:37	10 to the -6 . Now keep in mind that
23:40	X is equal to the hydroxide concentration . So we
23:45	could find the P . O . H . And
23:47	then we could find the ph of the solution .
23:53	So the P O . H . Is equal to
23:57	negative log of the hydroxide concentration , which is this
24:02	number . So notice that you see a six here
24:08	, the P . H . I mean , excuse
24:10	me , the ph is going to be somewhere between
24:11	five and six . So it's actually 5.34 . Now
24:24	that we have the P O . H , we
24:27	can calculate the ph which is 14 minus the P
24:30	. O . H . And it's going to be
24:36	about 8.66 . So that's the answer for this problem
24:40	. Now , you know how to calculate the ph
24:42	of the solution . If you're given a basic self
24:47	solution now , let's focus on percent ionization problems .
24:53	I'm actually mixing in some old videos that I had
24:56	in the past with this video . So what is
24:59	the formula in order to find the percent ionization ?
25:03	It's equal to X divided by the acid concentration .
25:08	Now , if you're dealing with a weak acid ,
25:10	X represents the amount of H . Plus that is
25:14	in equilibrium in a solution . And let's not forget
25:17	to multiply this by 100% . Now when you're dealing
25:21	with basis X represents the amount of hydroxide and instead
25:26	of H . A . Is going to be the
25:28	concentration of the base , that is the initial concentration
25:34	. So now let's go ahead and write a reaction
25:40	. So once a weak acid is in a solution
25:43	, let's say when it's dissolved in water , it's
25:45	going to be a nice and because it's weak ,
25:48	the reaction is going to be reversible . So we
25:50	need to put two arrows assets will always generate H
25:55	30 plus the hydrogen ion . We're going to get
25:57	the contract base fluoride . So now let's make a
26:01	nice table . Mhm . The initial amount of Hydrofluoric
26:06	acid is .75 and initially we don't have any hydrogen
26:13	ions or fluoride ions . So therefore the reaction has
26:16	no choice but to shift to the right , increasing
26:19	the products by X , decrease in the reactant by
26:23	X . So if we add the first two roles
26:27	, this is what we're going to get now .
26:30	K . A . The acid dissociation constant , like
26:34	any equilibrium constant is equal to the ratio of the
26:37	products over the reactant . Mhm . So okay Is
26:49	equal to 7.2 Times 10 to -4 H 30 plus
26:55	an f minus . They're both equal to X .
26:57	X times X is x squared and a chive Is
27:04	.75 -X . So basically we're using the values that
27:08	equilibrium when dealing with K . Now because today is
27:14	fairly small . We can ignore this X . X
27:19	is going to be relatively small compared to .75 .
27:26	So what we can do at this point is cross
27:28	multiplying one times X squared is x squared And .75
27:35	times K . And that's equal to 5.4 Times 10
27:43	to the -4 . Now let's take the square root
27:46	of both sides . So X is equal to 0.2
27:57	32 So .75 -1232 , That's about .07 . I
28:07	mean .7268 doesn't change that much , it changes a
28:11	little but we're gonna go with this estimation . So
28:16	now that we have the value of X we can
28:19	calculate the percent dissociation . Now keep in mind this
28:22	is not the exact answer but it's an estimation of
28:26	the answer to get the exact answer . You can
28:28	use the quadratic formula but we're just going to go
28:31	with the estimation . Mhm . So the percent organization
28:41	, It's going to be equal to X , which
28:43	is .0232 divided by the initial concentration of HF ,
28:50	Which is .75 . Yeah , so it's about 3.1%
00:0-1	.

Summarizer

DESCRIPTION:

This chemistry video explains how to calculate the pH of a weak acid and a weak base. It explains how to calculate the percent ionization of a weak acid using the acid dissociation constant Ka.

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PH of Weak Acids and Bases - Percent Ionization - Ka & Kb is a free educational video by The Organic Chemistry Tutor.

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