ACID BASE TITRATION

1. Objective
a. Student can prepare acid standart solution
b. Student can determine the normality of the acid/base using a standart solution
c. Student can determine the equivalence point using titration curve
2. Theory
An acidbase titration is the determination of the concentration of an acid or base
by exactly neutralizing the acid or base with an acid or base of known concentration.
This allows for quantitative analysis of the concentration of an unknown acid or base
solution. It makes use of the neutralization reaction that occurs between acids and bases.
(Wikipedia, 2009)

Titration Step by Step

Rinsing the Pipette
A pipette is used to measure and transfer a precise volume of liquid. You
rinse a pipette with the solution whose volume you are measuring. This ensures that
any drops that remain inside the pipette will form part of the measured volume.
1. Pour a sample of standard solution into a clean, dry beaker.
2. Place the pipette tip in a beaker of distilled water. Squeeze the suction bulb.
Maintain your grip while placing it over the stem of the pipette. (If your suction bulbs
have valves, your teacher will show you how to use them.)
3. Relax your grip on the bulb to draw up a small volume of distilled water.
4. Remove the bulb and discard the water by letting it drain out.
5. Rinse the pipette by drawing several millilitres of solution from the beaker into
it.
6. Rotate and rock the pipette to coat the inner surface with solution. Discard the
rinse.
7. Rinse the pipette twice in this way. It is now ready to fill with standard solution.

Filling the Pipette

1. Place the tip of the pipette below the surface of the solution.
2. Hold the suction bulb loosely on the end of the glass stem. Use the suction bulb
to draw liquid up just past the etched volume mark.
3. As quickly and smoothly as you can, slide the bulb off and place your index
finger over the end of the glass stem.
4. Gently roll your finger slightly away from end of the stem to let solution drain
slowly out.
5. When the bottom of the meniscus aligns with the etched mark, press your finger
back over the end of the stem. This will prevent more solution from draining
out.
6. Touch the tip of the pipette to the side of the beaker to remove any clinging
drop. The measured volume inside the pipette is now ready to transfer to an
Erlenmeyer flask or a volumetric flask.
Transfering the Solution
1. Place the tip of the pipette against the inside glass wall of the flask. Let the
solution drain slowly, by removing your finger from the stem.
2. After the solution drains, wait several seconds, then touch the tip to the inside
wall of the flask to remove any drop on the end. Note: You may notice a small
amount of liquid remaining in the tip. The pipette was calibrated to retain this
amount. Do not try to remove it.

Adding the Indicator

1. Add two or three drops of indicator to the flask and its contents. Do not add too
much indicator. Using more does not make the colour change easier to see. Also,
indicators are usually weak acids. Too much can change the amount of base
needed for neutralization. You are now ready to prepare the apparatus for the
titration.

Rinsing the Burette

Burette is used to accurately measure the volume of liquid added during a titration
experiment. It is a graduated glass tube with a tap at one end.
1. To rinse the burette, close the tap and add about 10 mL of distilled water from
a wash bottle.
2. Tip the burette to one side and roll it gently back and forth so that the water
comes in contact with all inner surfaces.
3. Hold the burette over a sink. Open the tap, and let the water drain out. While
you do this, check that the tap does not leak. Make sure that it turns smoothly
and easily.
4. Rinse the burette with 5 mL to 10 mL of the solution that will be measured.
Remember to open the tap to rinse the lower portion of the burette. Rinse the
burette twice, discarding the liquid each time.

Filling the Burette

1. Assemble a retort stand and burette clamp to hold the burette. Place a funnel in
the top of the burette.
2. With the tap closed, add solution until the liquid is above the zero mark. Remove
the funnel. Carefully open the tap. Drain the liquid into a beaker until the bottom
of the meniscus is at or below the zero mark.
3. Touch the tip of the burette against the beaker to remove any clinging drop.
Check that the portion of the burette that is below the tap is filled with liquid
and contains no air bubbles.
4. Record the initial burette reading in your notebook.
5. Replace the beaker with the Erlenmeyer flask that you prepared earlier. Place a
sheet of white paper under the Erlenmeyer to help you see the indicator colour
change that will occur near the end-point.

Reading the Burette

 1. A meniscus reader is a small white card with a thick black line on it. Hold the
card behind the burette, with the black line just under the meniscus. Record the
volume added from the burette to the nearest 0.05 mL.
(McGraw-Hill Reyerson. 1996. Chemistry 11.pdf)

Indicators and pH meters can be used to detrmine the equivalence point. A pH

meter will show a large pH change occurring at the equivalence point. If an indicators
is used, it must change color over a range thet includes the pH of the equivalence point.
pH meter can also be used to follow acid-base titration. In the acid-base titration, the
pH of the acidic solutions is slow, after added basic solution dropwise, the pH of the
mixture will increase up step by step, and will rise sharply when approaching the end
point titration. After exceeded the end point, the pH increasing becomes slowly again.
Thus, if the pH and the volume of titrant plotted in a curve, the end point of the titration
will lie on a straight line (which rises dramatically ). For a strong acid-strong base , the
theoretical and point occurs at pH = 7 and the equivalence point falls in the middle of
the upright curve.

