ACIDS AND BASES ACIDS AND BASES ACIDS AND BASES ACIDS AND BASES According to the Brönsted Brönsted Brönsted Brönsted-Lowry theory Lowry theory Lowry theory Lowry theory, an acid-base reaction is a proton exchange reaction. A Brönsted acid releases a proton . A Brönsted base accepts a proton. In Water, the proton H + interacts with a pair of free electrons located on the oxygen atom of a water molecule and the resulting cation H3O + is called the hydronium ion. The latter is hydrated and appears as a water cluster; for the sake of simplicity though it will be noted H3O + throughout this chapter. Similarly to any exchange process, as acid-base reaction requires two partners - In water, one of these partners is often the water molecule itself which can either accept or release a proton. - After, releasing a proton, an acid becomes a negatively charged particle which, in turn can associate with a proton. This negatively charged particle is termed the “conjugate base”. - After accepting a proton, a base becomes a positively charged particle which, in turn can release a proton. This particle is termed “conjugate acid”. Thus, a conjugate conjugate conjugate conjugate base A - is associated to any acid HA. They form an acid acid acid acid-base couple base couple base couple base couple noted HA/A - : Examples Reciprocally, a conjugate acid BH + is associated to any base B. They for an base-acid couple noted BH + /B) : Ion product of water (1/2) In pure water, a small part of the water molecules are ionized, which results in a light conduction of the current. Approximately two water molecules out of 10 9 are ionized (autoionization of water). The equilibrium constant equilibrium constant equilibrium constant equilibrium constant of this reaction is : The concentration of H2O can be considered as being constant in dilute solutions (H2O = 55.5 M) and its activity equal to 1, therefore : K i is temperature dependent and is called ion product of water. Thus in pure water, the concentrations of H3O + and OH - ions are the same equal. Such a solution is defined as being neutral.
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ACIDS AND BASESACIDS AND BASESACIDS AND BASESACIDS AND BASES According to the BrönstedBrönstedBrönstedBrönsted----Lowry theoryLowry theoryLowry theoryLowry theory, an acid-base reaction is a proton exchange reaction. A Brönsted acid releases a proton. A Brönsted base accepts a proton. In Water, the proton H+ interacts with a pair of free electrons located on the oxygen atom of a water molecule and the resulting cation H3O+ is called the hydronium ion. The latter is hydrated and appears as a water cluster; for the sake of simplicity though it will be noted H3O+ throughout this chapter.
Similarly to any exchange process, as acid-base reaction requires two partners - In water, one of these partners is often the water molecule itself which can either accept or release a proton. - After, releasing a proton, an acid becomes a negatively charged particle which, in turn can associate with a proton. This negatively charged particle is termed the “conjugate base”. - After accepting a proton, a base becomes a positively charged particle which, in turn can release a proton. This particle is termed “conjugate acid”. Thus, a conjugateconjugateconjugateconjugate base A- is associated to any acid HA. They form an acidacidacidacid----base couplebase couplebase couplebase couple noted HA/A- :
Examples Reciprocally, a conjugate acid BH+ is associated to any base B. They for an base-acid couple noted BH+/B) :
Ion product of water (1/2) In pure water, a small part of the water molecules are ionized, which results in a light conduction of the current. Approximately two water molecules out of 109 are ionized (autoionization of water).
The equilibrium constantequilibrium constantequilibrium constantequilibrium constant of this reaction is :
The concentration of H2O can be considered as being constant in dilute solutions (H2O = 55.5 M) and its activity equal to 1, therefore :
Ki is temperature dependent and is called ion product of water.
Thus in pure water, the concentrations of H3O+ and OH- ions are the same equal. Such a solution is defined as being neutral.
This concept has been introduced by the Danish chemist SørSørSørSørensenensenensenensen. It aims at characterizing the acidity strength of a solution. By definition, the pH of a solution is given by :
Mathematical reminder logarithms
p is the “power operator” often used in chemistry :
Following the same reasoning, the concentration of OH- ions is expressed by the pOH
Taking into account the properties of logarithms, we can rewrite as follows the ion product of water (Ki = 10-14 M2 at 298 K)
If the [H3O+] concentration increases, the[OH-] concentration decreases so that the ion product of water remains constant. The pH and the pOH values of diluted solutions generally lie between 0 and 14.
Examples : Calculate the pH and pOH of a solution having [H3O+] = 1 M ?
Calculate [H3O+] and [OH-] of a solution of pH = 8 ?
[H3O+] = 10 - pH = 10-8 M [OH-]= (10-14) / [H3O+] = (10 -14) / 10-8 = 10-6 M
Calculate the pH of human blood, knowing that [H3O+] = 4·10-8 M ?
pH = - log (4·10-8) = 7.4
Important note: the pH of a solution depends on the concentration of the ionized protons, and not on the total concentration of acid.
In pure water, the concentrations in [H3O+] and [OH-] are equal [H3O+] = [OH-] = 10-7 M at 298 K
There is an excess of protons in an acidic solution, [H3O+] > [OH-]
Conversely, there is a shortage of protons in a basic solution, [H3O+] < [OH-]
This can be visualized on the following pH scale :
The ionization equation of an acid HA :The ionization equation of an acid HA :The ionization equation of an acid HA :The ionization equation of an acid HA :
is characterized by a constant Ka, called acidity constantacidity constantacidity constantacidity constant (or dissociation constant, or ionization constant) :
The more the equilibrium is driven to the right, the stronger the acid is. Thus, Ka (or pKa) is a direct measure of the acid strength.
Notes : To avoid the use of acidity constants Ka expressed in terms of powers of 10, an acid-base couple HA/A- is characterized by its pKa. The stronger the acid is, the larger the Ka and the smaller the pKa (which can even be negative for very strong acids) !
Similarly, the associated equilibrium of a base B :Similarly, the associated equilibrium of a base B :Similarly, the associated equilibrium of a base B :Similarly, the associated equilibrium of a base B :
is characterized by constant Kb, called protonation constantprotonation constantprotonation constantprotonation constant :
The more the equilibrium is driven to the right, the stronger the base is, and thus Kb (or pKb) is a direct measure of the strength of the base.
