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Acid-Base Equilibria
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In this chapter, we study equilibrium problems as related to acids and bases as
well as their salts. The main point is to know how to calculate the hydrogen ion
concentration, [H+]. However, let us start with definitions of acids and bases and
then look at their equilibria.
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Acid-Base Theories
Four main attempts to define acids and bases are common in the literature of chemistry. Development of these attempts or theories usually followed a desire to explain the behavior of substances and account for their properties as related to having acidic or basic characteristics. Theories of acidity or basicity can be outlined below from oldest to most recent:
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1. Arrhenius theory: This theory is limited to water as a solvent where an acid is
defined as a substance which ionizes in water and donates a proton. A base is a substance that ionizes in water to give hydroxide ions. The hydrogen ion reacts
with water to give a hydronium ion while the base reacts with water to yield a hydroxide
ion.
HA + H2O H3O+ + A-
B + H2O BH+ + OH-
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2. Franklin Theory: This theory introduced the solvent concept where an acid was
defined as a substance that reacts with the solvent to produce the cation of the
solvent . The base is a substance that reacts with the solvent to yield the
anion of the solvent.
HA + EtOH EtOH2+ + A-
B + EtOH BH+ + EtO-
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3. Bronsted-Lowry Theory: Some solvents like hexane or benzene are non ionizable and the
Franklin theory can not be used to explain acidic or basic properties of substances. In Bronsted-
Lowry theory, an acid is defined as a substance that can donate a proton while a
base is a substance that can accept a proton. Also, an acid is composed of two components; a
proton and a conjugate base. For example
HOAc H+ + OAc-
Acetic acid is an acid which donates a proton and its proton is associated with a base that can accept the proton; this base is the acetate.
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4. Lewis Theory: Lewis introduced the electronic theory for acids and bases where in Lewis
theory an acid is defined as a substance that accepts electrons while a base is a
substance that donates electrons. Therefore, ammonia is a base because it donates electrons
as in the reaction
H+ + :NH3 H:NH3+
AlCl3 is an acid because it accepts electrons from
a base such as :OR2
AlCl3 + : OR2 = Cl3Al:OR2
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Lecture 21
Acid-Base Equilibria, Cont….
Strong Acids/Bases and Their Salts
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Acid-Base Equilibria in Water
Fortunately, we will only deal with aqueous solutions which means that water will always be our solvent. Water itself undergoes self ionization as follows
2 H2O H3O+ + OH-
K = [H3O+][OH-]/[H2O]2
However, only a very small amount of water does ionize and the overall water concentration will be constant.
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Therefore, one can write
Kw = [H3O+][OH-]
Kw is called autoprotolysis constant of water
or ion product of water, we will also refer to [H3O
+] as simply [H+] although this is not
strictly correct due to the very reactive nature of H+
Kw = [H+][OH-] = 10-14 at 25 oC.
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The pH Scale
In most cases, the hydrogen ion concentration is very small which makes it difficult to practically express a meaningful concept for such a small value. Currently, the pH scale is used to better have an appreciation of the value of the hydrogen ion concentration where:
pH = - log [H+]
We also know that kw = [H+][OH-] = 10-14 or
pH + pOH = 14
Therefore, calculation of either pH or pOH can be used to find the other.
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We are faced with different types of solutions that we should know how to calculate the pH or pOH for. These
include calculation of pH for
1. Strong acids and strong bases
2. Weak acids (monoprotic) and weak bases (monobasic)
3. Salts of weak acids and salts of weak bases
4. Mixtures of weak acids and their salts (buffer solutions)
5. Polyprotic acids and their salts and polybasic bases and their salts
We shall also look at pH calculations for mixtures of acids and bases as well as pH calculations for very
dilute solutions of the abovementioned systems.
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pH calculations
1. Strong Acids and Strong Bases
Strong acids and strong bases are those substances which are completely dissociated in water and dissociation is represented by one arrow pointing to right. Examples of strong acids include HCl, HBr, HI, HNO3, HClO4, and H2SO4
(only first proton). Examples of strong bases include NaOH, KOH, Ca(OH)2, as
well as other metal hydroxides.
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Find the pH of a 0.1 M HCl solution.