Indicators that change color at pH lower than 7(pH <7) are useful in determining
the equivalence point of strong acid/weak base titration. Methyl orange is an example
of this type. The equivalence point of a strong acid/weak base titration is acidic because
the salt forned is itself a weak acid. Thus the salt solution has a pH lower than 7.
Indicators that change color at pH higher than 7 (pH>7) are useful in determining the
equivalence point of weak acid/strong base titrations. Phenolphthalein is an example.
These reactions produce salt solutions whose pH is greater than 7. This occurs because
the salt formed is a weak base. The color transition of an indicator helps very little in
determining whether reactions between such acids and base are complete. In a titration,
successive additions of an aqueous base can be made to a measured volume of an
aqueous acid. As base is added, the pH changes from a low numerical value to a high
one. The change in pH occurs slowly at first, then rapidly through the equivalence point,
and then slowly again as the solution becomes more basic. (Holt, 2006)

3. Apparatus and Reagent

Apparatus : reagent :
Analytical balance oxalic acid crystals
100 ml volumetric flask distilled water
100 ml erlenmeyer 0,1 M NaOH
50 ml buret 0,1 M HCl
Clamps and statif 0,1 M CH3COOH
10 ml volumetric pipet PP indicators
PH meter / universal indicator
4. Procedure
a. Preparation of primary standart solution of H2C2O4(COOH)2.2H2O

Weigh accurately Dissolve with distilled

1,2607 gram of oxalic water in a 100 ml
acid crystal volumetric flask

b. Determine the concentraion of NaOH

c.
Clean buret with NaOH 10 mL oxalic acid solution
solution and filled with NaOH filled into two erlenmeyer.
solution up to the 0 mL mark Add 2-3 drops pf indicator PP

Drop down NaOH solution in

Record the volume of NaOH
the buret into the acid
used
solution, shake at the same
times

d. Determination of the concentration of HCl using secondary standard solution of NaOH

e.
Fill the buret with Provide two erlenmeyers and
standardized of NaOH fill with 10 mL of HCl anda
solution also add 2-3 drops of PP

Drop down NaOH solution in

Record the volume of NaOH
the buret into the acid
used
solution, shake at the same
times

f. Making titration curve

Put 10 mL of 0,1 M HCl into

the erlenmeyer and Measure the initial pH of
standardized NaOH into buret HCL
up to 0 mL mark

Do in the same way by Add 1 mL NaOH dropwise

replacing HCl with 0,1 M into the erlenmeyer, shake ,
CH3COOH measure the pH of the
solution
5. Observation Sheet
a. Preparation of primary standart solution of H2C2O4(COOH)2.2H2O
Mass of Oxalic acid = 1,2607 grams
Volume of solution = 10 mL
Molarity of Oxalic acid = 0,1 M
Normality of Oxalic acid = 0,2 N
b. Determination the concentration of NaOH
Volume of Oxalic acid = 10 mL
Volume of NaOH (1) = 39 mL
Volume of NaOH (2) = 40 mL
Average volume of NaOH = 39,5mL
c. Determination of the concentration of HCl using secondary standart solution
of NaOH
Volume of HCl 0,1 M = 10 mL
Volume of NaOH(1) 0,1 M = 15 mL
Volume of NaOH (2) 0,1 M = 15 mL
Volume of average NaOH = 15 mL
d. Relationship of the volume of titrant with pH in acid-base titration
No Volume of NaOH pH
added to 10 mL
0,118 M HCl
1 10 mL + 1 mL 1

2 10 mL + 2 mL 1

3 10 mL + 3 mL 1

4 10 mL + 4 mL 1

5 10 mL + 5 mL 1

6 10 mL + 6 mL 1

7 10 mL + 7mL 1

8 10 mL + 8 mL 1

9 10 mL + 9 mL 1

10 10 mL + 10 mL 2

11 10 mL + 11 mL 2

12 10 mL + 12 mL 2
 13 10 mL + 13 mL 5

14 10 mL + 14 mL 6

15 10 mL + 15 mL 7

No Volume of NaOH pH
added to 10 mL
0,1M CH3COOH
1 10 mL + 1 mL 3

2 10 mL + 2 mL 4

3 10 mL + 3 mL 4

4 10 mL + 4 mL 4

5 10 mL + 5 mL 4

6 10 mL + 6 mL 5

7 10 mL + 7 mL 5

8 10 mL + 8 mL 5

9 10 mL + 9 mL 5

10 10 mL + 10 mL 5

11 10 mL + 11 mL 5

12 10 mL + 12 mL 5

13 10 mL + 13 mL 5

14 10 mL + 14 mL 6

15 10 mL + 15 mL 6

6. Analysis Data
a. Preparation of primary standat solution of H2C2O4(COOH)2.2H2O

Molarity of Oxalic acid = mass/Mr. Volume

 =1,2607/1.2607 . 0.1
=0,1 M
Normality of oxalic acid = M . valence
= 0,1 . 2
= 0,2 N
b. Determine the cocentration of NaOH

V1 . N1 = V2 . N2
10 . 0,2 = 39,5 .N2
N2 = 10 . 0,2 / 39,5
N2 = 0,0506 N
M2 = 0,0506/1
= 0,0506 M

2NaOH + (COOH)2 (COONa)2 + 2H2O

Initial : 0,0506 0,1 - -

Reaction : 0,0506 0,0253 0,0253 0,0506
Remain : - 0,0747 0,0253 0,0506

c. Determination of the concentration of HCl using secondary standart solution of

