ACID BASE TITRATIONACID – BASE TITRATIONAn application method ofAn application method of
Inorganic Pharmaceutical Analysis
Lecturer : Dr Tutus GusdinarLecturer : Dr. Tutus GusdinarPharmacochemistry Research Group
School of Pharmacy INSTITUT TEKNOLOGI BANDUNGINSTITUT TEKNOLOGI BANDUNG
The application of neutralization reaction
l ld b d dNeutralization reaction could be used in determination of either natural or tried acid/basic analytes.
Water is commonly used as a good solvent because of its y gcheap cost, easy to be obtained and prepared, non‐toxic, and low coeficient of expansion (Aquous Titration).p ( q )
Several analytes cannot be titrated in water caused ofSeveral analytes cannot be titrated in water, caused of its low solubility or too weak acid/base properties, these should be titrated in non aquaous solvent (Non Aquaousshould be titrated in non‐aquaous solvent (Non‐Aquaous Titration).
Neutralization titration reagentsg
• Standard acid solution should be Sta da d ac d so ut o s ou d bestandardized with a Primary Standard Basessuch as Na‐carbonate TRIS atau THAM (trissuch as Na‐carbonate, TRIS atau THAM (tris hydroximethyl aminomethane), Na‐t t b t M i idtetraborate, Mercuric oxide.
• Standard basic solution (attention CO2 effect ( 2to distilled water) should be standardized with a Primary Standard Acid such as KH‐with a Primary Standard Acid such as KH‐Phtalate, Benzoic acid, Sulphamic acid, KH‐i d t S l h li ili idiodate, Sulphosalicilic acid.
Sulfuric acid molecular modelSulfuric acid molecular model
ACID – BASE INDICATORS
What is an acid‐base indicator? An acid‐base indicator is a weak acid or a eak base The undissociated form of theweak base. The undissociated form of the
indicator is a different color than the iogenic form of the indicator. An Indicator does not change color from pure acid to pure alkalinechange color from pure acid to pure alkaline at specific hydrogen ion concentration, but rather, color change occurs over a range of hydrogen ion concentrations This range ishydrogen ion concentrations. This range is termed the color change interval. It is expressed as a pH range.
How is an indicator used ?
Weak acids are titrated in the presence of indicators which change under slightly alkaline conditions.alkaline conditions.
W k b h ld b tit t d i thWeak bases should be titrated in the presence of indicators which change p gunder slightly acidic conditions.
What are some common acid‐base indicators?Several acid base indicators are listed below some moreSeveral acid‐base indicators are listed below, some more than once if they can be used over multiple pH ranges. Quantity of indicator in aqueous (aq ) or alcohol (alc )Quantity of indicator in aqueous (aq.) or alcohol (alc.) solution is specified. T i d d t i di t i l d th l blTried‐and‐true indicators include: thymol blue, tropeolin OO, methyl yellow, methyl orange, bromphenol blue, bromcresol green, methyl red, bromthymol blue, phenol red, neutral red, y , p , ,phenolphthalein, thymolphthalein, alizarin yellow, tropeolin O nitramine and trinitroben‐zoic acidtropeolin O, nitramine, and trinitroben zoic acid. Data in this table are for sodium salts of thymol blue, bromphenol blue, tetrabromphenol blue, bromcresol green, methyl red, brom‐, p , g , y ,thymol blue, phenol red, and cresol red.
Acidic : HIn + H2O H3O+ + In-
B i I + H O HI + OHBasic : In- + H2O HIn + OH-
[H O+][In ] [HIn][H3O+][In-] [HIn]Ka = ---------------- pH = pKa - log -------
[HIn] [In ][HIn] [In-]
End point colour depends to the most dominantEnd point colour depends to the most dominant concentration of such indicator form. Ex : if HIn is red and In- is yellow then :Ex : if HIn is red and In- is yellow, then :
At low pH in which [HIn] is dominant the ratio ofAt low pH in which [HIn] is dominant, the ratio of 10/1 (red). At high pH in which [In-] is dominant, the ratio of 1/10 (yellow) At medium pH in which [HIn] =ratio of 1/10 (yellow). At medium pH in which [HIn] = [In-], the ratio of 1 (orange).
