A 1.00 L solution is prepared by placing 0.10 moles of a weak acid (HA) and
0.050 moles of its conjugate base NaA. Determine the pH of the solution.
The Ka of the HA is 1.4 x 10-5 .
HA + H2O ↔ H3O+ + A-
Chemistry:
H O A 


Ka
I.
C.
E.

3
HA
Ka 
x0.050 x
0.10  x
5
1.4 x10
0.10
-x
0.10-x
0
+x
x
0.050
+x
0.050+x
Try dropping the +x and the –x terms (5 % rule).

x 0.050

0.10
X = [H3O+] = 2.8 x 10-5
pH = -log[x] = 4.55
A 1.00 L solution is prepared by placing 0.10 moles of a weak base (B) and
0.050 moles of its conjugate acid BHCl. Determine the pH of the solution.
The Kb of the B is 1.4 x 10-5 .
B + H2O ↔ OH- + BH+
Chemistry:
BH OH 


Kb
I.
C.
E.

B
Kb 
x 0.050 x 
0.10  x 
5
1.4 x10
0.10
-x
0.10-x
0
+x
x
0.050
+x
0.050+x
Try dropping the +x and the –x terms (5 % rule).

x 0.050

0.10
X = [OH-] = 2.8 x 10-5
pOH = -log[x] = 4.55 and pH = 9.45
A solution is 0.095 M in ascorbic acid and 0.055 M in sodium hydrogen
ascorbate. Determine the pH of the solution. Ka1 of the H2C6H6O6 is 6.8 x
10-5, Ka2 = 2.8 x10-12 .
Chemistry:
H O HC H O 

Ka
3
6
6
H 2C6 H 6O6 
Ka 
x 0.055 x 
0.095 x 
6.8 x105 
I.
C.
E.


6
H2C6H6O6 + H2O ↔ H3O+ + HC6H6O60.095
-x
0.095-x
0
+x
x
0.055
+x
0.055+x
Try dropping the +x and the –x terms (5 % rule).
x 0.055
0.095
X = [H3O+] = 1.17 x 10-4
pH = -log[x] = 3.94
A weak acid, HA (Ka = 1.6 x 10-5), is being titrated with a strong base, KOH.
Determine the pH of the solution resulting from the mixing of 30.00 mL of
0.10 M HA with 12.00 mL of 0.10 M KOH.
First we must determine
what is present after the
initial reaction between the
acid and the base.
first reaction: HA(aq) + KOH(aq) → H2O(l) + KA(aq) + XS-HA??
3
Find starting
moles of HA
and moles of
KOH.
Now, Let’s examine
the resulting
equilibrium.
Starting moles HA: 30.00 x10 L 0.10 mole
L
3
Starting moles KOH:
3
 3.0x10 mol HA

1molOH
3
12 .00 x10 L 0.10 mol KOH
 1.2x10 mol OH
L
1mol KOH
Therefore, we have 1.8 x 10-3 moles of HA left and
1.2 x 10-3 moles of its conjugate base (A-) formed.
We can now change these amounts to Molarity and
use them in our equilibrium expression.
A weak acid, HA (Ka = 1.6 x 10-5), is being titrated with a strong base, KOH.
Determine the pH of the solution resulting from the mixing of 30.00 mL of
0.10 M HA with 12.00 mL of 0.10 M KOH.
first reaction: HA(aq) + KOH(aq) → H2O(l) + KA(aq) + XS-HA??
3
Find starting
moles of HA
and moles of
KOH.
Now, Let’s examine
the resulting
equilibrium.
Starting moles HA: 30.00 x10 L 0.10 mole
L
3
Starting moles KOH:
3
 3.0x10 mol HA