NaOH

V1 . N1 = V2 . N2

15 . 0,0506 = 10 . N2

N2 = 15 . 0,0506 / 10
= 0,0759 N
M2 = 0,0759 / 1
= 0,0759 M

NaOH + HCl NaCl + H2 O

Initial : 0,0506 0,0759 - -

Reaction : 0,0506 0,0506 0,0506 0,0506
Remain : - 0,0253 0,0506 0,0506

d. Relationship of the volume of titrant with pH in acid-base titration

pH titration of NaOH HCl (theoritic)
1) VNaOH = 1 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
 n HCl = 0,759 mmol
n NaOH = 0,0506 mmol

HCl + NaOH NaCl + H2O

Initial : 0,759 0,0506 - -
Reaction : 0,0506 0,0506 0,0506 0,0506
Remain : 0,7084 - 0,0506 0,0506

MHCl = 0,7084 mmol/11 mL

=0,0644 M
[ H+ ] =M.a
= 0,0644 . 1
= 0,0644 M
pH = - log 0,0644
= 1,19

2) VNaOH = 2 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0.1012 mmol

HCl + NaOH NaCl + H2O

Initial : 0,759 0,1012 - -
Reaction : 0,1012 0,1012 0,1012
Remain : 0,6578 - 0,1012 0,1012

MHCl = 0,6578 mmol/ 12 mL

= 0,0548 M
[ H+ ] =M.a
= 0,0548 . 1
= 0,0548 M
pH = -log 0,0548
= 1,26
3) VNaOH = 3 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,1518 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,1518 - -
 Reaction :0,1518 0,1518 0,1518 0,1518
Remain :0,6072 - 0,1518 0,1518

MHCl = 0,6072 mmol/ 13 mL

= 0,0470 M
[ H+ ] =M.a
= 0,0470 . 1
= 0,0470 M
pH = -log 0,0470
= 1,33
4) VNaOH = 4 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759M
n HCl = 0,759 mmol
n NaOH = 0,2024 mmol

HCl + NaOH NaCl + H2O

Initial : 0,759 0,2024 - -
Reaction : 0,2024 0,2024 0,2024 0,2024
Remain :0,5566 - 0,2024 0,2024

MHCl = 0,5566 mmol/ 14 mL

= 0,039 M
[ H+ ] =M.a
=0,039 . 1
= 0,039 M
pH = -log 0,039
= 1,40
5) VNaOH = 5 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,253 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,253 - -
Reaction :0,253 0,253 0,253 0,253
Remain :0,506 - 0,253 0,253

MHCl = 0,506 mmol/ 15 mL

= 0,0337 M
[ H+ ] =M.a
 = 0,0337 . 1
= 0,0337 M
pH = -log 0,0337
= 1,47
6) VNaOH = 6 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,3036 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,3036 - -
Reaction :0,3036 0,3036 0,3036 0,3036
Remain :0,4554 - 0,3036 0,3036

MHCl = 0,4554 mmol/ 16 mL

= 0,0284M
[ H+ ] =M.a
= 0,0284 . 1
= 0,0284 M
pH = -log 0,0284
= 1,55
7) VNaOH = 7 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,3542 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,3542 - -
Reaction : 0,3542 0,3542 0,3542 0,3542
Remain :0,4048 - 0,3542 0,3542

MHCl = 0,4048 mmol/ 17 mL

= 0,024 M
[ H+ ] =M.a
= 0,024 . 1
= 0,024 M
pH = -log 0,024
= 1,62
8) VNaOH = 8 mL
VHCl = 10 mL
 MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,4048 mmol

HCl + NaOH NaCl + H2O

Initial : 0,759 0,4048 - -
Reaction :0,4048 0,4048 0,4048 0,4048
Remain : 0,3542 - 0,4048 0,4048

MHCl = 0,3542 mmol/ 18mL

= 0,01967 M
[ H+ ] =M.a
= 0,01967. 1
= 0,01967 M
pH = -log 0,01967
=1,70
9) VNaOH = 9 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,4554 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,4554 - -
Reaction :0,4554 0,4554 0,4554 0,4554
Remain :0,3036 - 0,4554 0,4554

MHCl = 0,3036 mmol/ 19 mL

= 0,0160M
[ H+ ] =M.a
= 0,0160 . 1
= 0,0160 M
pH = -log 0,0160
= 1,80
10) VNaOH = 10 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,506 mmol

HCl + NaOH NaCl + H2O

 Initial : 0,759 0,506 - -
Reaction : 0,506 0,506 0,506 0,506
Remain : 0,253 - 0,506 0,506

MHCl = 0,253 mmol/ 20 mL

= 0, 01265 M
[ H+ ] =M.a
= 0,01265. 1
= 0,01265 M
pH = log 0,01265
= 1,89
11) VNaOH = 11 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,5566 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,5566 - -
Reaction :0,5566 0,5566 0,5566 0,5566
Remain : 0,2024 - 0,5566 0,5566

MHCL = 0,2024 mmol/ 21 mL

= 0,009638 M
[ H+ ] =M.a
= 0,009638 . 1
= 0,009638 M
pH = log 0,009638
= 2,016
12) VNaOH = 12 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,6072 mmol

HCl + NaOH NaCl + H2O

Initial : 0,759 0,6072 - -
Reaction : 0,6072 0,6072 0,6072 0,6072
Remain : 0,1518 - 0,6072 0,6072

