here, we’ll show you how to calculate the ph and % ionization of a weak acid with a given...

Here, we’ll show you how to calculate the pH and % ionization of a weak acid with a given concentration and a known Ka value.

Ka to pH and Percent

Ionization

For the (a) part of this question, we’re asked to find the pH of a 0.25 M solution of ethanoic acid, CH3COOH.

a) Find the pH of 0.25 M CH3COOH

The “b” part of the question asks us to find the % ionization in 0.25 M ethanoic acid.


b) Find the % ionization of 0.25 M CH3COOH

We’ll start with the “a” part. We’re asked to find the pH of a 0.25 M solution of ethanoic acid, CH3COOH.


Whenever we’re asked to do a calculation with an acid, the first thing we always have to do is identify the acid as strong or weak.


Strong Acidor

Weak Acid?

We see by its position on the acid table (click) that ethanoic acid is a weak acid


Strong Acidor

Weak Acid?Weak Acid

When we’re doing calculations involving weak acids, we must use an ICE table.


Strong Acidor

Weak Acid?Weak Acid

ICE Table

We set up an ice table like this


[I][C][E]

We start by writing the equilibrium equation for ionization of ethanoic acid here.


[I][C][E]

3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl Equilibrium

Equation

Next, we draw borders in so that columns line up nicely with the substances in the equation.


[I][C][E]

3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl Equilibrium

Equation

Water is a liquid in the equilibrium equation so we can ignore the column below water. We’ll colour it blue here.


[I][C][E]

3 (aq) 3 (aq) 3 (aq2 ( ))CH COOH H O CH COOH O l

We’ll start out with the initial concentration row.


[I] 1[C]

1

[E]

1

3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl

The initial concentration of the ethanoic acid is 0.25 M, so will add 0.25 to this cell.


[I] 0.25 1[C]

1

[E]

1


No H3O+ or CH3COOminus was added, so we can consider their concentrations to be 0 before ionization.


[I] 0.25 1 0 0[C]

1

[E]

1


Now we’ll look at the (click) changes in concentration as the ionization occurs.


[I] 0.25 1 0 0[C]

1

[E]

1


Because there were no products initially, the equilibrium will (click) shift to the right during the ionization.


[I] 0.25 1 0 0[C]

1

[E]

1


Shift to the Right

As a shift to the right occurs, the concentrations of hydronium and ethanoate ions will both increase so we’ll write + signs here.


[I] 0.25 1 0 0[C]

1 +? +?

[E]

1


Shift to the Right

And the concentration of CH3COOH will decrease, so we’ll write a minus sign here.


[I] 0.25 1 0 0[C]

–x 1 +? +?

[E]

1


Shift to the Right

We’re not given any equilibrium concentrations, so we don’t know how much these will increase


[I] 0.25 1 0 0[C]

–x 1 +? +?

[E]

1


Shift to the Right

So we’ll write + x for both of them, as they both have a coefficient of 1 in the equilibrium equation.


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

1


Shift to the Right

Because the coefficient on CH3COOH is also 1, we can state that it will go down by x, so we write minus x here.


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

1


Shift to the Right

Now, we’ll look at the concentrations of everything (click) at equilibrium.


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

1


The hydronium and ethanoate ions will both be 0


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

0.25–x

1 x x


Plus x


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

0.25–x

1 x x


Which is equal to


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

0.25–x

1 x x


0 + x 0 + x

x.


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

0.25–x

1 x x


0 + x 0 + x

The concentration of CH3COOH started out as 0.25


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

0.25–x

1 x x


and went down by x,


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

0.25–x

1 x x


so its equilibrium concentration will be…


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

0.25–x

1 x x


0.25 minus x


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

0.25–x

1 x x


The question now is, How do we find x?


[I] 0.25 1 0 0[C]

–x 1 +x +x

[E]

0.25–x

1 x x


How do we find x ?

