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Results

Volume of NaOH used (mL) Titration 1 (pH) Titration 2 (pH)

0 3.09 3.14

1 4.43 4.16

2 8.95 8.87

3 10.09 10.19

4 10.41 10.47

5 10.59 10.58

6 10.72 10.79

7 10.76 10.87

8 10.89 10.95

9 10.96 11.01

10 11.02 11.04

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0

2

4

6

8

10

12

0 2 4 6 8 10 12

Titration 1 (pH)

Titration 1 (pH)

Half-equivalence point

E uivalence oint

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0

2

4

6

8

10

12

0 2 4 6 8 10 12

Titration 2 (pH)

Titration 2 (pH)

Half-equivalence point

E uivalence oint

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Calculations

Titration 1

By us i ng equa t i on ( 10) , pH a t ha l f - equ i va l ence po i n t = pKa = 5 .3

To ca l cu l a t e t he ac i d i on i za t i on cons t an t , by us i ng equa t i on ( 5 ) :

p Ka = l og Ka

5 .3 = - l og Ka

Ka = a r c l og - 5 .3

= 5 .01 x 10- 6

Titration 2

By us i ng equa t i on ( 10) , pH a t ha l f - equ i va l ence po i n t = pKa = 5 .3

To ca l cu l a t e t he ac i d i on i za t i on cons t an t , by us i ng eq ua t i on ( 5 ) :

p Ka = l og Ka

5 .3 = - l og Ka

Ka = a r c l og - 5 .3

= 5 .01 x 10- 6

Average Ka for titration 1 and titration 2 = (5. 01 x 1 0- 6

+ 5 .01 x 10- 6

)

2

= 5 .01 x 10- 6

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L i kewi s e , f r om t he r eac t i on equa t i on ,

HA ( aq ) + H 2 O ( l i q . ) H 3 O + ( aq ) + A - ( aq )

b ycons i de r i ng t he f o l l owi ng I CE t ab l e :

componen t s HA H 3 O+

A-

I n i t i a l

concen t r a t i on

0 . 1 0 0

Change i n

concen t r a t i on

0 .1 x + x + x

Concen t r a t i on a t

Equ i l i b r i um

0.1 - x x x

The concen t r a t i on o f H 3 O+

i s ca l cu l a t ed us i ng t he ave r age va l ue o f pH

r ead i ng f r om t he t i t r a t i on conduc t ed i n t he ea r l i e r expe r i men t .

pH = - l og [ H 3 O+]

3 .12 = - l og [ H 3 O+]

[ H 3 O+] = a r c l og ( - 3 .12)

= 7 .59 x 10 - 4

Thus , t he ac i d i on i za t i on cons t an t , Ka o f t he unknown monopr o t i c ac i d i s

d t e r mi ne by us i ng equa t i on ( 1 ) :

Ka = ( x ) ( x ) __

0 .1 (x)

= ( x )2_ __

0 .1 (x)

= ( 7 .59 x 10- 4

)2

0 .1 ( 7 .59 x 10 - 4 )

= 5 .80 x 10- 5

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Thus , t he pKa i s equa l t o :

p Ka = - l og Ka

= - l og ( 5 .80 x 10- 5

)

= 4 .24

SAMPLE ERROR CALCULATI ON

The unknown ac i d i s i den t i f i ed t o be po t as s i um hydr ogen ph t ha l a t e .

However , t heor e t i ca l l y , t he pKa va l ue f o r po t as s i um hydr ogen ph t ha l a t e i s

5 .5 , wher eas t he Ka va l ue co r r es ponds t o t he unknwon monopr o t i c ac i d

f r om t he t i t r a t i on i s 3 .16 x 10- 6

. Thus , t he pe r cen t age e r r o r s f o r each

t heor e t i ca l va l ue t o t he ca l cu l a t ed va l ue a r e :

Pe r cen t age e r r o r ( pKa) = 5 .50 5 .30 x 100%

5.50

= 3 .64 %

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Discussion

The experiment is aimed at determining the Ka value of an unknown acid which then the value

of Ka will be used to determine the unknown acid. Therefore a table of values of acids with their

respective Ka is used to compare the value calculated and to identify the acid. The first part in

determining the Ka is to determine the equivalence point from the graph obtained. Then the

value pKa is obtained by taking half of the value of equivalence point and the pKa is obtained.

The given solution was the revealed to be p o t a s s iu m h ydr o ge n p h t ha l a t e . After analyzing

the pKa value, it is similar to potassium hydrogen phthalate. The pKa value of the unknown acid

in the experiment is 5.3 whereas theoretical potassium hydrogen phthalates pKa value is 5.5.

With the main objective of the experiment is to determine the pKa value of the unknown acid so

as to identify the acid, however the precise value could not be obtained due to errors whilst

conducting the experiment. An error of 3.64 percent is obtained after comparing the theoretical

value with the experimental value is obtained. Therefore, a few errors or lack of awareness of theprecautions that must be considered when conducting the experiments may be the reasons that

lead to the errors in the value obtained.

First error that might affect the calculated values is done during when rinsing the burette with

distilled water. There are some water left inside the tube of the burette and this would affect the

concentration of the solution entering the burette. Hence, the titration curves might not have the

accurate values as the concentration of sodium hydroxide is not perfectly as set before the

experiment is conducted.

Secondly, the pH meter that has been used has never given the definite readings, as the valuesthat are shown are always changing in the region of about plus minus 0.1 pH. Thus, one can

never tell the accurate readings of pH values. The small difference in the pH would affect the

curve of the titration graph. Therefore, these will also affect the titration curves as well as pKa

values which correspond to the pH values at half equivalence point.

Besides that, the equivalence point is not necessarily being at pH of 7 as it occurs just when the

concentration of acid is equal to the concentration of base reacted in solution. Therefore, the final

pH depends on the major species of ions left in the solution after the reaction.