MHCL = 0,1518 mmol/ 22 mL

= 0,0069 M
 [ OH- ] =M.a
=0,0069 . 1
= 0,0069 M
pH = - log 0,0069
= 2,161
13) VNaOH = 13 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,6578 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,6578 - -
Reaction : 0,6578 0,6578 0,6578 0,6578
Remain : 0,1012 - 0,6578 0,6578

MHcl = 0,1012 mmol/ 23 mL

= 0,0044 M
[ H+ ] =M.a
=0,0044 . 1
= 0,0044 M
pH = - log 0,0044
= 2,35
=2,35

14) VNaOH = 14 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,7084 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,7084 - -
Reaction :0,7084 0,7084 0,7084 0,7084
Remain : 0,0506 - 0,7084 0,7084

MHcl = 0,0506 mmol/ 24 mL

= 0,02108 M
[ H+ ] =M.a
= 0,002108 . 1
= 0,00218 M
pH = - log 0,00218
 = 2,66

15) VNaOH = 15 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,759 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,759 - -
Reaction :0,759 0,759 0,759 0,759
Remain : - - 0.759 0,759

pH =7

16) VNaOH = 16 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,8096 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,8096 - -
Reaction :0,759 0,759 0,759 0,759
Remain : 0,0506 0,759 0,759

MNaOH = 0,0506 mmol/ 26 mL

= 0,001946 M
[ OH- ] =M.a
= 0,001946 . 1
= 0,001946 M
poH = - log 0,001946
= 2,71
pH = 14-2,71
= 11,29
17) VNaOH = 17 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,8602 mmol
 HCl + NaOH NaCl + H2O
Initial :0,759 0,8602 - -
Reaction :0,759 0,759 0,759 0,759
Remain : 0,1012 0,759 0,759

MNaOH = 0,1012 mmol/ 27 mL

= 0,003748 M
[ OH- ] =M.a
= 0,003748. 1
= 0,003748 M
poH = - log 0,003748
= 2.426
pH = 14-2.426
= 11,574
18) VNaOH = 18 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,9108 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,9108 - -
Reaction :0,759 0,759 0,759 0,759
Remain : 0,1518 0,759 0,759

MNaOH = 0,1518 mmol/ 28 mL

= 0,00542 M
[ OH- ] =M.a
= 0,00542 . 1
= 0,00542 M
poH = - log 0,00542
= 2,26
pH = 14-2,26
= 11,74
19) VNaOH = 19 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 0,9614 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 0,9614 - -
 Reaction :0,759 0,759 0,759 0,759

emain : - 0,2024 0,759 0,759

MNaOH = 0,2024 mmol/ 29 mL

= 0,006973 M
[ OH- ] =M.a
= 0,006973 . 1
= 0,006973 M
poH = - log 0,006973
= 2,156
pH = 14-2,156
= 11,844
20) VNaOH = 20 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1,012 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1,012 - -
Reaction :0,759 0,759 0,759 0,759
Remain : - 0,253 0,759 0,759

MNaOH = 0,253 mmol/ 30 mL

= 0,00843 M
[ OH- ] =M.a
= 0,00843 . 1
= 0,00843 M
poH = - log 0,00843
= 2,08
pH = 14-2,08
= 11,92
21) VNaOH = 21 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1,0626 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1,0626 - -
Reaction :0,759 0,759 0,759 0,759
 Remain : 0,3036 0,759 0,759

MNaOH = 0,3036 mmol/ 31 mL

= 0,0098 M
[ OH- ] =M.a
= 0,0098 . 1
= 0,0098 M
poH = - log 0,0098
= 2.0
pH = 14-2.00
= 12
22) VNaOH = 22 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1.6192 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1,1132 - -
Reaction :0,759 0,759 0,759 0,759
Remain : 0,3542 0,759 0,759

MNaOH = 0,3542 mmol/ 32 mL

= 0,011 M
[ OH- ] =M.a
= 0,011 . 1
= 0,011 M
poH = - log 0,011
= 1,96
pH = 14-1,96
= 12,04

23) VNaOH = 23 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1,1638 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1,1638 - -
Reaction :0,759 0,759 0,759 0,759
Remain : 0,4048 0,759 0,759
 MNaOH = 0,4048 mmol/ 33 mL
= 0,01226 M
[ OH- ] =M.a
= 0,01226 . 1
= 0,01226 M
poH = - log 0,01226
= 1.91
pH = 14-1.91
= 12.09
24) VNaOH = 24 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1,2144 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1,2144 - -
Reaction :0,759 0,759 0,759 0,759
Remain : 0,4554 0,759 0,759

MNaOH = 0,4554 mmol/ 34 mL

= 0,0134 M
[ OH- ] =M.a
= 0,0134 . 1
= 0,0134 M
poH = - log 0,0134
= 1,872
pH = 14-1,872
= 12,128
25) VNaOH = 25 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1,265 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1,265 - -
Reaction :0,759 0,759 0,759 0,759
Remain : 0,506 0,759 0,759

MNaOH = 0,506 mmol/ 35 mL

 = 0,01445 M
-
[ OH ] =M.a
= 0,01445. 1
= 0,01445 M
poH = - log 0,01445
= 1,84
pH = 14-1,84
= 12.16
26) VNaOH = 26 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1,3156 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1.3156 - -
Reaction :0,759 0,759 0,759 0,759
Remain : 0,5566 0,759 0,759