Indicator pH rangeIndicator pH range
Y ll l d l tiYellow coloured solution :
pHyellow = pKa + log 10/1 = 5 +1 = 6p yellow p g
Red coloured solution :
H K l 1/10 5 1 4pHred = pKa + log 1/10 = 5 ‐1 = 4
_____________________________
ΔpH = pHyellow – pHred = 2
pH range = 4 ‐ 6pH range 4 6
PHENOLPHTALEIN
H2In HIn-HIn
In-2
The first useful theory of indicator action was suggested by W 0stwald based upon the concept that indicators inby W. 0stwald based upon the concept that indicators in general use are very weak organic acids or bases.
The simple Ostwald theory of the colour change of indicators has been revised, and the colour changes are b li d t b d t t t l h i l di thbelieved to be due to structural changes, including the production of quinonoid and resonance forms; these may be illustrated by reference to phenolphthalein thebe illustrated by reference to phenolphthalein, the changes of which are characteristic of all phthalein indicators. In the presence of dilute alkali the lactone ring in (I) opens to yield (II), and the triphenylcarbinol structure (II) undergoes loss of water to produce the resonating ion (III) hi h i d If h l hth l i i t t d ith(III) which is red. If phenolphthalein is treated with excess of concentrated alcoholic alkali the red colour first produced disappears owing to the formation of (IV)produced disappears owing to the formation of (IV).
The chemical structure change of phenolphtalein indicator
PHENOL RED
H In+ HInH2In+(RED)
HIn(YELLOW)
In-(RED)
METHYL ORANGE( HELIANTHINE )
HIn+
In(RED)
In(YELLOW)
Acid‐Base indicators listNAME pH range pKa
(μ = 0,1 M)COLOUR CHANGE TYPE
Thymol Blue 1.2‐2.88.0‐9.6
1.658.90
Red‐YellowYellow‐Blue
Acid
Methyl Yellow 2.9‐4.0 Red‐Orange BasicMethyl Yellow 2.9 4.0 Red Orange Basic
Mehtyl Orange 3.1‐4.4 3.46* Red‐Orange Basic
Bromcresol Green 3.8‐5.4 4.66 Yellow‐Blue Acid
Methyl Red 4.2‐6.3 5.00* Red‐Yellow Basic
Bromcresol Violet 5.2‐6.8 6.12 Yellow‐Violet Acid
Bromthymol Blue 6.2‐7.6 7.10 Yellow‐Blue Acid
Phenol Red 6.8‐8.4 7.81 Yellow‐Red Acid
Cresol Violet 7 6 9 2 Yellow Violet AcidCresol Violet 7.6‐9.2 Yellow‐Violet Acid
Phenolphtalein 8.3‐10.0 Colourless‐Red Acid
Thymolphtalein 9.3‐10.5 Colourless– Blue Acid
Alizarin Yellow 10.0‐12.0 Yellow‐Violet Basic
0.1 M1 M
0.01 M
ALCALIMETRY
ACIDIMETRY
Mixed IndicatorWhen sharp indicator colour change could not be obtained at the end
point, a mixture of two indicators or an indicator mixture could be used.
INDICATOR COLOUR CHANGE
Methyl orange 1 gram violet – grey – greenIndigo carmine 2.5 gramDissolved in 1 liter of water
(acid) pH=4 (basic)
B l G 0 1 % 3 t dBromcresol Green 0.1 % 3 partMethyl Red 0.1 % 2 part
red – green(acid) pH=5.1 (basic)
Phenolphtalein 0 1 % 1 part green – pale blue – violetPhenolphtalein 0.1 % 1 partMethylene Green 0.1 % 2 part
green pale blue violet(acid) pH=8.8 (basic, pH>9)
Cresol Red 0.1 % 1 part yellow – pink – violetThymol Blue 0.1 % 3 part (acid) pH=8.2 (basic, pH>8.4)
Cresol Red 0.1 % 16 mlM th l R d 0 1 % 20 l
green – greenish grey – greenish violet –i l tMethyl Red 0.1 % 20 ml
Methylene Blue 0.2 % 4 mlviolet
pH=8.85 pH=8.35 pH=8.6 pH=8.8
Carbonate TitrationCarbonate Titration
CO32‐ + H3O+ HCO3
‐ + H2O pKa2 = 6 34CO3 + H3O HCO3 + H2O pKa2 = 6.34HCO3
‐ + H3O+ H2CO3 + H2O pKa1 = 10.36
By ΔpKa = 4.02 unit the equivalent point might be sharp, but Ka1 is too low and a sharp equivalent point could not be obtained.Carbonate titration should be titrated with a strong acid using phenolphtalein indicator acid (pH = 8.0‐9.6) :
H ½ [ K + K ] 8 35pHNaHCO3 = ½ [pKa1 + pKa2] = 8.35Methyl orange (pH = 3.1‐4.4) at the second equivalent point TE‐2, saturated solution of CO2 has pH = 3.9. The CO2 gaz could besaturated solution of CO2 has pH 3.9. The CO2 gaz could be removed by these two techniques :• Neutralization of sample using methyl orange indicator, or• Removing CO2 by boiling process of distilled water.
