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

## 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
03:59 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 .
05:29 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
07:35 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
24:40 . 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
28:28 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.

#### OVERVIEW:

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

This page not only allows students and teachers view PH of Weak Acids and Bases - Percent Ionization - Ka & Kb videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.

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