MNaOH = 0,5566 mmol/ 36 mL

= 0,01546 M
[ OH- ] =M.a
= 0,01546 . 1
= 0,01546 M
poH = - log 0,01546
= 1,81
pH = 14-1,81
= 12,19
27) VNaOH = 27 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1,3662 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1,3662 - -
Reaction :0,759 0,759 0,759 0,759
Remain : - 0,6072 0,759 0,759

MNaOH = 0,6072 mmol/ 37 mL

= 0,0164 M
[ OH- ] =M.a
 = 0,0164 . 1
= 0,0164 M
poH = - log 0,0164
= 1,78
pH = 14-1,78
= 12,22

28) VNaOH = 28 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1,4168 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1,4168 - -
Reaction :0,759 0,759 0,759 0,759
Remain : 0,6578 0,759 0,759

MNaOH = 0,6578 mmol/38 mL

= 0,0173M
[ OH- ] =M.a
= 0,0173 . 1
= 0,0173 M
poH = - log 0,0173
= 1,76
pH = 14-1,76
= 12,24
29) VNaOH = 29 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1,4674 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1,4674 - -
Reaction :0,759 0,759 0,759 0,759
Remain : 0,7084 0,759 0,759

MNaOH = 0,7084 mmol/39 mL

= 0,0181M
[ OH- ] =M.a
= 0,0181 . 1
 = 0,0181 M
poH = - log 0,0181
= 1,742
pH = 14-1,742
= 12,258
30) VNaOH = 30 mL
VHCl = 10 mL
MNaOH = 0,0506 M
MHCl = 0,0759 M
n HCl = 0,759 mmol
n NaOH = 1,518 mmol

HCl + NaOH NaCl + H2O

Initial :0,759 1,518 - -
Reaction :0,759 0,759 0,759 0,759
Remain : - 0,759 0,759 0,759

MNaOH = 0,759 mmol/40 mL

= 0,019M
[ OH- ] =M.a
= 0,019 . 1
= 0,019 M
poH = - log 0,019
= 1,72
pH = 14-1,72
= 12,28

From calculations, this is the table of NaOH with HCl.

No Volume of NaOH pH
added to 10 mL
0,0759 M HCl
1 1 mL 1,19

2 2 mL 1,26

3 3 mL 1,33

4 4 mL 1,40

5 5 mL 1,47

6 6 mL 1,55

7 7 mL 1,63
8 8 mL 1,70

9 9 mL 1,80

10 10 mL 1,89

11 11 mL 2,02

12 12 mL 2,16

13 13 mL 2,35

14 14 mL 2,66

15 15 mL 7

16 16 mL 11,29

17 17 mL 11,57

18 18 mL 11,74

19 19 mL 11,84

20 20 mL 11,92

21 21 mL 12

22 22 mL 12,04

23 23 mL 12,09

24 24 mL 12,12

25 25 mL 12,16

26 26 mL 12,19

27 27 mL 12,22

28 28 mL 12,24

29 29 mL 12,26

30 30 mL 12,28
 Volume of NaOH added to 10 mL 0,0759 M HCl
14

12

10

8
pH

0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Volume of NaOH added to 10 mL 0,0759 M

Titration curve from calculation ( theoritic )

Volume of NaoH added to 10 mL 0,0759 M

HCl
16
14
12
10
pH

8
pH theoritical
6
pH experiment
4
2
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Volume of NaoH added tp 10 mL 0,0759 M HcL

pH titration of NaOH CH3COOH (theoritic)

V CH3COOH = 10 mL
M CH3COOH = 0,1 M
MNaOH = 0,0506 M
1) VNaOH = 1 mL

nNaOH = 0,0506 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,0506 1 - -

Reaction : 0,0506 0,0506 0,0506 0,0506

Remain : - 0,9494 0,0506 0,0506

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,9494 mmol/ 0,0506 mmol
= 3,37. 10-4
pH = - log 3,37 . 10-4
= 4 log 3,37
= 3,47

2) VNaOH = 2 mL

nNaOH = 0,1012 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,1012 1 - -

Reaction : 0,1012 0,1012 0,1012 0,1012

Remain : - 0,8988 0,1012 0,1012

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,8988 mmol/ 0,1012 mmol
= 1,5984 . 10-4 M
pH = - log 1,5984 . 10-4
= 4 log 1,5984
 = 3,796

3) VNaOH = 3 mL

nNaOH = 0,1518 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,1518 1 - -

Reaction : 0,1518 0,1518 0,1518 0,1518

Remain : - 0,8482 0,1518 0,1518

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,8482 mmol/ 0,1518 mmol
= 1,002 . 10-4 M
pH = - log 1,002 . 10-4
= 4 log 1,002
= 3,999

4) VNaOH = 4 mL

nNaOH = 0,2024 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,2024 1 - -

Reaction : 0,2024 0,2024 0,2024 0,2024

Remain : - 0,7976 0,2024 0,2024

[ H+ ] = Ka . a/g
= 1,8 . 10-5. 0,7974 mmol/ 0,2024 mmol
= 7,09. 10-5 M
pH = - log 7,09 . 10-5
 = 5 log 7,09
= 4,14

5) VNaOH = 5 mL

nNaOH = 0,253 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,253 1 - -

Reaction : 0,253 0,253 0,253 0,253

Remain : - 0,747 0,253 0,253

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,747 mmol/ 0,253 mmol
= 5,32 10-5 M
pH = - log 5,32 10-5
= 5 log 5,32
= 4,27