Tit ti f b tTitration curve of carbonate
Phenolphtalein
pH
Methyl Orange
ml HClml HCl
Titration of Carbonate‐Bicarbonate Mixture
First end point (phenolphtalein) : completed neutralization ofNaOH Na CO is half neutralized HCO ‐ has not reacted yetNaOH, Na2CO3 is half‐neutralized, HCO3 has not reacted yet.
S d d i t ( th l ) ll f HCO ‐ t li dSecond end point (methyl orange) : all of HCO3‐ was neutralized,
small drops of titrant (HCl) can change the pH from 8 to 4 (could be corrected with indicator blanco).be corrected with indicator blanco).
The mixture of NaOH + NaHCO3 solution could not be preparedThe mixture of NaOH + NaHCO3 solution could not be prepared because of : HCO3
‐ + OH‐ CO32‐ + H2O
The reaction results should be a mixture of HCO3‐ + CO3
2‐ or the only CO3
2‐, depends on the relative amount of the compound content 3in a sample.
Titration curve of a mixture ofcarbonate + Bicarbonate
1312
OH- + H3O+ H2O
CO32- + H3O+ HCO3
- + H2O
pH
VV1 HCO3- + H3O+ H2CO3 + H2O
V2
ml HCl50 100
Titration of mixed two acidsLike as diprotic acid : [HX]initial = [HY]initialIf HX (with its Ka1) is strong acid and HY (with its Ka2) is weak acid,h f ibl i i d K K 4 ithen a feasible titration can occured at pKa1 – pKa2 > 4 unitFor unequal initial concentration, first equivalent point could be calculated by such following steps :calculated by such following steps :1) Charge balance [Na+] + [H3O+] = [OH‐] + [X‐] + [Y‐]2) [Na+] = acid formal concentration = [HX] + [X‐]2) [Na ] = acid formal concentration = [HX] + [X ]3) 1)+2) : [H3O+] = [OH‐] + [Y‐] – [HX]4) Substitute [OH‐], [Y‐] and [HX] from Kw, Ka1, Ka2 :4) Substitute [OH ], [Y ] and [HX] from Kw, Ka1, Ka2 :
[H3O+] = Kw/ [H3O+] + Ka2[HY]/[H3O+] ‐ [H3O+][X‐]/Ka15) [H3O+] = {Ka1 Kw + Ka1 Ka2 [HY]}/Ka1+[X‐]) [ 3 ] { 1 1 2 [ ]}/ 1 [ ]6) If Ka2[HY] >>>Kw and [X‐] >>Ka1 then
[H3O+] = Ka1Ka2[HY]/[X‐]7) pH = ½ (pKa1 + pKa2) – ½ log [HY]/[X‐]
Titration of mixed HCl + HAcHCl titrated first, pH is not influenced by H3O+ from HAc[Le Chatelier principle : excess proton will pressure weak acid [ p p p pdissociation]. This assumption is less valid at near the equivalent
point, caused of excess proton concentration increases. At first i l t i t th HCl h ll tit t d d th H d d lequivalent point the HCl has all titrated and the pH depends only
to HAc dissociation. After first equivalent point HAc titration will start In the next curveAfter first equivalent point HAc titration will start. In the next curve
this first equivalent point is not clear because of lower (not enough) ΔpH/ΔV. While the pH of HAc 0.067 M is about 3. B f H 4 hi i i i f ibl (b i i di )Because of pH< 4 this titration is not feasible (by using indicator).