Relationship between pRelationship between pRelationship between pRelationship between pKKKKaaaa and pand pand pand pKKKKbbbb : For a conjugated acid-basic couple :
therefore :
KKKKaaaa (HA) · (HA) · (HA) · (HA) · KKKKbbbb (A(A(A(A----) = ) = ) = ) = KKKKiiii = 10= 10= 10= 10----14141414 at 298 Kat 298 Kat 298 Kat 298 K
or pKa (HA) + pKb (A-) = 14
Strong bases A- are better H3O+ acceptors than weak bases and their conjugate acid HA is weak, that is its pKa is large. Strong acids HA are better H3O+ donors than weak acids and their conjugate base A- is weak, that is its pKb is large. In other words: Stronger acids have weaker conjugate bases and weaker acids have stronger conjugate bases. Stronger bases have weaker conjugate acids and weaker bases have stronger conjugate acids. STRONG AND WEAK ACIDS AND BASES An acid stronger than H3O+ is called strong acidstrong acidstrong acidstrong acid; it is totally dissociated in water and its pKa is negative (pKa 0).
An acid which is only partially dissociated in water is known as a weakweakweakweak acidacidacidacid and its pKa is > 0.
A base is strongstrongstrongstrong if it is completely protonated; its pKb is 0.
Examples The conjugate couple NH2-/NH3 has pKb = -20
The conjugate couple methanolate/méthanol CH3O-/CH3OH : pKb = -2
The hydroxide ions OH- resulting from the dissociation of sodium hydroxide NaOH or of potassium
hydroxide (KOH).
A base which is only partially protonated is a weakweakweakweak base and its pKb is > 0.
Examples: Imidazole is a weak base (C3H4N2/C3H5N2+ : pKb = 7,01)
Summarizing :Summarizing :Summarizing :Summarizing : Strengths of acids and bases in relation to the pKa scale of corresponding couples :
Solvent leveling effect and pSolvent leveling effect and pSolvent leveling effect and pSolvent leveling effect and pKKKKaaaa scalescalescalescale The strongest acid which can exist in water is H3O+. All the stronger acids react with H2O to yield H3O+. See for instance the hydrochloric acid : HCl dissociates completely in water to give the H3O+ and the very weak conjugate base Cl-. This can be interpreted as a leveling of HCl acidity by the solventleveling of HCl acidity by the solventleveling of HCl acidity by the solventleveling of HCl acidity by the solvent.
Similarly, the strongest base that can exist in water is OH-. Any stronger base reacts with H2O to form OH-. For instance, a strong base such as NH2- is completely replaced by OH- in aqueous medium. That is the solvent can also solvent can also solvent can also solvent can also level off the alkalinity of the solutionlevel off the alkalinity of the solutionlevel off the alkalinity of the solutionlevel off the alkalinity of the solution.
Note : This leveling effect can be understood by the simple statement: In solution, the strongest acid and the strongest base will always react to form a weaker conjugate base and a weaker conjugate acid.
RemindeRemindeRemindeReminder : a strong acid HA is completely dissociated (pr : a strong acid HA is completely dissociated (pr : a strong acid HA is completely dissociated (pr : a strong acid HA is completely dissociated (pKKKKa a a a 0000,,,, ]]]]G°G°G°G°dissdissdissdiss << 0)<< 0)<< 0)<< 0)
Note : the introduction of a strong acid HA into water drives equilibrium 2 towards the left ! If [HA]0 is the total concentration of acid initially introduced into the aqueous solution. Case (1) :Case (1) :Case (1) :Case (1) : [HA]0> 10-7 M : the H3O+ ions present in the solution come primarily from the dissociation of HA and :
Case (2) :Case (2) :Case (2) :Case (2) : [HA]0< 10-7 M : one cannot neglect the autoionization of water :
Examples : What is the pH of a 0.10 M aqueous solution of nitric acid (pKa = -1.44) ?
HNO3 is completely ionized in water : [HNO3]0 = [H3O+] =10-1 M and consequently pH = - log 10-1 = 1111
What is the pH of a 0.001 M aqueous solution of nitric acid ?
Reminder: a strong base B is completely protonated (pReminder: a strong base B is completely protonated (pReminder: a strong base B is completely protonated (pReminder: a strong base B is completely protonated (pKKKKbbbb 0000,,,, ]]]]G°G°G°G°assassassass << 0)<< 0)<< 0)<< 0) The following two equations have to be taken into account :
Note : the introduction of the strong base B into water drives equilibrium 2 towards the left ! If [B]0 is the total concentration of base initially introduced into the aqueous solution. Case (1) : [B]0 > 10-7 M : one can assume that the OH- come exclusively from the dissociation of H2O protons under the action of B.
Case (2) : [B]0< 10-7 M : the autoionization of water can no more be neglected.
Example: Calculate the pH of an aqueous solution of KOH 0.001 M ?
KOH is completely dissociated in water, thus : [OH-]0 =10-3 M pOH = -log (0,001) = 3 pH = 14 - pOH= 14 - 3 = 11
Reminder : a weak acid HA has a pReminder : a weak acid HA has a pReminder : a weak acid HA has a pReminder : a weak acid HA has a pKKKKaaaa in the range 0 in the range 0 in the range 0 in the range 0 ---- 11114444 ]]]]]]]]G°G°G°G°assassassass > 0)> 0)> 0)> 0). A weak acid is partially dissociated in water. At equilibrium, the species present in solution are HA, A-, H3O+ and HO- (arising from water autoionization). The following equilibria must be taken into consideration :
Calculation of the H3O+ concentration is quite complex, but approximations are often used, leading to welcome simplifications:
if the acid is concentrated enoughconcentrated enoughconcentrated enoughconcentrated enough (> 10-6 M), water autoionizationwater autoionizationwater autoionizationwater autoionization can be neglected
if the acid is sufficiently weaksufficiently weaksufficiently weaksufficiently weak, it is little dissociatedlittle dissociatedlittle dissociatedlittle dissociated and can be considered as un-dissociated Case (1) : the two aboveCase (1) : the two aboveCase (1) : the two aboveCase (1) : the two above----described conditions are met : [HA] described conditions are met : [HA] described conditions are met : [HA] described conditions are met : [HA] 10101010----6666 M and pM and pM and pM and pKKKKa a a a 2222
If the acid is not too diluted [HA] > 10[HA] > 10[HA] > 10[HA] > 10----6666 MMMM
If the acid is sufficiently weak (pKpKpKpKa a a a 2222), x will be very small with respect to [HA]0 and can be neglected in the
denominator
[HA]0 is the analytical (or total, or initial) concentration of the acid. Case (2) : [HA] < 10Case (2) : [HA] < 10Case (2) : [HA] < 10Case (2) : [HA] < 10----6666 M M M M andandandand////orororor ppppKKKKa a a a < 2< 2< 2< 2 When one or the two conditions defined above are not met, a full calculation has to be carried out.