6) VNaOH = 6 mL

nNaOH = 0,3036 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,3036 1 - -

Reaction : 0,3036 0,3036 0,3036 0,3036

Remain : - 0,6964 0,3036 0,3036

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,6964 mmol/ 0,3036 mmol
= 4,13 . 10-5 M
 pH = - log 4,13 . 10-5
= 5 log 4,13
= 4,38

7) VNaOH = 7 mL

nNaOH = 0,3542 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,3542 1 - -
Reaction : 0,3542 0,3542 0,3542 0,3542
Remain : - 0,6458 0,3542 0,3542

[ H+ ] = Ka . a/g
= 1,8 . 10-5 .. 0,6458 mmol/ 0,3542mmol
= 3,28. 10-5 M
pH = - log 3,28 . 10-5
= 5 log 3,28
= 4,48
8) VNaOH = 8 mL

nNaOH = 0,4048 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,4048 1 - -

Reaction : 0,4048 0,4048 0,4048 0,4048

Remain : - 0,5952 0,4048 0,4048
 [ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,5952 mmol/ 0,4048 mmol
= 2,646 10-5 M
pH = - log 2,646 10-5
= 5 log 2,646
= 4,57
9) VNaOH = 9 mL

nNaOH = 0,4554 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,4554 1 - -

Reaction : 0,4554 0,4554 0,4554 0,4554

Remain : - 0,5446 0,4554 0,4554

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,5446 mmol/ 0,4554 mmol
= 2,152 . 10-5 M
pH = - log 2,152 . 10-5
= 5 log 2,152
= 4,67

10) VNaOH = 10 mL

nNaOH = 0,506 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,506 1 - -

Reaction : 0,506 0,506 0,506 0,506

Remain : - 0,494 0,506 0,506
 [ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,494 mmol/ 0,506 mmol
= 1,75 . 10-5 M
pH = - log 1,75 . 10-5
= 5 log 1,75
= 4,76

11) VNaOH = 11 mL

nNaOH = 0,5566 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,5566 1 - -

Reaction : 0,5566 0,5566 0,5566 0,5566

Remain : - 0,4434 0,5566 0,5566

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,4434 mmol/ 0,5566 mmol
= 1,44 . 10-5 M
pH = - log 1,44. 10-5
= 5 log 1,44
= 4,84

12) VNaOH = 12 mL

nNaOH = 0,6072 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH CH3COOH + H2O

Initial : 0,6072 1 - -
 Reaction : 0,6072 0,6072 0,6072 0,6072

Remain : - 0,3928 0,6072 0,6072

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,3928 mmol/ 0,6072 mmol
= 1,16 . 10-5 M
pH = - log 1,16 . 10-5
= 5 log 1,16
= 4,94

13) VNaOH = 13 mL

nNaOH = 0,6578 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,6578 1 - -
Reaction : 0,6578 0,6578 0,6578 0,6578
Remain : - 0,3422 0,6578 0,6578
[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,3422 mmol/ 0,6578 mmol
= 9,36. 10-6 M
pH = - log 9,36 . 10-6
= 6 log 9,36
= 5,03

14) VNaOH =14 mL

nNaOH = 0,7084 mmol

n CH3COOH = 1 mmol
 NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,7084 1 - -
Reaction : 0,7084 0,7084 0,7084 0,7084
Remain : - 0,2916 0,7084 0,7084

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,2916 mmol/ 0,7084 mmol
= 7,41. 10-6 M
pH = - log 7,41 . 10-6
= 6 log 7,41
= 5,13

15) VNaOH = 15 mL

nNaOH = 0,759 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,759 1 - -

Reaction : 0,759 0,759 0,759 0,759

Remain : - 0,241 0,759 0,759

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,241 mmol/ 0,759 mmol
= 5,71 . 10-6 M
pH = - log 5,71 . 10-6
= 6 log 5,71
= 5,24

16) VNaOH =16 mL

nNaOH = 0,7084 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,8096 1 - -
Reaction : 0,8096 0,8096 0,8096 0,8096
Remain : - 0,1904 0,8096 0,8096

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,1904 mmol/ 0,8096 mmol
= 4,23. 10-6 M
pH = - log 4,23 . 10-6
= 6 log 4,23
= 5,37

17) VNaOH =17 mL

nNaOH = 0,8602 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,8602 1 - -
Reaction : 0,8602 0,8602 0,8602 0,8602
Remain : - 0,1398 0,8602 0,8602

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,1398 mmol/ 0,8602 mmol
= 2,93. 10-6 M
pH = - log 2,93 . 10-6
= 6 log 2,93
= 5,53
18) VNaOH =18 mL

nNaOH = 0,9108 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,9108 1 - -
Reaction : 0,9108 0,9108 0,9108 0,9108
Remain : - 0,0892 0,9108 0,9108

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,0892 mmol/ 0,9108 mmol
= 1,76. 10-6 M
pH = - log 0,098 . 10-6
= 6 log 1,76
=5,75

19) VNaOH =19 mL

nNaOH = 0,9614 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 0,9614 1 - -
Reaction : 0,9614 0,9614 0,9614 0,9614
Remain : - 0,0386 0,9614 0,9614

[ H+ ] = Ka . a/g
= 1,8 . 10-5 . 0,0386 mmol/ 0,9614 mmol
= 7,22. 10-7M
pH = - log 7,22 . 10-7
 = 7 log 7,22
= 6,14

