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Weak Acid Strong Base Titration

 

The titration of 50.0mL of 0.100M HC2H3O2 (Ka=1.8 x 10-5) with 0.100M NaOH is carried out in a chemistry laboratory. Calculate the pH of the solution after these volumes of the titrant have been added. Use these results to plot the titration curve.

   pH

Vol. NaOH

 

0.0mL    

 

12.5mL

  25.0mL
  37.5mL

 

49.9mL

 

50.0mL

 

50.1mL

 

75.0ml

 

 

 

0.0ml of 0.100M NaOH to  50.0ml of 0.100M  HC2H3O2

Part 1- No Base, only weak acid

 

     Set up the ice box using the molarity of the weak acid.

(Note the double arrow, use molarity)

 

HAc   

<==>  H+

+ Ac-

I

0.100M

 

--- ---
C -X   +X +X
E 0.100M-X   +X +X

Ka=

[H+][Ac-]
----------------------------------
[HAc]

Plug in the equilibrium values (E)

1.8x10-5=

[X][X]
---------------------------------
[0.100M-X]

If Ka x the molarity is >10-3, then X in the denominator will be too small and can be disregarded.

1.8x10-5=

[X][X]
---------------------------------
[0.100M]

Solve for X

X=[H+]=0.00134M

pH=-log[H+]=-log 0.00134M

pH=2.87

 

Part 2- Adding base to the weak acid

 

12.5ml of 0.100M NaOH to 50.0mL of 0.100M  HC2H3O2

Moles HAc=50.0mL x 0.100M=5.00mmol

Moles OH-=12.5mL x 0.100M=1.25mmol

Set up the ice box using the mmol.

(**Note only a one way arrow, use moles)

 

HAc  + 

OH- 

  Ac-

+H2O
I

5.00mmol

1.25mmol

------

 
C -1.25mmol -1.25mmol

+1.25mmol

 
E 3.75mmol 0

1.25mmol

 

Since at equilibrium we have an acid and its conjugate base. We made a buffer and can use the Henderson-Hasselbalch equation.

 pH=pKa+log [Base]
[Acid]

pKa=-log Ka=-log 1.8x10-5

pKa=4.74

 You do not need to convert the mmoles to Molarity, because the equation only requires the ratio. The mmole ratio is the same as the Molarity ratio.

 pH=4.74+log 1.25mmol
3.75mmol

pH=4.26

 

25.0ml of 0.100M NaOH to 50.0mL of 0.100M  HC2H3O2

Moles HAc=50.0mL x 0.100M=5.00mmol

Moles OH-=25.0mL x 0.100M=2.50mmol

Set up the ice box using the mmol.

(**Note only a one way arrow, use moles)

 

HAc  + 

OH- 

  Ac-

+H2O
I

5.00mmol

2.50mmol

------

 
C -2.50mmol -2.50mmol

+2.50mmol

 
E 2.50mmol 0

2.50mmol

 
 

This is the halfway point of the titration. When you consume 1/2 the acid, which produces the same number of moles of the base.

 pH=pKa+log [Base]
[Acid]

The ratio is of acid to base is 1:1.

log1=0

Therefore, pH=pKa

pH= pKa=4.74

 

37.5ml of 0.100M NaOH to 50.0mL of 0.100M  HC2H3O2

Moles HAc=50.0mL x 0.100M=5.00mmol

Moles OH-=37.5mL x 0.100M=3.75mmol

Set up the ice box using the mmol.

(**Note only a one way arrow, use moles)

 

HAc  + 

OH- 

  Ac-

+H2O
I

5.00mmol

3.75mmol

------

 
C -3.75mmol -3.75mmol

+3.75mmol

 
E 1.25mmol 0

3.75mmol

 

Since at equilibrium we have an acid and its conjugate base. We made a buffer and can use the Henderson-Hasselbalch equation.

 pH=pKa+log [Base]
[Acid]

pKa=-log Ka=-log 1.8x10-5

pKa=4.74

 You do not need to convert the mmoles to Molarity, because the equation only requires the ratio. The mmole ratio is the same as the Molarity ratio.

 pH=4.74+log 3.75mmol
1.25mmol

pH=5.22

 

 

49.9ml of 0.100M NaOH to 50.0mL of 0.100M  HC2H3O2

Moles HAc=50.0mL x 0.100M=5.00mmol

Moles OH-=49.9mL x 0.100M=4.99mmol

Set up the ice box using the mmol.