In the next step the weak acid titrated by strong base, the titration will be feasiblewill be feasible.
Example: Titration of 50 ml mixture of HCl 0.10 M and HAc 0.10 M with a solution of NaOH 0.10 M.with a solution of NaOH 0.10 M.
Titration curve of acids mixtureTitration curve of acids mixture
50 ml of HCl 0,10 M and HAc 0.10 M (Ka=1.10-5)titrated with NaOH 0.20 M
pH
50 ml of HCl 0,10 M and HX 0.10 M (Ka=1.10-8) titrated with NaOH 0.20 M
ml NaOH
TITRATION ERRORTITRATION ERROR
A feasible titration can be performed in a complete reaction at the end point,
resulting a sharp pH depletion (vertical curve).
A complete reaction should be obtained in a high p gvalue of K (equilibrium constant), great change of pH at near to end point to obtain easily aof pH at near to end point, to obtain easily a
high precision of the end point.
Titration of strong acid with strong base gives a very high K value :
H O+ + OH‐ H O K = 1/Kw = 1014H3O + OH‐ H2O K = 1/Kw = 10
A high ΔpH occurs at equivalent point, i.e. 5.40 unit pH for a ΔV = 0.10 ml. At this high ΔpH any indicator could be used for obtaining a high g gprecision of titration (ppm).
This is a feasible titration. But it is difficult to calculate exactly the K value for a feasible titration, because of the influence of ,analyte and titrant concentrations to the ΔpH. In certain condition, titration could be perform without high precision.
It is predicted about 99.9% and 99.99% of analyte
h t it ti d t t th i l t i tchanges to its reaction product at the equivalent point.
At this condition the K value could be predicted.
There is a limited ability of human eyes to observe indicator colour change at the equivalent point, a few drops of titrant could change pH of 1‐2 unit.
Example :
Titration of 50 ml HA 0.10 M with a strong base of 0.1 M. gCalculate Kminimum at an addition of 49.95 ml titran (complete reaction), when addition of 2 drops (0,10 ml) of titrant after
i l i h 2 00 iequivalent point changes pH at 2.00 units.
What is Kminimum at pH change of 1 unit ?
Influence of analyte and titrant concentration : Δ H d if th t ti f l tΔpH decreases if the concentration of analyte and titrant decrease.
Example :
Weak acid titration : low of Ka, high of pHequivalent point and low of
ΔpH. Increasing [HA] will decrease ΔpH.Increasing titrant volume should increase titration error.
(resulting a smaller value than true end point).
If [HA] is titrated at a smaller initial volume, then ΔpH increases, caused by little excessive titrant volume.y
If [titrant] increases then ΔpH increases, this diminues titrant Volume and increases titration error (higher than true end point).( g p )
In general:In general:
Precision (at ppm) could be produced by titration of weak acid/base solution of 0.05 M /(dissociation constant of 1x10‐6 M) with a titrant solution of 0 10 M (K = 1x108)titrant solution of 0.10 M (K = 1x10 ).
Salts of weak acid (Bronsted base) could be feasibly titrated by strong acid when itsfeasibly titrated by strong acid when its conjugated acid is too weak.
Example :p
id ( 9) i k fAn acid HA (Ka = 1x10‐9) is too weak for a titration with a base which has dissociation constant of its conjugate base A‐ = 1x10‐5, because of Ka x Kb = 1x10‐14.because of Ka x Kb 1x10 .
A‐ could be titrated by strong acid.
Same case for a weak base and its saltSame case for a weak base and its salt.
Titration error is the difference of reagent amount used between end point and equivalent point in % or o/oo of reactedend point and equivalent point, in % or o/oo of reacted compound equivalent quantity.
Titration of strong acid of 0.01 N with a strong base gives an error ofTitration of strong acid of 0.01 N with a strong base gives an error of + 0.1 % at pH 5 or 9; but this error will be + 0.01% at pH 6 or 8.