We start solving for x by writing the Ka expression for ethanoic acid

3 3a

3

2

a

2

a

H O CH COO

CH CO

0.25

0.25

OH

x

K

Kx

xK

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K


Using the equilibrium equation, we can see that the Ka expression is equal to the concentration of hydronium times concentration of ethanoate over the concentration of ethanoic acid.

3 3a

3

2

a

2

a

H O CH COO

CH CO

0.25

0.25

OH

x

K

Kx

xK

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K


The equilibrium concentrations of hydronium and ethanoate are both equal to x,

a3

3

2

3

a

2

a

CH COOH

0.25

0

H O CH COO

.25

K

Kx

x

xK

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K


so for their product in the Ka expression, we can substitute x times x or x squared.

a3

3

2

3

a

2

a

CH COOH

0.25

0

H O CH COO

.25

K

Kx

x

xK

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K


The concentration of ethanoic acid at equilibrium is 0.25 - x


3

3 3a

2

a

2

a

CH C

0

H O CH COO

0.25

.25

OOH

x

K

xK

xK

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K

So we’ll substitute 0.25 –x in for the concentration of CH3COOH


3

3 3a

2

a

2

a

CH C

0

H O CH COO

0.25

.25

OOH

x

K

xK

xK

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K

The degree of ionization for ethanoic acid is very low. So we make the assumption that x is insignificant compared to 0.25. This can be written as 0.25 – x is almost equal to 0.25


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

xK

x

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K

Assume0.25–x 0.25

We can check this assumption later when we determine the % ionization. In general the assumption is valid if the percent ionization is 5% or less.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

xK

x

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K

Assume0.25–x 0.25

Valid if the Percent Ionization

is5% or Less

Using this assumption will help us avoid having to use a quadratic equation. In Chemistry 12, this assumption is generally used here. When you use it, you Must Always state it.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

xK

x

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K

Assume0.25–x 0.25

Always STATE this assumption if

you’re using it!

so taking out the x on the bottom, we can state that Ka is approximately equal to x2 over 0.25


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

xK

x

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K

Rearranging this equation to solve for x2 gives us x2 = 0.25 times the Ka


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

xK

x

3 a

53

63

3

3

a

3

2

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log

0.2

2.12 10

p

5

H 2.67

x K

x K

2a

3

53

63

33

3

a

0.25

H O

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

0.2

pH 2.67

5K

x K

x

Taking the square root of both sides gives us x is equal to the square root of 0.25 times the ka.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

3

2a

a

53

63

33

3

0.25

0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH

H O

2.67

Kx

x K

At this point, we’ll remind ourselves that x is equal to the equilibrium concentration of hydronium.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

5

2a

63

33

3

a3

3

0.25

H O 0.25

H O 0.25

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

1.8 10

x K

x K

Looking on the acid table, we see that the Ka for ethanoic acid is 1.8 × 10 -5, so we substitute that for Ka in the equation.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

6

2a

3 a

3

3

33

3

5

0.25

H O 0.25

H O

H O

H O 2.12 10 M

pH

0

log 2.12 10

pH 2.

4.5

.25 1.8 10

10

67

x K

x K

0.25 times 1.8 × 10-5 is equal to 4.5 × 10-6, so we’ll substitute that in here, under the square root sign.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

2a

3 a

53

3

33

3

6

0.25

H O 0.25

H O 0.25 1.8 10

H O

2H O

pH log 2.12 10

pH 2.

4.5

67

. 10 M

10

12

x K

x K

The square root of 4.5 x 10-6 is 2.12 x 10-3. Because we now have a value for the concentration of hydronium, we’ll include the unit Molarity.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 10

H O 4.5 10

H O 2.12 10 M

pH

1.

log 2.12 10

pH 2.67

8

x K

x K

Both the value of Ka and the given concentration have 2 significant figures, which limits our final answer to 2 significant figures.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 10 M

pH log 2.1

2

2 10

pH 2.6

.1

7

2

x K

x K

We’ve written the concentration of hydronium with 3 significant figures here. We’ll use this and round to 2 significant figures at the end.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

2a

3 a

53

6

3

3

3

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5

pH 2.1

10

H O

log

pH 2.67

2 10

2.12 10 M

x K

x K

We’re asked for the pH. The pH is the negative log of the hydronium ion concentration, which is the negative log of 2.12 × 10-3.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

2a

3 a

53

63

3

33

0.25

H O 0.25

H O 0.25 1.8 10

H

pH 2.