Example: Calculate the pH of an acetic acid solution whose concentration is 0.5 M, knowing that pKa = 4.76 (Ka = 1.74·10-5). * Making use of the two approximations (case (1), since [CH3CO2H] > 10-6 M and pKa > 2) :
pH = 0.5 (4.76 + log (0.5)) = 2.53 In other words : [H3O+] = 10 - 2.53 = 2.95·10-3 M
[OH-] = 3.39·10-12 M [CH3CO2-] = 2.95·10-3 M and [CH3CO2H] = 0.5 - 2.95·10-3 = 0.5 M
Weak BaseWeak BaseWeak BaseWeak Base Reminder: the pReminder: the pReminder: the pReminder: the pKKKKbbbb ooooffff aaaa wwwweeeeaaaakkkk bbbbaaaasssseeee iiiissss iiiinnnn tttthhhheeee rrrraaaannnnggggeeee 0000 ttttoooo 11114444 ((((]]]]G°G°G°G°diss diss diss diss > 0).> 0).> 0).> 0). A weak base is partially protoned in water. At equilibrium, the species present in solution are B, BH+ as well as H3O+ and HO- arising from the autoionization of water. The following equilibria must be taken into consideration :
Calculation of [OH-] is complex, but approximations are often used to simplify it :
If the base is concentrated enoughconcentrated enoughconcentrated enoughconcentrated enough (> 10 -6) the autoionization of waterautoionization of waterautoionization of waterautoionization of water can be neglected
If the base is sufficiently weaksufficiently weaksufficiently weaksufficiently weak (pKb 2), it is not much protonatednot much protonatednot much protonatednot much protonated and can be considered as being almost un-protonated. Case (1) :Case (1) :Case (1) :Case (1) : the two conditions are met, [B] the two conditions are met, [B] the two conditions are met, [B] the two conditions are met, [B] 10101010----6666 M and pM and pM and pM and pKKKKbbbb 2222
If the base is not too much diluted [B] > 10[B] > 10[B] > 10[B] > 10----6666 M M M M
Each time that a molecule B is protonated, one OH- and one BH+ ions are produced :
If the base is sufficiently weak (pKb 2), x will be very small with respect to [B]0
[B]0 is the analytical (or total, or initial) concentration of the base. Case (2) : [B] < 10Case (2) : [B] < 10Case (2) : [B] < 10Case (2) : [B] < 10----6666 M M M M andandandand////orororor ppppKKKKb b b b < 2< 2< 2< 2 When one or the two conditions defined above are not met, a full calculation has to be carried out.
Example : Calculate the pOH of an ammonia solution whose concentration is 0.5 M, knowing that pKb = 4.76 (Kb = 1.74·10-5). *Making use of the two approximations (case (1) since [NH3] > 10-6 M and pKb > 2)
pOH = 0.5 (4.76 + log (0.5)) = 2.53 and pH = 14 - 2.53 =11.47 In other words: [OH-] = 10 -2.53 = 2.95·10-3 M [H3O+] = 3.39·10-12 M [NH4+] = 2.95·10-3 M and [NH3] = 0.5 - 2.95·10-3 = 0,5 M
Buffer Solution:Buffer Solution:Buffer Solution:Buffer Solution: Utilization: The pH of human blood must stay in a relatively narrow range around pH = 7.4 (7.35-7.45) in order to ensure cellular survival. It can be maintained within these limits thanks to buffer sysbuffer sysbuffer sysbuffer systemstemstemstems utilizing plasmatic carbonates, phosphates or proteins for example. The buffer ability of blood, in addition to its role in the metabolism, can also minimize the consequences of an accidental ingestion of acid (or base). A buffer solutionbuffer solutionbuffer solutionbuffer solution is a solution able to absorb a certain quantity of acid or base without undergoing a strong variation in pH. A buffer solution is a mixture of a weak acid HA and its conjugate base A- (usually added under the form of the sodium or potassium salt, NaA or KA). Alternatively a mixture of a weak base B and of its conjugate acid BH+ is also a buffer solution. Note that a solution of a weak acid or a weak base is by itself a buffer solution, but its capacity is quite limited, so that the addition of the conjugate particle is necessary to the preparation of a practical buffer solution.
Examples : Weak acids and their conjugate base
CH3CO2H / CH3CO2- H2CO3 / HCO3-
Weak bases and their conjugate acid
NH3 / NH4+ H2PO42-/ H3PO4
HendersonHendersonHendersonHenderson----Hasselbalch Hasselbalch Hasselbalch Hasselbalch RRRRelationshipelationshipelationshipelationship A buffer solution is in fact a solution containing two acid-base couples.
When either a strong base or a strong acid is added to the buffer solution, the resulting pH can be calculated provided the following conditions are valid:
then :
Example : Acetic buffer CH3CO2H / CH3CO2- (pKa = 4,76)
Calculate the pH of a mixture containing 100 ml of acetic acid (CH3CO2H) 0.15 M and 200 ml of
sodium acetate (NaCH3CO2) 0.25 M ?
Calculate the pH of a mixture containing 200 ml of acetic acid 0.25 M and 100 ml of sodium
acetate 0.15 M ?