20) VNaOH =20 mL

nNaOH = 1,012 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2 O

Initial : 1,012 1 - -
Reaction : 1 1 1 1
Remain : 0,012 - 1 1

MNaOH = 0,012 mmol/ 30 mL

= 0,0004M
[ OH- ] =M.b
=0,0004 . 1
= 0,0004 M
pH = 14 pOH
= 14 log 0,0004
= 14 3,39
= 10,61
21) VNaOH =21 mL

nNaOH = 1,0626 mmol

n CH3COOH = 1 mmol

NaOH + CH3COOH NaCH3COO + H2O

Initial : 1,0626 1 - -
Reaction : 1 1 1 1
Remain : 0,0626 - 1 1

MNaOH = 0,0626 mmol/ 31 mL

=M
[ OH- ] =M.b
=0,0004 . 1
= 0,0004 M
pH = 14 pOH
= 14 log 0,0004
 = 14 3,39
= 10,61

No Volume of NaOH pH
added to 10 mL
0,1M CH3COOH
1 0 mL 2,9

2 1 mL 3,86

3 2 mL 4,2

4 3 mL 4,47

5 4 mL 4,7

6 5 mL 4,99

7 6 mL 5,1

8 7 mL 5,4

9 8 mL 5,8

10 9 mL 11,3

11 10 mL 11,88

12 11 mL 12,1

13 12 mL 12,1

14 13 mL 12,3

15 14 mL 12,8
 Volume of NaOH added to 10 mL of 0,1 M CH3COOHl
14 12.8
12.3
11.88 12.1 12.1
12 11.3

10
pH theoritical

8
5.8
5.1 5.4
6 4.7 4.99
4.2 4.47
3.86
4 2.9

0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
pH 2.9 3.86 4.2 4.47 4.7 4.99 5.1 5.4 5.8 11.3 11.88 12.1 12.1 12.3 12.8
Volume of NaOH

pH

Titration curve from calculation ( theoritic )

Chart Title
18

16

14

12
Axis Title

10

8 pH theoritic
pH based on experiment
6

0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
pH theoritic 2.93.94.24.54.7 5 5.15.45.8 11 12 12 12 12 13
pH based on experiment 3 4 4 4 4 5 5 5 5 5 5 5 5 6 16
 Volume of NaOH added to 10mL of 0,1
CH3COOH
14
12
10
8
pH

6
pH based on experiment
4
pH theoritical
2
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Volume of NaOH added to 10mL of 0,1 CH3COOH

7. Discussion
a. Preparation of primary standard solution of H2C2O4(COOH)2.2H2O
In this experiment, we have to calculate the molarity and the normality of oxalic
acid. The mass of oxalic acid = 1,2607 grams and volume of solution =100 ml.
Because we have the mass and the volume of oxalic acid so we can find the molarity
of the solution by the formula :
Molarity of Oxalic acid = mass/Mr. Volume
=1,2607/1.2607 . 0.1
= 0,1 M

Based on the formula, we know the molarity of oxalic acid solution is 0,1 M. To
calculate the normality of oxalic acid solution we can use the formula :
Normality of oxalic acid = M . valence
= 0,1 . 2
= 0,2 N

So the normality of oxalic acid solution is 0,2 N.

Oxalic acid is used as primary standart solution in this titration. Oxalic acid is used
as primary standard solution because its stable. So, we can find its molarity and
normality by calculating it. The molarity and normality of oxalic acid are used to
determine the molarity of NaOH solution. Standart solution is a solution that its
consentration is known and usually like acid and base and its concentration is
constant.
 b. Determine the concenration of NaOH
(COONa)2 + 2H2O 2NaOH + (COOH)2

We determine the concentration of NaOH by titration between NaOH solution and

oxalic acid solution as primary standard solution. We use 10 ml of 0,1 M oxalic
acid and 0,1 M NaOH.
The equivalence point of titration is marked by colour change. The colour of this
solution will change from transparent into pink. And when this solution has been
changed into pink, the titration must be stopped. Because the equivalence point
have left.
The first volume of NaOH solution is 39 mL and the second volume is 40. Then
the average of volume NaOH is 39,5 mL. After we get this volume we can
determine the concentration of NaOH by this formula :
V1 . N1 = V2 . N2
10 . 0,2 = 39,5 .N2
N2 = 10 . 0,2 / 39,5
N2 = 0,0506 N
M2 = 0,0506/1
= 0,0506 M

2NaOH + (COOH)2 (COONa)2 + 2H2O

Initial : 0,0506 0,1 - -

Reaction : 0,0506 0,0253 0,0253 0,0506
Remain : - 0,0747 0,0253 0,0506

From the calculation we know that the concentration of NaOH is 0,0506 M

NaOH solution is used as secondary standard. It is used to determine the
concentration of the other acid solution.
c. Determination of the concentration of HCl using secondary standard solution of
NaOH
In the third experiment, we use NaOH as the secondary solution to determine the
concentration of HCl. Because the basic solution ( NaOH ) as secondary standard
solution can be used to determine the concentration of the other acid solution ( HCl
). In this titration we use 10 ml HCl 0,1 M and 0,0506 M NaOH solution.

The equivalence point of titration is marked by colour change. The colour of this
solution will change from transparent into pink. And when this solution has been
changed into pink, the titration must be stopped. Because the equivalence point
have left.