O 4.5 10

H O 2.12 10

6

M

p log 2.

73

1H 1

6

2 0

6

x K

x K

And that comes out to 2.67366.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

2a

3 a

53

6

3

3

3

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5

log 2.1

10

2

H O 2.12 10 M

pH 1

pH . 7

0

2 6

x K

x K

But we round this to 2.67 in order to give us 2 decimal places in this pH value, or 2 significant figures in our final answer.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.1

pH 2.6

2 1

7

0

x K

x K

So now we have the final answer for Part “a”. The pH of 0.25M ethanoic acid is equal to 2.67


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K

Part “b” of this question asks us to find the percent ionization of 0.25 M ethanoic acid.


2a

3 a

53

6

3

3

3

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

pH log 2.1

H O 2.12 10 M

2 10

pH 2.67

x K

x K

At one point in our calculations, we had determined that the hydronium ion concentration in this solution is 2.12 x 10-3 molar.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

x

xK

2a

3 a

53

6

3

3

3

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

pH log 2.1

H O 2.12 10 M

2 10

pH 2.67

x K

x K

We can use this to help us find the percent ionization


2a

3 a

53

6

3

3

3

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

pH log 2.1

H O 2.12 10 M

2 10

pH 2.67

x K

x K

We’ll make a note of it up here.


33H O 2.12 10 M

The formula for percent ionization is the hydronium ion concentration divided by the initial concentration of the acid times 100 percent.


33H O 2.12 10 M

3

3

H O% Ionization 100%

aci

100%0.25 M

0.

d

2.1

8 %

2

5

10 M

initial

We’ll substitute 2.12 × 10-3 M in for the concentration of hydronium

3

3% Ionization 100%

acid

H O

100%0.25 M

0.8

2.12 1

5%

0 M

initial

33 2.12O 0 MH 1


And 0.25 M in for the initial concentration of ethanoic acid.

3

3

aci

0.25


2.12 10 M10

M0%

0.85

d

%

initial


We can cancel the unit Molarity and we multiply by 100%

3

3


aci

100

d

2.12 10 M

0.25 M0. 5%

%

8

initial


And we get 0.85%

3

3


acid

2.12 10 M100%

0.0.

M85%

25

initial


So now we’ve answered question “b”. The percent ionization of 0.25 M ethanoic acid is 0.85%. Notice it’s quite small, less than 1% of the ethanoic acid molecules have ionized.

3

3


acid

2.12 10 M100%

0.25% Ioniza

Mtion 0.85%

initial


remember during the calculations for pH, we assumed that x was insignificant compared to 0.25 M. We said this was valid if the % ionization is 5% or less.


3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

xK

x

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K

Assume0.25–x 0.25

Valid if the Percent Ionization is 5% or

less

We have just determined that the % ionization is 0.85 %, which is much less than 5%.

3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

xK

x

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K

Assume0.25–x 0.25


less

Percent Ionization= 0.85%


So this verifies that our assumption was definitely valid!

3 3a

3

2

a

2

a

H O CH COO

CH COOH

0.25

0.25

K

xK

xK

x

2a

3 a

53

63

33

3

0.25

H O 0.25

H O 0.25 1.8 10

H O 4.5 10

H O 2.12 10 M

pH log 2.12 10

pH 2.67

x K

x K

Assume0.25–x 0.25


less

Percent Ionization= 0.85%


here, we’ll show you how to calculate the ph and % ionization of a weak acid with a given...

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