Special case: One mixes equivalent volumes of the acid and of its conjugate base; for instance, 200 ml of acetic acid 0.25 M and 200 ml of sodium acetate 0.15 M
We compare here additions of acid or base to a buffer solution and to pure water. Let’s start from one liter of a buffer solution 1 M in acetic acid, CHLet’s start from one liter of a buffer solution 1 M in acetic acid, CHLet’s start from one liter of a buffer solution 1 M in acetic acid, CHLet’s start from one liter of a buffer solution 1 M in acetic acid, CH3333COCOCOCO2222H and 0.3 M in sodium acetate, NaCHH and 0.3 M in sodium acetate, NaCHH and 0.3 M in sodium acetate, NaCHH and 0.3 M in sodium acetate, NaCH3333COCOCOCO2222
Upon addition of 100 ml of HCl 1 M :
The equilibrium is completely driven to the right, therefore [CH3CO2H] = (1 + 0.1) / 1.1 = 1.0 M and [CH3CO2-] = (0.3 - 0.1) / 1.1 = 0.18 M. pH = 4.76 + log (0.18 / 1) = 4.024.024.024.02.
Upon addition of 100 ml of NaOH 1 M :
The equilibrium is completely driven to the right, therefore [CH3CO2H] = (1 - 0.1) / 1.1 = 0.82 M and [CH3CO2-] = (0.3 + 0.1) / 1.1 = 0.36 M. pH = 4.76 + log (0.36 / 0.82) = 4.404.404.404.40. The pKa of the weak acid must not differ more than one unit from the targeted pH value. Then simply use the Henderson-Hasselbalch equation.
Example: Preparation of a buffer solution with pH = 5.2.
pKa (CH3CO2H / CH3CO2-) = 4.76, so that this mixture acetic acid (CH3CO2H) / sodium acetat (CH3CO2-) can be used.
According to the Henderson-Hasselbalch relationship, pH = pKa + log ([A-] / [AH])
To prepare one liter of this buffer solution: mix 2.75 moles of acetate and 1 mole of acetic and complete to 1 liter with distilled water
The capacity of the buffer can be adjusted increasing or decreasing the acid (or base) concentration (keepinThe capacity of the buffer can be adjusted increasing or decreasing the acid (or base) concentration (keepinThe capacity of the buffer can be adjusted increasing or decreasing the acid (or base) concentration (keepinThe capacity of the buffer can be adjusted increasing or decreasing the acid (or base) concentration (keeping [Ag [Ag [Ag [A----]/[AH] = ]/[AH] = ]/[AH] = ]/[AH] = 2.75!).2.75!).2.75!).2.75!). Indeed, when most of the weak acid (or of it conjugate base) is transformed into its conjugate base (weak acid) base, the buffer solution becomes ineffective.
TITRATIONSTITRATIONSTITRATIONSTITRATIONS 5.1 Definitions and principle The neutralization of an acid (or a base) by a base (or an acid) results in a salt and water :
Examples :
The reaction is complete; in fact, in water one has :
Same remark as above :
AcidAcidAcidAcid----base titrationbase titrationbase titrationbase titration is a quantitative method for the determination of the concentration of an acid (or base) by reacting it with a standard solution of base (or acid), that is a solution with a concentration which is exactly known. To warrant complete reaction, titrations are generally carried out with solutions of strong bases or acids. The implied reactions are as follows:
The principle consists in finding the exact volume necessary to neutralize a given volume (known exactly) of the acid (or base) solution of unknown concentration. By definition, neutralization corresponds to the moment when the moles of OH- (or of H3O+) added to the acid (base) solution exactly match the moles of H3O+ (or OH-) released by the acid (or the base) in solution. This is called the equivalence pointequivalence pointequivalence pointequivalence point:
AdviceAdviceAdviceAdvice:::: 1) Always write the neutralization equation, especially if a polyacid or a polybase is titrated. 2) The concentration of the titrant solution must be known very precisely, as must be known the volume added to reach the equivalence point. Examples :
What is the volume of a solution of HCl 0.3 M necessary to neutralize 50 ml of NaOH 0.2 M ?
1) Equation of neutralization :
2) Initial number of moles of base : nb = (50·10-3 l)·(0.2 M) = 0.01 mol) 3) Volume of HCl needed to neutralize the base: nb = na = ca·Va Va = nb / ca = (0.01 mol) / (0.3 M) = 33.33·10-3 l (33.33 ml)
If 30 ml of a NaOH solution 0.2 M are needed to neutralize 10 ml of a solution of HCl, whatis the concentration of the latter ?
The variation of pH versus the volume of added titrant is a titration curve. Such a curve can be recorded (by an automatic potentiometer) or be established point by point. The curve is sigmoïd and presents a significant variation of pH at the equivalence point, which allows an easy determination of the latter.
Note :Note :Note :Note : The more diluted is the acid (or bases) the smaller the pH jump.
During the titration of a weak acid by a strong base or conversely, one forms a buffer solution. The buffer effect is reflected in the small variation of pH upon addition of the initial volumes of the strong base. The pH at half-equivalence corresponds to
Determination of the equivalence pointDetermination of the equivalence pointDetermination of the equivalence pointDetermination of the equivalence point: experimental me: experimental me: experimental me: experimental methodsthodsthodsthods:::: a) Graphical methoda) Graphical methoda) Graphical methoda) Graphical method During the titration of an acid by a base the pH of the solution is recorded versus the volume of added base and the equivalence point can be determined from the graph pH = Vtitrant taking advantage of the approximate symmetry of the curve. For example, if 25 ml of HCl 0.1 M are titrated by NaOH 0.1 M, the equivalence point occurs at pH = 7.00.
b) Colored indicatorsb) Colored indicatorsb) Colored indicatorsb) Colored indicators A simpler method to determine the equivalence point consists in using a colored indicator. It is a weak acid (or base) whose conjugate base (acid) changes color in a particular pH range, close to its pKa An indicator is generally noted HindHindHindHind (acid form) and Ind Ind Ind Ind ---- (conjugate base). In solution, the indicator is involved in a proton transfer equilibrium :
Example : phenolphtalein (pKa = 9.2)
The color changes are usually seen when the ratios of acid to base concentrations are smaller or larger than 10:
The lower limit of perception thus corresponds to
Similarly, the higher limit of perception corresponds to
The pH range into which an indicator changes color is therefore
Example : : for phenolphthalein whose pKa is 9.2, this range is between pH = 8.2 and pH = 10. Important note :Important note :Important note :Important note : A precise determination of the equivalence point during a titration requires that its pH lies in the middle of the pH range in which the indicator changes color.
