ABSTRA CT / SUMMARY
An experiment is conducted to identify an unknown monoprotic acid by obtaining the pKa value form titration curves. 10 m i l l i l i t r e s o f t h e u n k n o w n a c i d i s t i t r a t e d w i t h a pp r o x i m a t e l y 0 . 1 M o f s o d i u m h y d r o x i d e s o l u t i o n . T h r e e t r i a l s a r e c o n d u c te d a n d t h e r e f o r e t h r e e t i t r a t i o n c u r v e s a r e p l o t t e d . T h e v a l u e s o f pK a a n d K a o f t h e u n k n o w n a c i d f r o m t h e t i t r a t i o n p r o c e s s a r e 4 . 5 3 7 an d 2 . 9 0 x 1 0 - 5 respectively. Whereas, the pKa and Ka values obtai ned from using t h e i n i t i a l p H o f t h e u n k n o w n a c i d a r e 4 . 8 1 a n d 1 . 5 3 x 1 0- 5 respectively. Theoretically, the pKa and Ka values of the un known monoprotic acid, which is believed to be of acetic acid are 4.75 and 1.76 x 10- 5. Therefore, an error of 4.48 percents of the p Ka value and 64.77 percents of the Ka value from the titration process are calculated. Meanwhile, an error of 1.26% and 13.1% of pKa an d Ka v a l u e s o b t a i n e d f r o m u s i n g t h e i n i t i a l p H o f t h e u nk n o w n a c i d i s calculated. Therefore, there is only a slight di fference from the theoretical value compared to the values from th etitration process. Hence, the unknown monoprotic acid i s identified as acetic acid and the value determined from using the initial pH valu e of the acid is a more accurate method. The experiment is co mpleted and successfully conducted.
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IN
TRODUCT I O N
Acid-base titration is a method of neutralization process which p r o v i d e s i n f o r m a t i o n r e g a r d i n g t h e p r o p e r t i e s a s w el l a s n a t u r e o f either acid or base applied d uring the process. Likewise, it is useful to determine the molecular mass and pKa values of the substances. The end-point of such titration can be monitored by using indicators or as what is used in this experi ment, a pH meter with electrodes.
A titration curve is a graph of measured pH values obtained f r o m p H m e t e r r e a d i n g s v e r s u s v o l u m e o f t i t r a n t b e in g a d d e d i n millimetres. Basically, the most important infor mation that is supposed to be obtained from acid-base titration is the equivalence point. The equivalence point i s theoretically reached when the number of moles of base being add ed into the Erlenmeyer flask is equal to the number of moles of aci d. It occurs in the titration curve in the region where there is a large and noticeable change in pH with a relatively small change in volu me of titrant.
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A I MS / OBJECT I V E S
The experiment is conducted to achi eve main objective, which is to identify the unknown monoprotic acid by comparing the calculated average Ka values from the data obtained from this experiment with the Ka values of some common a cids found in a general chemistry book. In order to get the Ka values for the unknown acid, it can be obtained by performing titration process based on the pH value. Beside s that, the Ka value can also be determined by using the initial pH of the unkno wn monoprotic acid.
THEORY
n this experiment, we will be dealing with monoprotic acid. Based on Brønsted and Lo wry, an acid is a proton donor whereas a base is a proton acceptor. This port rays a very important idea to u n d e r s t a n d i n g m o n o p r o t i c a n d p o l y p r o t i c a c i d s a n d ba s e s s i n c e monoprotic, as a matter of fact, is basically referred to the transfer o f o n e p r o t o n . O n t h e c o n t r a r y , p o l y p r o t i c c o r r e s p on d s t o t h e t r a n s f e r o f m o r e t h a n o n e p r o t o n . T h e r e f o r e , m o n o p ro t i c a c i d i s a n acid that can donate one proton while polyprotic acid is an acid that can donate more than one proton. To be more precise, monoprotic I
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acid can release one proton per molecule and hence, have only one equivalence point.