HCl is a strong acid that completely dissociates in water, therefore we have
HCl H+ + Cl-
H2O H+ + OH-
[H+]Solution = [H+]from HCl + [H+]from water
However, [H+]from water = 10-7 in absence of a common ion, therefore it will be much less in
presence of HCl and can thus be neglected as compared to 0.1 ( 0.1>>[H+]from water)
[H+]solution = [H+]HCl = 0.1MpH = -log 0.1 = 1
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We can always look at the equilibria present in water to solve such questions, we have
only water that we can write an equilibrium constant for and we can write:
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Kw = (0.1 + x)(x)
However, x is very small as compared to 0.1 (0.1>>x)
10-14 = 0.1 x
x = 10-13 M
Therefore, the [OH-] = 10-13 M = [H+]from water
Relative error = (10-13/0.1) x 100 = 10-10%
[H+] = 0.1 + x = 0.1 + 10-13 ~ 0.1 M
pH = 1
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Find the pH of a 1x10-5 M HNO3 solution.
Nitric acid is a strong acid which means that it dissociates completely in water. Therefore,
[H+]from Nitric acid = 1x10-5 M
We can now set the equilibria in water as above
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Kw = (10-5 + x)(x)
However, x is very small as compared to 10-5 (10-5>>x)
10-14 = 10-5 x
x = 10-9 M
Therefore, the [OH-] = 10-9 M = [H+]from water
Relative error = (10-9/10-5) x 100 = 0.01 %
[H+] = 10-5 + 10-9 ~ 10-5 M
pH = - log 10-5 = 5
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Find the pH of a 10-7 M HCl solution.
SolutionHCl is a strong acid, therefore the [H+]from HCl =10-7 M
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Kw = (10-7 + x)(x)
Let us assume that x is very small as compared to 10-7
(10-7>>x)
10-14 = 10-7 x
x = 10-7 M
Therefore, the [OH-] = 10-7 M = [H+]from water
Relative error = (10-7/10-7) x 100 = 100 %
Therefore, the assumption is invalid and we have to solve the quadratic equation. Solving the quadratic
equation gives:
[H+] = 7.62x10-7 M
pH = 6.79
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Find the pH of a 10-10 M HCl solution.
HCl H+ + Cl-
H2O H+ + OH-
[H+]Solution = [H+]from HCl + [H+]from water
Before equil10-100
EquationH2OH+OH-
After equil10-10 + xx
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Assume 10-10>>xX = 10-4 , it is clear that the assumption is
invalid. In fact x is much larger than 10-10.Therefore reverse assumption and
assume x>>10-4
10-14 = (10-10 + x)(x)X = 10-7
RE = (10-7/10-10)*100% = 0.1%The assumption is valid and [H+] = [OH-] =
10-7 M pH = 7
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Calculate the pH of the solution resulting from mixing 50 mL of 0.1 M HCl with 50 mL of 0.2 M
NaOH.
When HCl is mixed with NaOH neutralization takes place where they react in a 1:1 mole ratio.
Therefore, find mmol of each reagent to see if there is an excess of either reagent
mmol HCl = 0.1 x 50 = 5 mmol
mmol NaOH = 0.2 x 50 = 10 mmol
mmol NaOH excess = 10 – 5 = 5 mmol
[OH-] = 5/100 = 0.05 M
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Find [H+] which is equal to [OH-]water [H+] = 10-14/0.05 = 2*10-13M
[OH-]water = 2*10-13M
The hydroxide ion concentration from the base is high enough to neglect the
contribution from water.
pOH = 1.3 and pH = 14 – pOH = 14 – 1.3 = 12.7
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It is better to use the equilibrium
10-14 = (0.05 + x)(x)Assume 0.05>>xX= 2*10-13, RE = very small[OH-] = 0.05 M, and pOH = 1.3pH = 14 – 1.3 = 12.7
Before equil00.05
EquationH2OH+OH-
After equilx0.05 + x
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Find the pH of a solution prepared by mixing 2.0 mL of a strong acid at pH 3.0 and 3.0 mL of a strong base
at pH 10.
Solution
First, find the concentration of the acid and the base
pH = 3.0 means [H+] = 10-3.0
pH = 10 means [H+] = 10-10 M or [OH-] = 10-4 M
Now find the number of mmol of each
mmol H+ = 10-3 x 2.0 = 2.0x10-3 mmol
mmol OH- = 10-4 x 3.0 = 3.0 x 10-4 mmol
mmol H+ excess = 2.0x10-3 – 3.0x10-4 = 1.7x10-3
[H+] = 1.7x10-3/5 = 3.4x10-4 M
pH = - log 3.4x10-4 = 3.5
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2 .Salts of Strong Acids and Bases
The pH of the solution of salts of strong acids or bases will remain constant (pH = 7) where the following arguments apply:
NaCl dissolves in water to give Cl- and Na+. For the pH to change, either or both ions should react with water. However, is it possible for these ions to react with water? Let us see:
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Cl- + H2O HCl + OH- (wrong equation)Now, the question is whether it is
possible for HCl to form as a product in water!! Of course this will not happen as HCl is a strong acid which is 100%
dissociated in water. Therefore, Cl- will not react with water but will stay in
solution as a spectator ion. The same applies for any metal ion like K+ where
if we assume that it reacts with water we will get:
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K+ + H2O KOH + H+ (wrong equation)
Now, the question is whether it is possible for KOH to form as a product in water!! Of
course this will not happen as KOH is a strong base which is 100% dissociated in
water. Therefore, K+ will not react with water but will stay in solution as a spectator ion. It is clear now that neither K+ nor Cl- react with water and the hydrogen ion concentration of solutions of salts of strong acids and bases comes from water dissociation only and will
be 10-7 M (pH =7).