Bromothymol blue (6.0-7.6) is most adequate for the titration of strong acids (bases) by strong bases (acids since the pH at neutralization is 7). For the titration of a weak base by a strong acid (pH at the equivalence point < 7), Helianthin is recommended, whereas for the titration of a weak acid by a strong base (pH at the equivalence point > 7), phenolphthalein is often used.
Titration ofTitration ofTitration ofTitration of a strong acid by a strong basea strong acid by a strong basea strong acid by a strong basea strong acid by a strong base:::: Titration of 50 ml of HCl (1) 0.1 M by NaOH Titration of 50 ml of HCl (1) 0.1 M by NaOH Titration of 50 ml of HCl (1) 0.1 M by NaOH Titration of 50 ml of HCl (1) 0.1 M by NaOH (2)(2)(2)(2) 0.1 M. 0.1 M. 0.1 M. 0.1 M. 1) pH calculation
For For For For V2222 = 0= 0= 0= 0 pH = pH = pH = pH = ----log [Hlog [Hlog [Hlog [H3333OOOO++++] ] ] ] pH = -log (0.1) = 1
For For For For V2222 > 0 > 0 > 0 > 0 The pH is governed by the strong acid, thus pH = The pH is governed by the strong acid, thus pH = The pH is governed by the strong acid, thus pH = The pH is governed by the strong acid, thus pH = ----log [Hlog [Hlog [Hlog [H3333OOOO++++] ] ] ]
ex : V2 = 20 ml = 0.020 l pH = -log (4.3·10-2) = 1.37 For For For For V2222 = = = = Veqeqeqeq
[H[H[H[H3333OOOO++++] = [OH] = [OH] = [OH] = [OH----] = 1.00·10] = 1.00·10] = 1.00·10] = 1.00·10----14141414 pH = 7.00pH = 7.00pH = 7.00pH = 7.00 For For For For V2 2 2 2 > > > > Veqeqeqeq The pH is governed by the base, thus pH = 14 + log [OHThe pH is governed by the base, thus pH = 14 + log [OHThe pH is governed by the base, thus pH = 14 + log [OHThe pH is governed by the base, thus pH = 14 + log [OH----] ] ] ]
ex : V2= 70 ml = 0.070 l pH = 12.22
2) Titration curve
3) Remarks Indicators: The change of color must occur at the equivalence point. In this case bromothymol blue is adequate with a pKa = 7.1 and a pH range into which it changes color of 6.1 – 7.6. Influence of the dilution: the equivalence point occurs at the same pH, but the pH jump is smaller.
The precision of the titration is better the more concentrated the solutions are, since the pH jump is then larger. Titration of a strong base by a strong acid The reasoning is exactly the same as above. If one plots pOH versus the volume of added acid,The reasoning is exactly the same as above. If one plots pOH versus the volume of added acid,The reasoning is exactly the same as above. If one plots pOH versus the volume of added acid,The reasoning is exactly the same as above. If one plots pOH versus the volume of added acid, one gets the same one gets the same one gets the same one gets the same titration curves as in the case of the titration of a strong acid by a strong base. If one plots pH versus V(HA), the graph titration curves as in the case of the titration of a strong acid by a strong base. If one plots pH versus V(HA), the graph titration curves as in the case of the titration of a strong acid by a strong base. If one plots pH versus V(HA), the graph titration curves as in the case of the titration of a strong acid by a strong base. If one plots pH versus V(HA), the graph obtained is the mirror curve.obtained is the mirror curve.obtained is the mirror curve.obtained is the mirror curve.
Titration of a Titration of a Titration of a Titration of a weak acid by a strong base :weak acid by a strong base :weak acid by a strong base :weak acid by a strong base : Titration of 50 ml of CHTitration of 50 ml of CHTitration of 50 ml of CHTitration of 50 ml of CH3333COCOCOCO2222H (1) 0,1 M by NH (1) 0,1 M by NH (1) 0,1 M by NH (1) 0,1 M by NaOH aOH aOH aOH (2)(2)(2)(2) 0.1 M.0.1 M.0.1 M.0.1 M.
1) pH calculation
For For For For VVVV2222 = 0= 0= 0= 0 The pH is that oThe pH is that oThe pH is that oThe pH is that of a weak base, thus pH = ½ (pf a weak base, thus pH = ½ (pf a weak base, thus pH = ½ (pf a weak base, thus pH = ½ (pKKKKaaaa ---- log [CHlog [CHlog [CHlog [CH3333COCOCOCO2222H]) H]) H]) H]) pH = ½ (4.76 - log (0.1)) = 2.88
For For For For VVVVeq eq eq eq > > > > VVVV2 2 2 2 > 0 > 0 > 0 > 0
The pH is still governed by the quantity of weak acid remaining in solution.The pH is still governed by the quantity of weak acid remaining in solution.The pH is still governed by the quantity of weak acid remaining in solution.The pH is still governed by the quantity of weak acid remaining in solution.