(**Note only a one way arrow, use moles)

 

HAc  + 

OH- 

  Ac-

+H2O
I

5.00mmol

4.99mmol

------

 
C -4.99mmol -4.99mmol

+4.99mmol

 
E 0.01mmol 0

4.99mmol

 

Since at equilibrium we have an acid and its conjugate base. We made a buffer and can use the Henderson-Hasselbalch equation.

 pH=pKa+log [Base]
[Acid]

pKa=-log Ka=-log 1.8x10-5

pKa=4.74

 You do not need to convert the mmoles to Molarity, because the equation only requires the ratio. The mmole ratio is the same as the Molarity ratio.

 pH=4.74+log 4.99mmol
0.01mmol

pH=7.44

 

 

Part 3- The equivalence point.

Where the moles of acid= moles of base

 

50.0ml of 0.100M NaOH to 50.0mL of 0.100M  HC2H3O2

Moles H+=50.0mL x 0.150M=7.50mmol

Moles OH-=50.0mL x 0.150M=7.50mmol

Set up the ice box using the mmol.

(**Note only a one way arrow, use moles)

 

HAc  + 

OH- 

  Ac-

+H2O
I

5.00mmol

5.00mmol

------

 
C -5.00mmol -5.00mmol

+5.00mmol

 
E

0

0

5.00 mmol

 

The acetate ion will reestablish equilibrium by reacting with the water. The reaction reverses. (Note the double arrow, use Molarity)

Ac-   

+H2O

<==> HAc

+  OH-

Find the concentration of the acetate ion using the total volume.

(add the ml of acid and mL of the base)

 Molarity= mol = mmol
liters mL
 [Ac-]= 5.00mmol
50.0ml + 50.0ml

[Ac-]=0.05M

Set up the ice box using the molarity of the acetate ion.

 

Ac-

H2O

<==> HAc

+  OH-
I

0.05M

 

------

 
C -X  

+X

+X

E

0.05M-X

 

+X

+X

Write the equilibrium expression.

(Note=K is now Kb..there is OH-)

Kb=

[HAc][OH-]
----------------------------------
[Ac-]

Solve for Kb

Kb=

Kw
----------------------------------
Ka

Kb=

10-14

=5.56 x 10-10

----------------------------------
1.8 x 10-5

Plug values into the Kb expression

Kb=

[HAc][OH-]
----------------------------------
[Ac-]

5.56 x 10-10=

[X][X]
----------------------------------
[0.05-X]

If Kb x the molarity is >10-3, then X in the denominator will be too small and can be disregarded.

5.56 x 10-10=

[X][X]
----------------------------------
[0.05]

Solve for X

X=[OH-]=5.27 x 10-6

Solve for pOH

pOH=-log[OH-]=-log 5.27 x 10-6M

pOH=5.28

Solve for pH

pH=14-5.28=8.72

 

Part 4- Beyond the equivalence point. Excess base.

 

50.1ml of 0.100M NaOH to 50.0mL of 0.100M  HC2H3O2

Moles H+=50.0mL x 0.100M=5.00mmol

Moles OH-=50.1mL x 0.100M=5.01mmol

Set up the ice box using the mmol.

(**Note only a one way arrow, use moles)

 

HAc  + 

OH- 

  Ac-

+H2O
I

5.00mmol

5.01mmol

------

 
C -5.00mmol -5.00mmol

+5.00mmol

 
E

0

0.01mmol

5.00mmol

 

There are 2 bases present the weak base Ac- and the strong base OH-. Ignore the Ac-.

Find the concentration of the hydroxide ion using the total volume.  (add the ml of acid and mL of the base)

 Molarity= mol = mmol
liters mL
  [OH- ]= 0.01mmol =0.000 1M
50.1mL +50.0mL

pOH=-log[OH-]

pOH=-log[0.000 1M]=4.00

pH=14-4.00=10.00

 

75.0ml of 0.100M NaOH to 50.0mL of 0.100M  HC2H3O2

Moles H+=50.0mL x 0.100M=5.00mmol

Moles OH-=75.0mL x 0.100M=7.50mmol

Set up the ice box using the mmol.

(**Note only a one way arrow, use moles)

 

HAc  + 

OH- 

  Ac-

+H2O
I

5.00mmol

7.50mmol

------

 
C -5.00mmol -5.00mmol

+5.00mmol

 
E

0

2.50mmol

5.00mmol

 

There are 2 bases present the weak base Ac- and the strong base OH-. Ignore the Ac-.

Find the concentration of the hydroxide ion using the total volume.  (add the ml of acid and mL of the base)

 Molarity= mol = mmol
liters mL
  [OH- ]= 2.50mmol =0.02M
75.0mL +50.0mL

pOH=-log[OH-]

pOH=-log[0.02M]=1.70

pH=14-1.70=12.30

 

   pH

Vol. NaOH

 2.87

0.0mL    

 4.26

12.5mL

 4.74 25.0mL
 5.22 37.5mL

 7.44

49.9mL

 8.72

50.0mL

 10.00

50.1mL

 12.30

75.0ml

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