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Lecture 22
Acid-Base Equilibria, Cont…
Weak acids/Bases and their Salts
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3. Weak Acids and BasesA weak acid or base is an acid or base that
partially (less than 100%) dissociated in water. The equilibrium constant is usually small and, in most cases, one can use the concepts mentioned in the equilibrium calculations section discussed previously with application of the assumption that the amount dissociated is negligible as compared to original concentration. This assumption is valid if the equilibrium constant is very small and the concentration of the acid or base is high enough.
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Calculate the pH and pOH for a 0.10 M acetic acid solution. Ka= 1.75x10-5
Solution
[H+]solution = [H+]HOAc + [H+]water
However, in absence of an acid the dissociation of water is extremely small and in presence of an acid dissociation of water becomes negligible due to the common ion effect. Therefore, we can neglect the [H+]water (which is equal to [OH-]) in presence of an
acid since the hydroxide ion concentration is insignificant in an acid solution, therefore we can write
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[H+]solution = [H+]HOAc
The first point is to write the equilibrium where
HOAc H+ + OAc-
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Ka = [H+][OAc-]/[HOAc]
Ka = x * x / (0.10 – x)
Ka is very small. Assume 0.10 >> x
1.75*10-5 = x2/0.10
x = 1.3x10-3
Relative error = (1.3x10-3/0.10) x 100 = 1.3%
The assumption is valid and the [H+] = 1.3x10-3 M
pH = 2.88
pOH = 14 – 2.88 = 11.12
Now look at the value of [OH-] = 10-11.12 M = 7.6x10-12 M =
[H+]from water.
Therefore, the amount of H+ from water is negligible
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Calculate the pH and pOH for a 1.00x10-3 M acetic acid solution. Ka = 1.75x10-5
Solution
The first point is to write the equilibrium where
HOAc H+ + OAc-
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Ka = [H+][OAc-]/[HOAc]
Ka = x * x / (1.00*10-3 – x)
Ka is very small. Assume 1.00*10-3 >> x1.75*10-5 = x2/1.00*10-3
x = 1.32x10-4
Relative error = (1.32x10-4/1.00x10-3) x 100 = 13.2%
The relative error is more than 5% therefore, the assumption is invalid and we have to use the quadratic equation to solve the problem.
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Find the pH and pOH of a 0.20 M ammonia solution. Kb = 1.8x10-5.
Solution
The same treatment above can be used to solve this problem where:
[OH-]solution = [OH-]ammonia + [OH-]water
[H+] = [OH-]water
However, in absence of a base the dissociation of water is extremely small and in presence of the base the dissociation of water becomes negligible due to the common ion effect.
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Therefore, we can neglect the [H+] in presence of ammonia since the hydrogen ion
concentration is insignificant in basic solution, therefore we can write
OH-]solution = [OH-]ammonia
NH3 + H2O NH4+
+ OH-
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Kb = [NH4+][OH-]/[NH3]
1.8*10-5 = x * x / (0.20 – x)
kb is very small that we can assume that 0.20>>x.
We then have:
1.8*10-5 = x2 / 0.2
x = 1.9x10-3 M
Relative error = (1.9x10-3 /0.2) x 100 = 0.95%
The assumption is valid, therefore:
[OH-] = 1.9x10-3 M, [H+] = 5.3x10-12 M = [OH-]water
which is very small.
pOH = 2.72
pH = 11.28
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4. Salts of Weak Acids and Basesa. Salts derived from weak acids/bases and strong
bases/acidsImagine that an acid is formed from two species a
hydrogen ion and a conjugate base attached to it. The acid is said to be strong if its conjugate base is weak while a weak acid has a strong conjugate base. Therefore, we can fairly recognize conjugate bases like Cl-, NO3
-, and ClO4- as weak bases that do not
react with water and thus will not change the pH of water (pH = 7). On the other hand, conjugate bases derived from weak acids are strong bases which react with water and alter its pH.