1molOH
3
12 .00 x10 L 0.10 mol KOH
 1.2x10 mol OH
L
1mol KOH
Therefore, we have 1.8 x 10-3 moles of HA left and
1.2 x 10-3 moles of its conjugate base (A-) formed.
We can now change these amounts to Molarity and
use them in our equilibrium expression.
HA + H2O ↔ H3O+ +
A0
0.0286
I. 0.0429
+x
+x
C. -x
E. 0.0429-x
x
0.0286+x
A weak acid, HA (Ka = 1.6 x 10-5), is being titrated with a strong base, KOH.
Determine the pH of the solution resulting from the mixing of 30.00 mL of
0.10 M HA with 12.00 mL of 0.10 M KOH.
Now, Let’s examine
the resulting
equilibrium.
Therefore, we have 1.8 x 10-3 moles of HA left and
1.2 x 10-3 moles of its conjugate base (A-) formed.
We can now change these amounts to Molarity and
use them in our equilibrium expression.
H O    A _ 
 3   
Ka 
HA 
5
1.6x10 
x0.0286 x
0.0429 x
x = [H3O+] = 2.4 x 10-5
pH = -log [2.4 x 10-5] = 4.62
HA + H2O ↔ H3O+ +
A0
0.0286
I. 0.0429
+x
+x
C. -x
E. 0.0429-x
x
0.0286+x
Try dropping the -x and +x terms. If the
value of x comes out to less than 5% of
0.0286, dropping the term is justified.
A weak acid, HA (Ka = 1.6 x 10-5), is being titrated with a strong base, KOH.
Now try: Determine the pH of the solution resulting from the mixing of
30.00 mL of 0.10 M HA with 22.00 mL of 0.10 M KOH.
First we must determine
what is present after the
initial reaction between the
acid and the base.
first reaction: HA(aq) + KOH(aq) → H2O(l) + KA(aq) + XS-HA??
3
Find starting
moles of HA
and moles of
KOH.
Now, Let’s examine
the resulting
equilibrium.
Starting moles HA: 30.00 x10 L 0.10 mole
L
3
Starting moles KOH:
3
 3.0x10 mol HA

1molOH
3
22 .00 x10 L 0.10 mol KOH
 2.2x10 mol OH
L
1mol KOH
Therefore, we have 0.8 x 10-3 moles of HA left and
2.2 x 10-3 moles of its conjugate base (A-) formed.
We can now change these amounts to Molarity and
use them in our equilibrium expression.
A weak acid, HA (Ka = 1.6 x 10-5), is being titrated with a strong base, KOH.
Determine the pH of the solution resulting from the mixing of 30.00 mL of
0.10 M HA with 12.00 mL of 0.10 M KOH.
first reaction: HA(aq) + KOH(aq) → H2O(l) + KA(aq) + XS-HA??
3
Find starting
moles of HA
and moles of
KOH.
Now, Let’s examine
the resulting
equilibrium.
Starting moles HA: 30.00 x10 L 0.10 mole
L
3
Starting moles KOH:
3
 3.0x10 mol HA

1molOH
3
22 .00 x10 L 0.10 mol KOH
 2.2x10 mol OH
L
1mol KOH
Therefore, we have 0.8 x 10-3 moles of HA left and
2.2 x 10-3 moles of its conjugate base (A-) formed.
We can now change these amounts to Molarity and
use them in our equilibrium expression.
HA + H2O ↔ H3O+ +
A0
0.0423
I. 0.0154
+x
+x
C. -x
E. 0.0154-x
x
0.0423+x
A weak acid, HA (Ka = 1.6 x 10-5), is being titrated with a strong base, KOH.
Determine the pH of the solution resulting from the mixing of 30.00 mL of
0.10 M HA with 12.00 mL of 0.10 M KOH.
Now, Let’s examine
the resulting
equilibrium.
Therefore, we have 1.8 x 10-3 moles of HA left and
1.2 x 10-3 moles of its conjugate base (A-) formed.
We can now change these amounts to Molarity and
use them in our equilibrium expression.
H O    A _ 
 3   
Ka 
HA 
5
1.6x10 
x0.0423 x
0.0154 x
x = [H3O+] = 2.4 x 10-5
pH = -log [5.825 x 10-5] = 5.23
HA + H2O ↔ H3O+ +
A0
0.0423
I. 0.0154
+x
+x
C. -x
E. 0.0154-x
x
0.0423+x
Try dropping the -x and +x terms. If the
value of x comes out to less than 5% of
0.0286, dropping the term is justified.