When a weak monoprotic acid (HA) is dissol ved in water, only c e r t a i n o f i t s m o l e c u l e s w i l l d i s s o c i a t e t o y i e l d hy d r o n i u m i o n s , H3O+, and A- ions. At this point, the reactio n has reached dynamic equilibrium. Consider the following reaction: HA (aq) + H 2 O ( l i q . ) < í > H 3 O + ( a q ) + A - ( a q )
U n d e r s u c h e q u i l i b r i u m c o n d i t i o n s , t h e t o t a l c o n c e nt r a t i o n s o f e a c h s p e c i e s r e m a i n c o n s t a n t , e v e n t h o u g h t h e s p e c ie s i n s o l u t i o n a r e c o n s t a n t l y d i s s o c i a t i n g a n d r e c o m b i n i n g . T h e i on i z a t i o n c o n s t a n t of the weak monoprotic acid is used to characterize the a cid, and is calculated by using the following equation: K a = [ H 3 O + ] [ A - ] . . . . . . . . . . . . . .. .. . . . . . . . . . . ( 1 ) [HA] n this expression, Ka refers to acid i onization constant. Strong a c i d s t y p i c a l l y d i s s o c i a t e c o m p l e t e l y , a n d t h e r e f o re w o u l d h a v e a K a value of greater than 1. Weak acids have Ka values much smaller than 1, typically less than 10-4 to be more precise. For instance, the K a v a l u e o f p r o p a n o i c a c i d i s 1 . 3 x 1 0- 5 a n d i t s p K a v a l u e i s 4 . 8 7 4 . I n p r i o r i t i z i n g t h e c o n v e n i e n c e , s c i e n t i s t s o f t e n u se t h e p K a v a l u e s of weak acid, as it allows them to deal with whole numbers. I
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By assuming a 1:1 molar ratio of acid : base, the volume as well as concentration of sodium hydroxide, which is the base can be used to determine the number of moles of acid present: b a s e = mass b a s e . . . . . . . . . . . . . . . . . . . . . . ( 2 ) M r b a s e acid = base x 1 mol of acid ..................(3) 1 mol of base
The pH of a solution is related to th e hydronium ions, H3O+, concentration by the equation: p H = í l o g [ H 3 O + ] . . . . . . . . . . . . . . .. .. . . . . ( 4 ) and the pKa of an acid is si mply: p K a = í l o g K a . . . . . . . . . . . . . . . . . . . . .. .. . . . . . ( 5 ) By considering equation (1), take ±log on both sides: -log Ka = -log [H3O +] ± log [A-] ..................................(6) [HA]
T h u s , b y s u b s t i t u t i n g e q u a t i o n ( 6 ) i n t o e q u a t i o n ( 5) : pKa = - log [H3O + ] ± log [A -] ..................................(7) [HA]
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and therefore, by substituting equation (4) i nto equation (7): pKa = pH ± log [A -] ..................................(8) [HA]
However, since pKa is constant and pH vari es,hence: pH = pKa + log [A -] ..................................(9) [HA] Therefore, when [A-] = [HA], thus: p H a t h a l f - e q u i v a l e n c e p o i n t = p K a . . . . . . . . . . . . . . . . .. .. . . . . . . ( 1 0 )
This is the midway point to the equi valence point. A titration c u r v e i s n e e d e d t o a n a l y z e t h e a c i d i o n i z a t i o n c o n st a n t , K a v a l u e s o f the unknown monoprotic acid to be iden tified in this experiment. The t i t r a t i o n c u r v e r e p r e s e n t s t h e p H v a l u e s a t p a r t i c ul a r v o l u m e o f sodium hydroxide being added. Thus, pKa values can be read d i r e c t l y f r o m t i t r a t i o n c u r v e . F r o m t h e t i t r a t i o n da t a c o l l e c t e d f r o m the experiment conducted, three titration curves of pH versus volume of base added in millilitres are plotted. On each curve, the volume of base at equivalence point i s clearly marked. Next, the volume of base at half-equivalence poin t at each curves are determine and clearly marked. The pH values at that particular halfequivalence point are then extrapolated. The pH val ues can now be used to calculate the acid ionizatio n constant and the average value is determined. In addition, the volume of base titrated at the equivalence point is used to cal culate the concentration of the 6
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unknown monoprotic acid. The average is determined as well. The average acid ionization constant is then used to identify the unknown monoprotic acid.
