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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M 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.
a) Find the pH of 0.25 M CH3COOH
Strong Acidor
Weak Acid?
We see by its position on the acid table (click) that ethanoic acid is a weak acid
a) Find the pH of 0.25 M CH3COOH
Strong Acidor
Weak Acid?Weak Acid
When we’re doing calculations involving weak acids, we must use an ICE table.
a) Find the pH of 0.25 M CH3COOH
Strong Acidor
Weak Acid?Weak Acid
ICE Table
We start by writing the equilibrium equation for ionization of ethanoic acid here.
a) Find the pH of 0.25 M CH3COOH
[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.
a) Find the pH of 0.25 M CH3COOH
[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.
a) Find the pH of 0.25 M CH3COOH
[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.
a) Find the pH of 0.25 M CH3COOH
[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.
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1[C]
1
[E]
1
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
No H3O+ or CH3COOminus was added, so we can consider their concentrations to be 0 before ionization.
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
1
[E]
1
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
Now we’ll look at the (click) changes in concentration as the ionization occurs.
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
1
[E]
1
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
Because there were no products initially, the equilibrium will (click) shift to the right during the ionization.
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
1
[E]
1
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
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.
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
1 +? +?
[E]
1
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
Shift to the Right
And the concentration of CH3COOH will decrease, so we’ll write a minus sign here.
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +? +?
[E]
1
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
Shift to the Right
We’re not given any equilibrium concentrations, so we don’t know how much these will increase
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +? +?
[E]
1
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
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.
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
1
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
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.
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
1
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
Shift to the Right
Now, we’ll look at the concentrations of everything (click) at equilibrium.
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
1
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
The hydronium and ethanoate ions will both be 0
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
0.25–x
1 x x
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
Plus x
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
0.25–x
1 x x
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
Which is equal to
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
0.25–x
1 x x
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
0 + x 0 + x
x.
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
0.25–x
1 x x
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
0 + x 0 + x
The concentration of CH3COOH started out as 0.25
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
0.25–x
1 x x
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
and went down by x,
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
0.25–x
1 x x
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
so its equilibrium concentration will be…
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
0.25–x
1 x x
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
0.25 minus x
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
0.25–x
1 x x
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
The question now is, How do we find x?
a) Find the pH of 0.25 M CH3COOH
[I] 0.25 1 0 0[C]
–x 1 +x +x
[E]
0.25–x
1 x x
3 (aq) 2 ( ) 3 (aq) 3 (aq)CH COOH H O H O CH COOl
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
a) Find the pH of 0.25 M CH3COOH
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
a) Find the pH of 0.25 M CH3COOH
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
a) Find the pH of 0.25 M CH3COOH
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
a) Find the pH of 0.25 M CH3COOH
The concentration of ethanoic acid at equilibrium is 0.25 - x
a) Find the pH of 0.25 M 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
So we’ll substitute 0.25 –x in for the concentration of CH3COOH
a) Find the pH of 0.25 M 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
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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
a) Find the pH of 0.25 M CH3COOH
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
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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
a) Find the pH of 0.25 M CH3COOH
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.
b) Find the % ionization of 0.25 M CH3COOH
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.
b) Find the % ionization of 0.25 M CH3COOH
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
b) Find the % ionization of 0.25 M CH3COOH
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.
b) Find the % ionization of 0.25 M CH3COOH
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.
b) Find the % ionization of 0.25 M CH3COOH
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
b) Find the % ionization of 0.25 M CH3COOH
And 0.25 M in for the initial concentration of ethanoic acid.
3
3
aci
0.25
H O% Ionization 100%
2.12 10 M10
M0%
0.85
d
%
initial
b) Find the % ionization of 0.25 M CH3COOH
We can cancel the unit Molarity and we multiply by 100%
3
3
H O% Ionization 100%
aci
100
d
2.12 10 M
0.25 M0. 5%
%
8
initial
b) Find the % ionization of 0.25 M CH3COOH
And we get 0.85%
3
3
H O% Ionization 100%
acid
2.12 10 M100%
0.0.
M85%
25
initial
b) Find the % ionization of 0.25 M CH3COOH
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
H O% Ionization 100%
acid
2.12 10 M100%
0.25% Ioniza
Mtion 0.85%
initial
b) Find the % ionization of 0.25 M CH3COOH
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.
a) Find the pH of 0.25 M CH3COOH
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
Valid if the Percent Ionization is 5% or
less
Percent Ionization= 0.85%
a) Find the pH of 0.25 M CH3COOH
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
Valid if the Percent Ionization is 5% or
less
Percent Ionization= 0.85%
a) Find the pH of 0.25 M CH3COOH
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