Lower titration error should be obtained in a titration of carboxilic acid (Ka > 10‐5) with a strong base.
If Ca = acid analytical concentration and Cb = standad base analytical
concentration, then at the equivalent point Ca = Cb, at the other
points : Ca – Cb = + or – as titration error.
During titration : Ca = [HA] + [A‐] and Cb = [A‐] + [OH‐] – [H+]g
If weak base solution (near equivalent point) and [H+] are neglected,
hence Ca = Cb = [A‐] + [HA] = [A‐] + [OH‐] [HA] = [OH‐][ ] [ ] [ ] [ ] [ ] [ ]
At equivalent point a pure cmpound NaA dissociatesA + H O HA + OH‐A‐ + H2O HA + OH‐
If acetic acid titration (Ka = 1 8 x 10‐5) without a significant volumeIf acetic acid titration (Ka = 1.8 x 10 5) without a significant volume Change, error at end point + 1 from equivelent point,At the equivalent point [HA] = [OH‐] = 7 5 x 10‐6 MAt the equivalent point [HA] [OH ] 7.5 x 10 M
pH = 8.8
If equivalent point occured at pH = 9.88 (or pH = 9.9) hence [HA] = 7 x 10‐7 M or [OH‐] = 76 x 10‐5
Titration error = Cb – Ca = {[OH‐] – [HA]}/Ca x 100%= 0.07%.
If colour change at the equivalent point is clear and sharp, and the indicator is very good, titration error could be neglected.
Titration of Heavy Metal Cations
An aquaous solution of heavy metal is a base (Bronsted), titrated
with a strong base form an insoluble salt base.with a strong base form an insoluble salt base.
Al3+ + 3 OH‐ Al(OH)3 hi i iAl + 3 OH Al(OH)3white precipitateCu2+ + 2 OH‐ Cu(OH)2 blue precipitate
The end point could be obtained before equivalent point caused of
the early precipitated salt base It is beter to performe this with anthe early precipitated salt base. It is beter to performe this with an
indirect titration technique.
Titration of Borate (Borax)
A borate (borax) dissolved in water could form a half‐neutralized
b dboric acid
Na2B4O7 + 5 H2O 2 H3BO3 + 2 H2BO3‐ + 2 Na+
2 H2BO3‐ + 2 H+ 2 H3BO3
A sharp end point should be obtained, this is an appropriate primary
standard base for a standardization of HCl.
Titration of phosphoric acidTitration of phosphoric acid with NaOH is a mono/di‐protic (not triprotic)
E i t 1 H ½ ( K + K ) 4 66Eq point‐1 : pH = ½ (pKa1 + pKa2) = 4.66Indicator bromcresol green or methyl yellow. End point in detected using pure NaH2PO4 as control compound.
Eq point‐2 : pH = ½ (pKa2 + pKa3) = 9.7Indicator phenolphtalein or thymol blue colour change occures in basic solutionIndicator phenolphtalein or thymol blue, colour change occures in basic solution before equivalent point. Indicator timolphtalein would be better because of pH =9.6
Third ionization product of phosphoric acid has Ka = 5 x 10‐13
Solution of Na PO is strong basic and Eq Point‐3 is never obtained exceptSolution of Na3PO4 is strong basic and Eq. Point 3 is never obtained, except trivalent phosphate ion is removed such by CaCl2 precipitation after Eq.Point‐22 Na2HPO4 + 3 CaCl2 Ca3(PO4)2 + 4 NaCl + 2 HCl
Titration of carbonic acid (hydrated CO2)
As a diprotic acid : pH = ½ (pKa1 + pKa2) = 8.40
At Eq Point‐1 the CO2 could be titrated as monoprotic acid with a
solution of NaOH using phenolphtalein or thymol blue (or its
mixture), colour change is not sharp, needs control solution
(pure NaHCO3 + indicator at same quantity as for sample).
Second ionization product, as a weak diprotic acid is too week for a
direct titration. Carbonate ion has to be removed (precipitation) by
addition of excess Ba(OH)2H2CO3 + Ba(OH)2 BaCO3 + 2H2O
NaHCO3 + Ba(OH)2 BaCO3 + NaOH + H2O
Back titration with an acid standard solution could use phenol‐
phtalein or thymol blue as the indicator without any filtering
process.