APPARAT US / REAGE N T S
~ 5 0 - m i l l i l i t r e b u r e t t e , 1 0 - m i l l i l i t r e g r a d u a t e d c yl i n d e r s , 5 0 a n d 2 5 0 millilitre beakers, Erlenmeyer flask, retort stand and burette clamp, p H m e t e r , s p a t u l a , a n a l y t i c a l b a l a n c e , f i l t e r f u n n el , m a g n e t i c s t i r r e r , p e l l e t s o f s o d i u m h y d r o x i d e , d i s t i l l e d w a t e r , 1 0 m il l i l i t r e s o f u n k n o w n monoprotic acid.
EXPER I M E N TAL PROCEDURE
1. An approximately 2.0 grams of pellets of sodium hydroxide, NaOH is weighted to the nearest fo ur decimal points and dissolved in 500 millilitres of distilled water.
2 . A b u r e t t e i s c l e a n e d , r i n s e d a n d f i l l e d wi wi t h
N
aOH solution. 7
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3 . 1 0 m i l l i l i t r e s o f u n k n o wn wn m o n o p r o t i c a c i d i s p r e p a r e d a n d t h e n transferred into the Erlenmeyer flask.
4 . T h e f l a s k i s t h e n p l a c e d o n a s t i r p l a t e a n d a ma ma g n e t i c s t i r r e r i s i n s e r t e d i n t o t h e f l a s k a n d l e t t o s t i r t h e s o l u t i on .
5 . A p H me m e t e r i s c a l i b r a t e d u s i n g b u ff ff e r s o l u t i o n b e f o r e t h e electrode is being rinsed well with distilled water and blotted dry.
6. The pH electrode is inserted into the flask and the position of the electrode is adjusted so that the magnetic stirrer does n ot hit it. The acid is titrated and following increments: At pH 1 until 5.5 At pH 5.5 until 10.5 At pH 10.5 until 12.5 7.
the pH readi ng as recorded with the : 1 millilitre of N a O H a t a t i m e : 0 . 5 mi mi l l i l i t r e o f N a O H a t a t i m e : 1 millilitre of NaOH at a time
8. The titration is continued to at least a pH of 12. 9 . T h e d a t a o f p H r e a d i n g s a n d t h e v o l u m e o f N a O H s o l u t i o n a d d e d and titrated are recorded and tabulated. 10. A titration curve of pH versus volume of NaOH solution titrated is plotted. Based on the titration curve, pKa value of th e acid is calculated and the acid is identified.
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RESULTS Mass of sodium hydroxide pellets dissol ved : 2.0823 grams Vol. NaOH
pH
pH
pH
Average
(mL)
(Trial 1)
(Trial 2)
(Trial 3)
pH
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 7.5 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0
2.89 4.08 4.23 4.42 4.68 4.85 5.17 5.89 6.33 11.17 11.55 11.73 11.83 11.90 11.93 12.00 12.04
2.87 3.70 4.09 4.32 4.55 4.79 5.04 5.46 6.11 10.63 11.34 11.52 11.64 11.72 11.77 11.80 11.84
2.98 3.67 4.05 4.26 4.53 4.72 5.02 5.43 5.90 10.56 11.26 11.52 11.67 11.77 11.82 11.87 11.90
2.91 3.82 4.12 4.33 4.59 4.79 5.08 5.59 6.11 10.79 11.38 11.63 11.71 11.80 11.84 11.89 11.93
Volume of NaOH at equivalence point (mL) pH at half-equivalence point § pKa Acid ionization constant, Ka
7.5951 4.537 2.90 x 10-5
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Titration curve of pH versus Volume of NaOH (mL)
pH 13 12 11 10 9
8 7 6 5
Vol. of NaOH (mL)
4 3 2
1 0 0
2
4
6
8
10
12
14
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SAMPLE CALCULAT I O N
By using equation (2): b a s e = mass b a s e Mr base = 2.0823 g 39.997 g/mol = 0.0521 mol 10 | P a g e
From the reaction equation, number of moles of base is equi valent to the number of moles of acid. Thus, acid = base = 0.0521 mol By using equation (10), pH at half-equivalence p oint = pKa = 4.535 To calculate the acid ioni zation constant, by using equation (5): pKa = ílog Ka 4.537 = -log Ka Ka = arc log -4.535 = 2.90 x 10-5 Likewise,from the reaction equation, HA (aq) + H 2 O ( l i q . ) < í > H 3 O + ( a q ) + A - ( a q ) by considering the following ³ICE´ table: components nitial concentration Change in concentration Concentration at Equilibrium I