V1 . N1 = V2 . N2
 0,0506 = 10 . N2

N2 = 15 . 0,0506 / 10
= 0,0759 N
M2 = 0,0759 / 1
= 0,0759 M

NaOH + HCl NaCl + H2O

Initial : 0,0506 0,0759 - -

Reaction : 0,0506 0,0506 0,0506 0,0506
Remain : - 0,0253 0,0506 0,0506

From the calculation, we know that the concentration of HCl is 0,0759 M

d. Making titration curve

1) Titration between NaOH and HCl
From the empirical curve, equivalence point of this titration is 7. Its appropriate
with theoretical curve. After calculating, from the theoritical curve, the
equivalence point of the titration is also 7. In the theory, the result pH of strong
acid-strong base is 7. Our group is supposed to add more than 15 mL of 0,1 M
into 0,1 M HCl. In this graph, curve begins at strong acid (HCI) and ends at high
PH of strong base. This equivalence point occurs at volume of NaOH is 15 ml
(in the theoritical curve) and 15 ml NaOH solution in the empirical curve. So,
the equivalence point in the theoritical curve equals the equivalence point in the
empirical curve.
2) Titration between NaOH and CH3COOH
On this experiment we use 10 mL volume of o,1 M CH3COOH. The equivalent
point at 8,7 mL volume of NaOH, and pH 8,9. And base on experiment, the
located of equivalent point is at 9,8 mL of volume NaOH and pH 8,9. There is
experiment error around 1,1 mL because the indicator which we used is
universal indicator, not pH meter. And the buret which we use has so old, so the
volume is not acurate.

8. Conclusion
1. The standard solution is a solution which has known of its concentration and
usually as acid or base (its has constant concentration ). In this experiment Oxalic
acid as standart solution. Because oxalicc acid is stable.
2. Molarity of NaOH is 0,0506 M and normality is 0,0506 N. And then, the molarity
of HCI is 0,0759 M and normality is 0,0759 N
3. The pH of equivalence point of titration between strong acid and strong base is 7,
while for weak acid and strong base is more than 7.

9. Suggestion
 1. We must know the procedure well before doing the experiment
2. We must use the apparatus carefully especially buret
3. We have to look the solution until it reach the equivalence point
4. We should do the experiment quietly and carefully.

10. Reference
McGraw-Hill Reyerson. 1996. Chemistry 11.pdf.
http://en.wikipedia.org/wiki/Acid-base_titration
Holt,Rinehart and Winston.2006.Modern Chemistry. United States of America:
National Science Teachers Association
 Question :

a. Determine the molarity of Oxalic Acid as primary standard solution

b. Determine molarity NaOH solution
c. Determine molarity HCI solution
d. Based on the data d above, draw a titration curve on graph paper by plotting pH and
volume of NaOH added.
e. From HCI-NaOH titration curve, what is the pH of the mixture empirical/graphics
equivalent?
f. From the titration curve of acetic acid-NaOH, determine Ka of acetic acid
g. What is the approximate pH of the mixture equivalence CH3COOH-NaOH
graphically/empirical ?
h. What is pH from CH3COOH-NaOH mixture equivalent theoretically, if Ka
CH3COOH= 1,8.10-5

Answer :

a. Mass of Oxalic acid = 1,2607 grams

Volume of solution = 10 mL
Molarity of Oxalic acid = 0,1 M
Normality of Oxalic acid = 0,2 N

Molarity of Oxalic acid = mass/Mr. Volume

=1,2607/1.2607 . 0.1
=0,1 M

b. Volume of Oxalic acid = 10 mL

Volume of NaOH (1) = 39 mL
Volume of NaOH (2) = 40 mL
Average volume of NaOH = 39,5mL

V1 . N1 = V2 . N2
10 . 0,2 = 39,5 .N2
N2 = 10 . 0,2 / 39,5
N2 = 0,0506 N
M2 = 0,0506/1
= 0,0506 M
c. Volume of HCl 0,1 M = 10 mL
Volume of NaOH(1) 0,1 M = 15 mL
Volume of NaOH (2) 0,1 M = 15 mL
Volume of average NaOH = 15 mL

V1 . N1 = V2 . N2

15 . 0,0506 = 10 . N2

N2 = 15 . 0,0506 / 10
= 0,0759 N
M2 = 0,0759 / 1
= 0,0759 M
d. Graph

Volume of NaOH added to 10 mL 0,0759 M HCl

14

12

10

8
pH

0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Volume of NaOH added to 10 mL 0,0759 M
 Volume of NaOH added to 10 mL of 0,1 M CH3COOHl
14 12.8
12.3
11.88 12.1 12.1
12 11.3

10
pH theoritical

8
5.8
5.1 5.4
6 4.7 4.99
4.2 4.47
3.86
4 2.9

0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
pH 2.9 3.86 4.2 4.47 4.7 4.99 5.1 5.4 5.8 11.3 11.88 12.1 12.1 12.3 12.8
Volume of NaOH

pH

e. pH of the mixture empirical/graphics equivalent is 7 (neutral)

f. [OH-] = . []

0,000289 = 10-14 / Ka . 0,045

Ka = 0,045 . 10-14 / 0,000289
Ka = 1,8 . 10-5
g. The pH = 8,9
h. n1 = n2

1 = 0,115 . V2
V2 = 1 : 0,115
=8,7

[OH-] = . []

[OH-]2 = 10-14 / Ka 1/22
Ka = 1,75 . 10-5