BaCO3.
H CO N HCO
NaHCO3Na2CO3
H2CO3 + NaHCO3
pH NaHCO3 + Na2CO3
ml NaOH0.1 N50 100
Titration curve of 100 ml of carbonic acid 0.05 M with NaOH 0.1N
SSummary
Choice of indicatorChoice of indicator
Strong acid and strong base. g gFor 0.1 M or more concentrated solutions,
any indicator may be used which has a range
between the limits pH 4.5 and pH 9.5. Withbetween the limits pH 4.5 and pH 9.5. With
0.01 M solutions, the pH range is somewhat
smaller (5.5‐8.5). If carbon dioxide is present,
either the solution should be boiled while stilleither the solution should be boiled while still
acid and the solution titrated when cold, or an
indicator with a range below pH 5 should be
employedemployed.
Weak acid and a strong base.gThe pH at the equivalence point is calculated from the equation:calculated from the equation:pH = ½ pKw + ½ pKa – ½ pCThe pH range for acids with Ka > 10-5 is 7-10 5; for weaker acids (Ka >10-6) the7 10.5; for weaker acids (Ka >10 ) the range is reduced (8-10). The pH range 8-10.5 will cover most of the exampleslikely to be encountered; this permits thelikely to be encountered; this permits the use of thymol blue, thymolphthalein,
h l hth l ior phenolphthalein.
Weak base and strong acid. gThe pH at the equivalence point is computed from the equation:computed from the equation:pH = ½ pKw – ½ pKb + ½ pCThe pH range for bases with Kb > 10-5 is 3-7 and for weaker bases (Kb > 10-6)3 7, and for weaker bases (Kb > 10 )3-5. Suitable indicators will be methyl red, methyl orange, methyl yellow,bromocresol green, and bromophenolbromocresol green, and bromophenol blue.
Weak acid and weak base.There is no sharp rise in the neutralisation curve and generally no simple indicator cancurve and, generally, no simple indicator can be used. The titration should therefore be avoided if possible The approximate pH atavoided, if possible. The approximate pH at the equivalence point can be computed from h ithe equation :pH = ½ pKw + ½ pKa – ½ pKb
It is sometimes possible to employ a mixed indicator whichhibit l h li it d H fexhibits a colour change over a very limited pH range, for
example, neutral red-methylene blue for dilute ammonia solution and acetic (ethanoic) acidsolution and acetic (ethanoic) acid.
Polyprotic acids (or mixtures of acids, yp ( ,with dissociation constants K1, K2, andK ) and strong basesK3) and strong bases.
The first stoichiometric end point is given approximately bypp y ypH = ½ (pK1 + pK2)
The second stoichiometric end point is given approximately byapproximately bypH = ½ (pK2 + pK3)
Anion of a weak acid titrated with a strong acid.The pH at the equivalence point is givenThe pH at the equivalence point is given bypH = ½ pKw – ½ pKa – ½ pC
Cation of a weak base titrated with a strong base.The pH at the stoichiometric end point isThe pH at the stoichiometric end point is given by
H ½ K ½ Kb ½ CpH = ½ pKw – ½ pKb – ½ pC
As a general rule wherever an indicator does not give a sharp endAs a general rule, wherever an indicator does not give a sharp end point, it is advisable to prepare an equal volume of a comparison solution containing the same quantity of indicator and of the final
d d h f h i i i h l iproducts and other components of the titration as in the solution under test, and to titrate to the colour shade thus obtained.
In cases where it proves impossible to find a suitable indicator (and this will occur when dealing with strongly coloured solutions) then titration may be possible by an electrometric method such as y p yconductimetric, potentiometric or amperometric titration.
In some instances spectrophotometric titration may be feasibleIn some instances, spectrophotometric titration may be feasible.
It should also be noted that if it is possible to work in a non-aqueous solution rather than in water then acidic and basic properties maysolution rather than in water, then acidic and basic properties may be altered according to the solvent chosen, and titrations which are difficult in aqueous solution may then become easy to perform.Thi d i id l d f th l i f i t i lThis procedure is widely used for the analysis of organic materials but is of very limited application with inorganic substances.
Endd