For For For For VVVVeqeqeqeq = = = = VVVV2222 TTTThe weak acid is completely neutralized. The main species present is its conjugate base, which governs the pH value. he weak acid is completely neutralized. The main species present is its conjugate base, which governs the pH value. he weak acid is completely neutralized. The main species present is its conjugate base, which governs the pH value. he weak acid is completely neutralized. The main species present is its conjugate base, which governs the pH value. The solution is basic, the source of OHThe solution is basic, the source of OHThe solution is basic, the source of OHThe solution is basic, the source of OH---- being the acetate CHbeing the acetate CHbeing the acetate CHbeing the acetate CH3333COCOCOCO2222H which protonates and gives equal quantities of H which protonates and gives equal quantities of H which protonates and gives equal quantities of H which protonates and gives equal quantities of CHCHCHCH3333COCOCOCO2222H and OHH and OHH and OHH and OH----....
ex : V2 = 50 ml = 0.050 l pH = 8.73 FFFFor or or or VVVV2222> > > > VVVVeqeqeqeq The pH is imposed by the excess of strong base, thus pH = 14 + log[OHThe pH is imposed by the excess of strong base, thus pH = 14 + log[OHThe pH is imposed by the excess of strong base, thus pH = 14 + log[OHThe pH is imposed by the excess of strong base, thus pH = 14 + log[OH----]]]]
ex : V2= 70 ml = 0.070 l pH = 12.22 2) Titration curve
3) Remarks • If the end point is determined with a colored indicator, its pH must be around 8.7 (typically between 7.7 and 9.7). A
good possibility is phenolphthalein (8.2-10.0). • Influence of the concentration: contrary to the case where a strong aid is titrated by a strong base, dilution results in a
variation of the pH of the end point, as well as in a smaller pH jump.
4) Influence of the pKa on the pH jump at the equivalence point In the vicinity of the equivalence point, the variation of pH becomes smaller if the acid is weaker (larger value of pKa). The reaction between the acid and the base is less complete if the acid is weaker.
Titration of a weak base by a strong acidTitration of a weak base by a strong acidTitration of a weak base by a strong acidTitration of a weak base by a strong acid The reasoning is the same as previously; if the pH is plotted against V(acid), the obtained graph is the mirror curve with respect to the latter situation. Similar points can be made as to the choice of the colored indicator and the influence of dilution and pKb.
pH DETERMINATIONSpH DETERMINATIONSpH DETERMINATIONSpH DETERMINATIONS 1. MixtuMixtuMixtuMixtures of strong acids res of strong acids res of strong acids res of strong acids
Example : What is the pH of a solution obtained by mixing 250 ml of HClO4 0.4 M and 350 ml of HCl 0.1 M ?
2. Mixture of a strong acid and a weaMixture of a strong acid and a weaMixture of a strong acid and a weaMixture of a strong acid and a weak acidk acidk acidk acid
Example : What is the pH of a solution obtained by mixing 500 ml of CH3CO2H 0.03 M and 200 ml of HCl 0.04 M ?
5. Mixture of a strong base and a weak base Mixture of a strong base and a weak base Mixture of a strong base and a weak base Mixture of a strong base and a weak base
Example : What is the pH of a solution obtained by mixing 500 ml of NH3 0.03 M and 200 ml of NaOH 0.04 M ?
Salt solutions:Salt solutions:Salt solutions:Salt solutions: The dissolution of a salt results in a total dissociation into solvated ions. The resulting concentrations in anions and cations depends upon the chemical formula of the salt. These ions can behave as acids or bases. The pH of the solution is higher than 7 if the salt provides ions with base properties and lower than 7 if these ions are acidic. 1. Acidic ionsAcidic ionsAcidic ionsAcidic ions In solution all the cations which are conjugate acids of weak bases behave like acids, lowering the pH of the solution (ex : NH4+). Similarly, the small metal cations bearing a multiple charge (2+, 3+) behave like Lewis acids and react with OH- ions (Example : Fe3+=> Fe(OH)3).
Some anions behave as acids; they usually arises from the dissociation of polyacids (Example : H2PO4-). 2. Basic ions Basic ions Basic ions Basic ions In solution all the anions which are conjugate bases of weak acids behave like bases, increasing the pH of the solution (Example : HCO2-). Some anions coming from the deprotonation of polyacids can accept a proton in water and therefore behave as bases. (Example : PO43-). No cation behaves as a basic ion because the positive charge exerts a strong repulsion when a proton approaches it. 3. "Neutral" ions"Neutral" ions"Neutral" ions"Neutral" ions In solution the anions which are conjugate bases of strong acids, are very weak bases and can be regarded as neutral with respect to acid-base properties (ex : ClO4-). In solution the cations which are conjugate acids of strong bases are very weak acids and can be regarded as neutral with respect to acid-base properties (ex : Na+). Monovalent metal cations (with charge +1) as well as metal ions from groups 1 and 2 in the Periodical Table are very weak Lewis acids. They do not favor proton dissociation of water molecules (ex : Ca2+).
6.16.16.16.1.3.4 Salts having two neutral ions.3.4 Salts having two neutral ions.3.4 Salts having two neutral ions.3.4 Salts having two neutral ions Dissolution of sodium chloride NaCl (a salt formed upon reaction of HCl and NaOH) results in a total dissociation into Na+ and Cl-. These ions do not interact with water because the cation Na+ and the anion Cl- are neutral with respect to acid-base properties (NaOH is a strong base, HCl a strong acid).
The H3O+ ions in solution only arises from the water autoionization :
The pH of a solution of a salt containing two neutral ions is equal to 7.The pH of a solution of a salt containing two neutral ions is equal to 7.The pH of a solution of a salt containing two neutral ions is equal to 7.The pH of a solution of a salt containing two neutral ions is equal to 7. 6.1.3.5 Salt having one acid ion6.1.3.5 Salt having one acid ion6.1.3.5 Salt having one acid ion6.1.3.5 Salt having one acid ion Such a salt results from the reaction :
Example : With A- being a neutral anion, one can consider that only two equilibria take place :
If one takes into account the two usual approximations (Ka (BH+) ≦ 10-2 and [salt] > 10-6 M), one obtains :
The pH of a solution of a salt containing an acid ion is lower than 7The pH of a solution of a salt containing an acid ion is lower than 7The pH of a solution of a salt containing an acid ion is lower than 7The pH of a solution of a salt containing an acid ion is lower than 7
Example : What is the pH of an ammonium chloride solution NH4Cl 0.3 M (pKb = 4.76)
pKb (NH3) = 4.76 pKa (NH4+) = 9.24 [H3O+] = [NH3] and [NH4+] = 0.3 - [H3O+] ]]0.3 M
pH = 0.5·(9.24 - log (0.3)) pH = 4.9 Note: This is equivalent to calculating the pH of a solution of weak acidof concentration 0.3 M and pKa = 9.24.