HA 0.1
H3 O + 0
A 0
0.1 ± x
+ x
+ x
0.1 - x
x
x
the concentration of H3O+ is calculated using the average value of pH reading from the titration conducted in th earli er experiment. 11 | P a g e
pH = -log [H3O +] 2.91 = -log [H3O +] [H3O+] = arc log (-2.91) = 1.23 x 10-3 Thus, the acid ionization constant, Ka of the unknown monoprotic acid is dtermine by using equation (1) : Ka
=
x.x__ 0.1 ± x = x 2 ___ 0.1 ± x = (1.23 x 10-3)2 0 . 1 ± 1 . 2 3 x 1 0- 3 = 1.53 x 10-5
Thus, the pKa is equal to: pKa = -log Ka = -log (1.53 x 10-5) = 4.81
SAMPLE ERROR CALCULAT I O N T h e u n k n o w n m o n o p r o t i c a c i d i s i d e n t i f i e d t o b e a c et i c a c i d . H o w e v e r , t h e o r e t i c a l l y , t h e p K a v a l u e f o r a c e t i c a ci d i s 4 . 7 5 , w h e r e a s t h e K a v a l u e c o r r e s p o n d s t o t h e u n k n w o n m o no p r o t i c a c i d from the titration is 1.76 x 10- 5. Thus, the percentage errors for each theoretical value to the calculated value are: Percentage error (pKa) = 4.75 ± 4.537 x 100% 4.75 12 | P a g e
= 4.48 %
Percentage error (Ka) = 1.76 x 10-5 - 2.90 x 10- 5x100% 1.76 x 10-5 = 64.77 % For the identifying the unknown acid using the pH value in order to determine the Ka value,the value is sl ightly different from the one obtained from the titration process. The percentage error is calculated to be: P e r c e n t a g e e rr rr o r ( p K a ) = 4 . 7 5 ± 4 . 8 1 x 1 0 0 % 4.75 = 1.26% Percentage error (Ka)
= 1.76x10-5 ± 1.53x10- 5x100% 1 . 7 6 x 1 0- 5 = 13.1%
D I SCUSS I O N A lot of information is required in order to assure that the identity of the unknown acid is conclusive. Notice that one equivalence point is obtained , therefore one pKa value. After analyzing it as well the Ka value, it is concluded to be remarkably 13 | P a g e
s i m i l a r t o a c e t i c a c i d . T h e u n k n o w n ¶ s K a v a l u e i s 2 . 9 0 x 1 0- 5 whereas acetic acid¶s Ka is merely di fferent which value is 1 . 7 6 x 1 0 - 5 . A s f o r t h e p K a v a l u e o f t h e u n k n o w n a c i d i s c a l c ul a t e d to be 4.537 whereas acetic acid¶s p Ka value is 4.75.
The main objective of the experiment i s to determine the Ka value of the unknown monoprotic ac id so as to identify the acid. However, an error of 64.77 percents after comparing the theoretical v a l u e w i t h t h e e x p e r i m e n t a l v a l u e i s o b t a i n e d . T h e re f o r e , a f e w m i s t a k e s o r l a c k o f a w a r e n e s s o f t h e p r e c a u t i o n s t ha t m u s t b e considered when conducting the e xperiments may be the reasons that lead to the erroneous calculations.
First error that might affect the calculated values is done d u r i n g w e i g h i n g t h e p e l l e t s o f s o d i u m h y d r o x i d e . T he w e i g h t o f a n empty beaker should have been con sidered as well. Then, it must be subtracted from the weight of the beaker containi ng the pellets. H e n c e , t h e t i t r a t i o n c u r v e s m i g h t n o t h a v e t h e a c c ur a t e v a l u e s a s the concentration of sodium hydroxide is n ot perfectly 0.1 M.
S e c o n d , t h e p H m e t e r t h a t h a s b e e n u s e d h a s n e v e r gi v e n t h e definite readings, as the values that are sho wn are always changing r a p i d l y . T h u s , o n e c a n n e v e r t e l l t h e a c c u r a t e r e a di n g s o f p H v a l u e s . T h e r e f o r e , t h e s e w i l l a l s o a f f e c t t h e t i t r a t i o n c u rv e s a s w e l l a s p K a v a l u e s w h i c h c o r r e s p o n d t o t h e p H v a l u e s a t h a l f e qu i v a l e n c e p o i n t .
