Experiment 6
pH Measurements
A. Purpose
1. To review the concepts of pH measurements.
2. To learn the use of a pH meter.
3. To determine (measure) the pH of different solutions and compare the
measurements with the theoretical prediction.
B. Theoretical Background
Let us consider different types of solutions and go over the calculation of the hydrogen
ion concentration [H+] (and subsequently the pH) for each of them.
1. Strong Acid: pH of a 0.100 M HCl solution
HCl
Initially
After complete
dissociation
0.100
_

 H+
Cl
+
_
_
0.100
0.100
Thus [H+] = 0.100 M  pH = log10 [H+] = 1.00
2. Strong Base: pH of a 0.100 M NaOH solution
NaOH
Initially
After complete
dissociation

 Na+
+
OH
0.100
_
_
_
0.100
0.100
Thus [OH] = 0.100 M  pOH = log10 [OH] = 1.00
We know that: pH + pOH = 14.00  pH = 14.00 – 1.00 = 13.00
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3. Weak Acid: pH of a 0.100 M CH3COOH (HAc) solution

HAc
H+
Ac
+
Initially
0.100
_
_
Change
x
+x
+x
+x
+x
0.100 – x
At
equilibrium
For HAc, the acid dissociation constant Ka is:
Ka 

1.75 105 =
[H  ][Ac  ]
[HAc]
; Ka = 1.75 105
x2
0.100  x
x << 0.100 (Ka very small)  it is negligible with respect to 0.100

1.75 106 = x2  x = 1.32  103
Thus [H+] = 1.32  103 M  pH = log10[H+] = 2.88
4. Weak Base: pH of a 0.100 M of NH3 solution
NH3
+ H2O
Initially 0.100
Change
x
At
equilibrium 0.100 – x

NH 4
OH
+
_
_
+x
+x
+x
+x
For NH3, the base dissociation constant Kb is:
Kb 
[NH 4 ][OH  ]
; Kb = 1.75 105
[NH 3 ]
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
1.75  105 =
x2
0.100  x
x << 0.100  it is negligible with respect to 0.100
 1.75  106 = x2  x = 1.32  103
Thus [OH] = 1.32  103 M  pOH = log10[OH] = 2.88
We know that: pH + pOH = 14.00  pH = 14.00 – 2.88 = 11.12
5. Salt of a Strong Acid and a Strong Base: pH of a 0.100 M NaCl solution
NaCl
Initially
After complete
dissociation


Na+
+
Cl
0.100
_
_
_
0.100
0.100
We can see that NaCl is not a source of H+ or OH, thus [H+] is obtained solely from
the dissociation of water:
H2O

H+
Initially
Equilibrium
+
OH
_
+x
_
+x
The water dissociation constant Kw is:
Kw = [H+][OH]; Kw = 1.00  1014
 1.00 1014 = x2  x = 1.00  107
Thus [H+] = [OH] = 1.00  107 M  pH = log10[H+] = 7.00
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6. Salt of a Weak Acid: pH of a 0.100 M NaAc solution
NaAc
Initially


Na+
0.100
_
_
0.100
After complete
dissociation
Ac
+
_
0.100
Now, we should consider the hydrolysis of Ac
Ac + H2O

HAc
Initially 0.100
Change
OH
+
_
x
At
equilibrium 0.100 – x
_
+x
+x
+x
+x
The base dissociation constant for the acetate ion, Kb is:
K
[HAc][OH  ]
; Kb = w = 5.71  1010

Ka
[Ac ]
Kb 

5.71  1010 =
x2
0.100  x
x << 0.100  it is negligible with respect to 0.100
 5.71  1011 = x2  x = 7.56  106
Thus [OH] = 7.56  106 M  pOH = log10[OH] = 5.12
We know that: pH + pOH = 14.00  pH = 14.00 – 5.12 = 8.88
7. Salt of a Weak Base: pH of a 0.100 M NH4Cl solution
NH4Cl
Initially
After complete
dissociation


NH 4
+
Cl
0.100
_
_
_
0.100
0.100
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Now, we should consider the hydrolysis of NH 4
NH 4
+ H2O

H3O+
+
NH3
simplified as:
NH 4 
Initially
0.100
Change
At
equilibrium
H+
+
NH3
_
_
x
+x
+x
0.100 – x
+x
+x
The acid dissociation constant for the ammonium ion Ka is:
Kw
[H  ][NH 3 ]
Ka 
; Ka =
= 5.71  1010

[NH 4 ]
Kb

5.71 
1010 =
x2
0.100  x
x << 0.100  it is negligible with respect to 0.100
 5.71  1011 = x2  x = 7.56 106
Thus [H+] = 7.56 106 M  pH = log10[H+] = 5.12
8. Salt of a Weak Acid and a Weak Base: pH of a 0.100 M NH4Ac solution
NH4Ac
Initially
After complete
dissociation
0.100
_