6.1.3.6 Salt having a basic ion6.1.3.6 Salt having a basic ion6.1.3.6 Salt having a basic ion6.1.3.6 Salt having a basic ion Such a salt results from the reaction :
Example : Two equilibria take place :
If one takes into account the two usual approximations (Ka (HA) ≦ 10-2 and [salt] > 10-6 M), one obtains :
The pH of a solution of a salt containing a basic ion is larger than 7The pH of a solution of a salt containing a basic ion is larger than 7The pH of a solution of a salt containing a basic ion is larger than 7The pH of a solution of a salt containing a basic ion is larger than 7
Example : What is the pH of a sodium acetate solution NaCH3CO2 0.3 M (pKa = 4.76)
pKa (CH3CO2H) = 4.76 pKb (CH3CO2-) = 9.24 [OH-] = [CH3CO2H] and [CH3CO2-] = 0.3 - [OH-] = 0.3 M
pOH = 0.5·(9.24 - log (0.3)) pOH = 4.9 and pH = 14 - 4.9 = 9.1 Note: This is equivalent to calculating the pH of a weak base 0.3 M and of pKb = 9.24.
6.1.3.7 Salt having one acid ion and one basic ion 6.1.3.7 Salt having one acid ion and one basic ion 6.1.3.7 Salt having one acid ion and one basic ion 6.1.3.7 Salt having one acid ion and one basic ion
Such a salt results from the reaction :
Example : Three interdependent equilibria take place :
If one takes into account the two usual approximations, one obtains :
Examples : What is the pH of ammonium salt solutions 0.3 M, pKa(NH4+ / NH3) = 9.24 ?
Ammonium format HCO2NH4, pKb(HCO2H / HCO2-) = 10.25
A chemical compound able to react with both an acid or a base is amphoteric.A chemical compound able to react with both an acid or a base is amphoteric.A chemical compound able to react with both an acid or a base is amphoteric.A chemical compound able to react with both an acid or a base is amphoteric.
Water is amphoteric. The two acid-base couples of water are H3O+/H2O and H2O/HO- It behaves sometimes like an acid, for example
And sometimes like a base :
Hydrogenocarbonate HCO3- is also amphoteric, it belongs to the two acid-base couples H2CO3/HCO3- and HCO3-
/CO32-
The pH of a solution of an amphoteric compound is calculated according to :
PolyacidsPolyacidsPolyacidsPolyacids A majority of simple acids, such as hydrochloric acid (HCl), nitric acid (HNO3), acetic acid (CH3COOH) are monoacids since they release only one mole of H3O+ ions per mole of acid. Representative of diacids are sulfuric acid (H2SO4), phtalic acid C6H4(COOH)2, carbonic acid (H2CO3). Some triacid are also known: phosphoric acid (H3PO4), citric acid HOC(CH2COOH)3, and so on. For polyacids, each dissociation equilibrium is characterized by its own acidity constant. Sulfuric acid (1/2)Sulfuric acid (1/2)Sulfuric acid (1/2)Sulfuric acid (1/2) What is the pH of a solution of sulfuric acid 0.1 M ?
The first dissociation sulfuric acid is complete (strong acid, pKa1< 0, yielding hydrogenosulfate) while the second dissociation is incomplete (weak acid, pKa2 = 1.92, yielding sulfate ions). As a consequence, two simultaneous equilibria occur :
The total dissociation of 0.1 M sulfuric acid releases 0.1 M of H3O+ ions and as many hydrogenosulfate ions HSO4- :
The dissociation HSO4- into SO42- is partial and the following equilibrium must be taken into account:
The carbonic acid, of empirical formula H2CO3 , cannot be isolated pure (it decomposes). In fact it forms when carbon dioxide CO2 is dissolved into water Both protons are weakly acidic :
What is the pH of a solution of carbonic acid ? The hydrogenocarbonate ion HCO3- is amphoteric, thus
independently of the concentration in carbonic acid ! Phosphoric acid (1/2)Phosphoric acid (1/2)Phosphoric acid (1/2)Phosphoric acid (1/2) Phosphoric acid H3PO4 and its derived ions play a major role in the biochemical mechanisms of energy transport and exchange: for instance, the conversion of ATP (adenosine triphosphate) into ADP (adenosine diphosphate) involves the exo-energetic dissociation of a phosphate ion. What are the concentrations of all the species present in a solution of H3PO4 0.1 M ?
In water, the following equilibria take place :
Adding to the preceding three equilibria the water autoionization equilibrium and the mass conservation and electroneutrality equations, yields a system of 6 equations with 6 unknown variables ! For a dilute solution of a polyacid, if the first acidity constant Ka1 is much larger than the second one Ka2, one can calculate the pH taking only the first dissociation into account :
This is the case here: the difference between pKa1 and pKa2 (and subsequently pKa3) is large enough to assume that [H3O+] arises primarily from the first dissociation :
Biological applicationsBiological applicationsBiological applicationsBiological applications In the laboratory, buffer solutions are used to calibrate pH-meters and to control the pH of solutions in which pH-dependent chemical reactions take place. Similarly, life processes, for instance those involving enzymatic reactions, can only take place in a narrow range of pH. Since some of these reactions either consume or produce hydronium ions, it is essential that the pH be strictly controlled by buffer systems. For example, imagine that an enzyme has a carboxylic acid group and is active only when this group is deprotonated (under its carboxylate form). In order to avoid any inhibition of the enzymatic reactivity, a buffer system must maintain a sufficiently basic and constant pH in the physiological solution
When acidity is low, the enzyme has an optimum activity.
When acidity increases, the carboxylate binds a proton and the enzyme becomes inactive.
In addition, we have seen in the introduction that both temperature and pH influence the three-dimensional structure of
proteins and thus the reactivity of enzymes (the optimum values of pH generally lie between 6 and 8 in the human organism). For example, trypsin, which is a digestive enzyme located in the intestine, has an optimum activity at pH=8. Under more acidic conditions, this enzyme is denatured, like the majority of the enzymes. On the other hand, pepsin which is a digestive enzyme located in the stomach, has an optimum activity at pH ranging between 1 and 2; it is thus perfectly adapted to the acidic environment of the stomach.