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B e s i d e s t h a t , t h e e q u i v a l e n c e p o i n t i s n o t n e c e s s a ri l y b e i n g a t pH of 7 as it occurs just when the c oncentration of acid is equal to the concentration of base reacted in sol ution. Therefore, the final pH depends on the major species of ion s left in the solution after the reaction.
n a d d i t i o n , t h e p H e l e c t r o d e m i g h t h a v e c o m e i n t o co n t a c t w i t h the magnetic stirrer. Therefore, a problem might be encountered during the readings of pH values on the pH meter. I
C O N CLUS I O N
T h e e x p e r i m e n t a l v a l u e s o f p K a a n d K a o f t h e u n k n o wn a c i d are 4.537 and 2.90 x 10-5 respectively. Whereas, the pKa and Ka v a l u e s o b t a i n e d f r o m u s i n g t h e i n i t i a l p H o f t h e u nk n o w n a c i d a r e 4.81 and 1.53 x 10-5 respectively. Theoretically, the pKa and Ka values of the unknown monoprotic a cid, which is believed to be of acetic acid are 4.75 and 1.76 x 10-5. Therefore, an error of 4.48 percents of the pKa value and 64.77 percents of the Ka value from the titration process are calculated. Meanwhile, an e rror of 1.26% and 13.1% of pKa and Ka values obtained from using the initial pH of the unknown acid is calcul ated. Therefore, there is only a slight d i f f e r e n c e f r o m t h e t h e o r e t i c a l v a l u e c o m p a r e d t o th e v a l u e s f r o m t h e t i t r a t i o n p r o c e s s . H e n c e , t h e u n k n o w n m o n o p r o t i c ac i d i s i d e n t i f i e d 15 | P a g e
as acetic acid and the value deter mined from using the initial pH value of the acid is a more accurate method.
RECOMME N D A T I O N S
T h e r e a r e a f e w r e c o m m e n d a t i o n s , a n d p r e c a u t i o n s t ha t h a v e to be considered during the experi ments in order to get an accurate value and readings of data.
Firstly, the standard solution that is used should be a hundred percent pure and stable at room temperatures. Thus, it i s more preferable to use a dried standard material before weighing and diluted.
Secondly, in order to be more conclusive in identifying the u n k n o w n m o n o p r o t i c a c i d , t h e m o l e c u l a r w e i g h t o f t he a c i d s h o u l d b e considered as well. This then can b e used to compare it with the t h e o r e t i c a l v a l u e o f m o l e c u l a r w e i g h t o f a c e t i c a c id . T h e r e f o r e , i t i s more preferable to obtain the mass of solid acid and then only it is diluted and titrated.
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REFERE N C E S
~ S t e v e n L . M u r o v , 2 0 0 4 , E x p e r i m e n t s i n G e n e r a l C h e m i s t r y , 4t h Edition, United States: Thomson/Brooks/Cole. ~ http://en.wikipedia.org/wiki/Acid-base_titration ~ http:// chemlab.com/Chemistry_3_5_ _Monoprotic_and_Polyprotic_Acids_-_Chemistry ~ http://www.titrationexperiment.com/html/
APPE N D I C E S
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T a b l e 1 : W e a k A c i d s , K a , a n d p K a v a l u e s Acid Acetic Ammonium Benzoic Carbonic Chloroacetic Citric
Formic Phosphoric
y
HA CH3COOH NH4+ C6H5C O O H H2CO3 HCO3CH2ClCOOH C6O7H8 C6O7H72C6O7H6 HCOOH H3P O4 H2P O4 2HP O4
ACH3C OONH3 C6H5COOHCO3CO3 2CH2C l C O O C6O7H7 C6O7H6 23C6O7H5 HCOO H2P O4-2 HP O4 3PO4
Ka 1.76 x 10-5 5.6 x 10-10 6.46 x 10-5 4.3 x 10-7 4.8 x 10-11 -3 1.4 x 10 7.41 x 10-4 1.74 x 10-5 -7 3.98 x 10 1.77 x 10-4 7.52 x 10-3 -8 6.23 x 10 -13 2.2 x 10
pKa 4.75 9.25 4.19 6.37 10.32 2.85 3.13 4.76 6.40 3.75 2.12 7.21 12.67
Refer to the attachment provided on the ne xt page.
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