NH 4
_
0.100
+
Ac
_
0.100
Here we have a source of H+ from NH 4 and a source of OH– from Ac
It can be shown that, in this case, pH 
1
(pK a1  pK a2 )
2
36
Where, pKa1 is that for HAc = 4.76, pKa2 is that for NH4+= 9.24.
 pH =
4.76  9.24
= 7.00
2
9. Buffer solutions
A buffer solution is one that can resist small additions of acid or base without
appreciably changing its pH. Buffer solutions are used to maintain the pH of
specific chemical reactions at a nearly fixed value, such as notably in
biochemical metabolitic reactions.
The composition of a buffer solution involves the presence of a weak acid and
the salt of its conjugate base (such as a mixture of HAc and NaAc) or a weak
base and the salt of its conjugate acid (such as a mixture of NH3 and NH4Cl).
a. Equimolar solution: 50.00 mL of 0.500 M HAc + 50.00 mL of 0.500
M NaAc
Initial molarity of HAc in the solution is: MHAc 
Initial molarity of NaAc is:
MNaAc 
50.00  0.500
 0.250 M
100.00
50.00  0.500
 0.250 M
100.00
From the Henderson-Hasselbäch equation:
pH  pKa  log 10
[NaAc]
[HAc]
 pH = 4.76 + log101.00  pH = 4.76
b. 70.00 mL of 0.500 M HAc + 30.00 mL of 0.500 M NaAc
Molarity of HAc in the solution is: MHAc 
70.00  0.500
 0.350 M
100.00
Molarity of NaAc in the solution is: MNaAc 
30.00  0.500
 0.150 M
100.00
From the Henderson-Hasselbäch equation:
pH  pKa  log 10
[NaAc]
[HAc]
37
 pH = 4.76 + log10 0.428  pH = 4.39
c. 30.00 mL of 0.500 M HAc + 70.00 mL of 0.500 M NaAc
Molarity of HAc in the solution is: MHAc 
30.00  0.500
 0.150 M
100.00
Molarity of NaAc in the solution is: MNaAc 
70.00  0.500
 0.350 M
100.00
From the Henderson-Hasselbäch equation:
pH  pKa  log 10
[NaAc]
[HAc]
 pH = 4.76 + log10 2.33  pH = 5.12
C. Chemicals and Setup
pH meter (Cyberscan 10), 3 in 1 combination pH electrode (Eutech), electrode stand.
Buffer solutions (pH = 4.00, 7.00, 10.00), 0.100 M HCl, 0.100 M NaOH, 0.100 M
HAc, 0.100 M NH3, 0.100 M NaCl, 0.100 M NaAc, 0.100 M NH4Cl, 0.100 M NH4Ac,
0.500 M HAc and 0.100 M NaAc.
Two 50 mL burets, 50 mL beaker, 10 mL graduated pipet, 100 mL beaker, two 250 mL
beakers.
D. Procedure

(Work in pairs throughout this experiment)
Calibration of the pH meter
Learn, with the help of your lab instructor, the principles and procedure for the
calibration of the pH meter instrument.

pH measurements
1. Measure the pH of a 0.100 M HCl solution.
2. Measure the pH of a 0.100 M NaOH solution.
3. Measure the pH of a 0.100 M HAc solution.
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4.
5.
6.
7.
8.
9.
Measure the pH of a 0.100 M NH3 solution.
Measure the pH of a 0.100 M NaCl solution.
Measure the pH of a 0.100 M NaAc solution.
Measure the pH of a 0.100 M NH4Cl solution.
Measure the pH of a 0.100 M NH4Ac solution.
Buffer solutions: Obtain from the storeroom two burets and a 100.0 mL
volumetric flask. Collect in two 250 mL clean and dry beakers 200 mL of
0.500 M HAc and 0.500 M NaAc solutions. Label the burets and fill one of
them with the HAc solution and the other with the NaAc solution.
 Equimolar (50-50) mixture: Solution A
a. Empty from the two burets 50.00 mL of HAc and 50.00 mL of NaAc
respectively, into a clean 100 mL beaker. Mix well.
b. Pour 30 mL of this buffer solution into a 50 mL beaker. Measure the pH.
c. Add to this solution 0.5 mL of 1.00 M HCl and read its pH after good
stirring.
d. Add another 0.5 mL of HCl, stir well and read the pH.
e. Pour another 30 mL of the buffer solution into a 50 mL beaker. Measure
the pH.
f. Add to this solution 0.5 mL of 1.00 M NaOH and read its pH after good
stirring.
g. Add another 0.5 mL of NaOH, stir well and read the pH.

70-30 mixture: Solution B
Repeat steps a through g but using 70.00 mL of HAc and 30.00 mL of
NaAc in step a.

30-70 mixture: Solution C
Repeat steps a through g but using 30.00 mL of HAc and 70.00 mL of
NaAc in step a.
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E. CHEM. 203 Lab Report
NAME : ____________________________
EXPERIMENT No. : ______________
TITLE : ____________________________________
DATE : _________________
Partner : ________________________
 pH Measurements
Solution
0.100 M HCl
0.100 M NaOH
0.100 M HAc
0.100 M NH3
0.100 M NaCl
0.100 M NaAc
0.100 M NH4Cl
0.100 M NH4Ac
Measured pH
Theoretical pH
% Error
 pH of a Buffer solution and its variation upon addition of strong acid or
strong base
Solution
Solution A (equimolar)
Solution A + 0.5 mL of 1.00 M HCl
Solution A + 1.0 mL of 1.00 M HCl
Solution A + 0.5 mL of 1.00 M NaOH
Solution A + 1.0 mL of 1.00 M NaOH
Measured pH
Solution
Solution B (70-30)
Solution B + 0.5 mL of 1.00 M HCl
Solution B + 1.0 mL of 1.00 M HCl
Solution B + 0.5 mL of 1.00 M NaOH
Solution B + 1.0 mL of 1.00 M NaOH
Measured pH
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Solution
Solution C (30-70)
Solution C + 0.5 mL of 1.00 M HCl
Solution C + 1.0 mL of 1.00 M HCl
Solution C + 0.5 mL of 1.00 M NaOH
Solution C + 1.0 mL of 1.00 M NaOH
Measured pH
Comment on the variation of the pH of each of the above buffer solutions, upon addition
of acid or base. Which is the strongest buffer?
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