When the acidity is low, the enzyme has optimum activity.
When acidity increases, conformational changes take place and the enzyme becomes inactive.
Phosphate buffers (1/2)Phosphate buffers (1/2)Phosphate buffers (1/2)Phosphate buffers (1/2) The main intracellular buffer, as well as in renal medium, is the phosphate buffer, which is present in urine. The main intracellular buffer, as well as in renal medium, is the phosphate buffer, which is present in urine. The main intracellular buffer, as well as in renal medium, is the phosphate buffer, which is present in urine. The main intracellular buffer, as well as in renal medium, is the phosphate buffer, which is present in urine.
At the pH of blood (7.4) it is the second acid-base couple H2PO4-/HPO42- which acts as buffer. Its capacity is maximum when the pH is close to the pKa2 of the H2PO4-/HPO42- couple.
At this pH (7.2), the other species H3PO4 and PO43- have negligible concentrations. Tampon phosphate (2/2)Tampon phosphate (2/2)Tampon phosphate (2/2)Tampon phosphate (2/2)
For a solution 0.15 M of phosphate buffer at pH = 7.2 :
If an enzymatic reaction consumes 0.02 M of H3O+, ions, the buffer compensates for this loss (the equilibrium is driven to the right) :
At the end of the reaction :
henceforth, a very small pH change.
In absence of buffer the pH would have increased from 7.2 to 12; enzymatic activity is impossible under such basic conditions. Bicarbonate (1/2)Bicarbonate (1/2)Bicarbonate (1/2)Bicarbonate (1/2)
The pH of blood is controlled, among others, by the carbonic acid/hydrogenocarbonate (bicarbonate) buffer. In mammals, it is estimated that 50% of the buffer potential of blood is due to this system. Blood is maintained at pH 7.4 (physiological pH) and the pH range compatible with life lies between 7 and 7.6; this represents a variation of the concentration in H3O+ ions between 10-7 and 2.5·10-8. Carbonic acid conveyed by blood is in constant exchange between its gaseous form CO2 and its hydrated form H2CO3. Carbon dioxide CO2 is partly eliminated by breathing, which causes a decrease in [H2CO3. In addition, HCO3- and H2CO3 are excreted with urine. The concentration of each one of these species is therefore controlled and this buffer has the advantage of being able to function in an open system.
Normally, in the blood plasma, [HCO3-] / [H2CO3] = 20/1. However, as wastes from the cellular metabolism mainly contain organic acids (for example, lactic acid), the strong concentration in HCO3- ions allows the organism to absorb most of these acids and to fight against disorders (diseases or burns) producing an excess of acid. Acidosis and alkalosis :Acidosis and alkalosis :Acidosis and alkalosis :Acidosis and alkalosis : When the concentration in HCO3- ions increases, the pH increases. One speaks of alkalosisalkalosisalkalosisalkalosis when it reaches a value
higher than the viable limit, that is when pH > 7.6. Conversely, when the pH decreases below the lower physiological limit (pH< 7), one speaks of acidosisacidosisacidosisacidosis. These variations of the acid-base equilibria can have very serious effects on health. They are usually caused by metabolic disorders such as diabetes or renal failures (for example, when the excretion of H2PO4- ions is inhibited). In case of hyperventilation, following an emotional shock, an excessive quantity of carbon dioxide CO2 is eliminated, resulting in alkalosis. On the other hand, acidosis can be caused by a broncho-pneumonia, if a too small quantity of CO2 is
eliminated. It can also follow a diarrhoea (significant loss of alkaline salts by the intestine), whereas significant vomiting leads to alkalosis. Protein bufferProtein bufferProtein bufferProtein buffer
One understands by protein buffer the set of proteinic groups which behave as amphoteric compounds. In plasma, proteins play a minor role. At plasmatic pH, the carboxylic acid functions of the proteins are deprotonated (carboxylate form), which ensure a buffer effect. To calculate the corresponding pH, one has to take an average pKa into account.
Hemoglobin bufferHemoglobin bufferHemoglobin bufferHemoglobin buffer Hemoglobin (Hb) features several sites able to play the role of a buffer; one thus considers a average pKa, which depends on the exact form of Hb (oxygenated or not) :
ZwitterioZwitterioZwitterioZwitterionic form (1/2)nic form (1/2)nic form (1/2)nic form (1/2)
In aqueous solution, acid-base equilibria are as follows (assuming that the side chain R does not bear any ionizable group) :
Isoelectric pointIsoelectric pointIsoelectric pointIsoelectric point Amino acids, under their zwitterionic form behave as amphoteric particles; the pH of their solutions is given by :
This pH is called isoelectric pHisoelectric pHisoelectric pHisoelectric pH because the zwitterion is overall neutral. It is noted pHi. Distribution diagram of the speciesDistribution diagram of the speciesDistribution diagram of the speciesDistribution diagram of the species
Separation of amino acids by electrophoresis
Each amino acid has a specific isoelectric pH (pHi the pH at which amino acids are electrically neutral). At pH differing from pHi, amino acids bear an overall charge and therefore migrate under the effect of an electric field. Thus, working at a fixed pH (buffered solution) allows one to separate various amino acids by electrophoresis.
When an amino acid is placed in an electric field, it migrates towards the electrode of opposed polarity, while the neutral (zwitterionic) molecules do not migrate. Therefore : When pH > pHi the amino acid bears a negative total charge: it migrates towards the positive electrode (anode). When pH < pHi the amino acid bears a positive total charge: it migrates towards the negative electrode (cathode). When pH = pHi the amino acid is in its zwitterionic, neutral form: it does not migrate and remains at the starting point. The pH of the solution is adjusted in such a way that an optimal separationoccurs :
![Colour maps of acid–base titrations with colour …...almost all acid–base titrations in aqueous solutions a proper indicator is available [13]. Of course, acid base titrations](https://static.documents.pub/doc/80x56/5e6ef47fc39e8d412360ae3c/colour-maps-of-acidabase-titrations-with-colour-almost-all-acidabase-titrations.jpg)