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Examples of strong conjugate bases include OAc-, NO2
-, CN-, etc.. The same applies for
bases where weak bases are weak because their conjugate acids are strong which means
they react with water and thus alter its pH.
One important piece of information with regards to salts of weak acids and bases is that we have to find their ka or kb as the equilibrium
constants given in problems are for the parent acid or base. Let us look at the following
argument for acetic acid:
HOAc H+ + OAc-
Ka = [H+][OAc-]/[HOAc]
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For the conjugate base of acetic acid (acetate) we have
OAc- + H2O HOAc + OH-
Kb = [HOAc][OH-]/[OAc-]
Let us multiply ka times kb we get
Ka kb = [H+][OH-] = kw , or
Ka kb = kw
Therefore, if we know the ka for the acid we can
get the equilibrium constant for its conjugate base since we know kw. We can find ka for the
conjugate acid by the knowledge of the equilibrium constant of the parent base.
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Find the pH of a 0.10 M solution of sodium acetate. Ka = 1.75x10-5
Solution
[OH-]solution = [OH-]acetate + [OH-]water
[OH-]water = [H+]
Since the hydrogen ion concentration is very small in a solution of a base, we can neglect [OH-]water and we then have
[OH-]solution = [OH-]acetate
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OAc- + H2O HOAc + OH-
Kb = kw/ka
Kb = 10-14/1.75x10-5 = 5.7x10-10
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Kb = [HOAc][OH-]/[OAc-]
Kb = x * x/(0.10 – x)
Kb is very small and we can fairly assume that
0.10>>x
5.7x10-10 = x2/0.1
x = 7.6 x 10-6
Relative error = (7.6x10-6/0.10) x100 = 7.6x10-3%
The assumption is valid.
[OH-] = 7.6x10-6 M
[H+] = 1.3x10-9 M = [OH-]water
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The relative error in neglecting OH- from water = (1.3x10-9/7.6x10-6) x 100 = 0.017%
This validate our assumption at the beginning of the solution that [OH-]acetate >> [OH-]water
pOH = 5.12
pH = 14 – 5.12 = 8.88
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Calculate the pH of a 0.25 M ammonium chloride solution. Kb = 1.75x10-5
Solution
[H+]solution = [H+]ammonium + [H+]water
However, in absence of an acid the dissociation of water is extremely small and in presence of an acid dissociation of water becomes negligible due to the common ion effect. Therefore, we can neglect the [H+]water in presence of an acid since the hydroxide ion concentration is insignificant in an acid solution
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therefore we can write
[H+]solution = [H+]ammonium
The first point is to write the equilibrium where
NH4+ H+ + NH3
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Ka = 10-14/1.75x10-5 = 5.7x10-10
Ka = [H+][NH3]/[NH4+]
Ka = x * x / (0.25 – x)
Ka is very small. Assume 0.25 >> x
5.7*10-10 = x2/0.25
x = 1.2x10-5
Relative error = (1.2x10-5/0.25) x 100 = 4.8x10-3 %
The assumption is valid and the [H+] = 1.2x10-5 M
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Now look at the value of [OH-] = 10-14/1.2x10-5 = 8.3x10-10 M = [H+]from water.
Therefore, the amount of H+ from water is negligible when compared to that from the
acid. The relative error for neglecting the H+ from water = (8.3x10-10/1.2x10-5) x 100 = 6.9x10-3%
pH = 4.92pOH = 14 – 4.92 = 9.08
We should remember that dissociation of water is negligible in presence of an acid or base.
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b. Salts derived from weak acids/bases and weak bases/acids
In this case, both ions are strong conjugates that react with water and may affect the pH of the solution.
For example, look at the calculation of pH for a 0.1 M NH4CN (kb, NH3 = 1.8*10-5, ka,
HCN = 6*10-10)
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NH4+ H+ + NH3 ka = 5.7*10-10
CN- + H2O HCN + OH- kb = 1.7*10-5
Comparing the equilibrium constants suggests that the lower one is much larger than the first. This suggests that:
1. The first equilibrium can be neglected.
2. The solution will be basic.
Since the solution is basic, water dissociation can be assumed to be neglected as well, due to common ion effect.
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It is therefore justified to assume that the only important equilibrium is:
CN- + H2O HCN + OH- kb = 1.7*10-5
Before equil00
EquationCN- + H2OHCNOH-
After equil0.1 - xxx
Kb = [HCN][OH-]/[CN-]
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1.7*10-5 = x * x / (0.10 – x)
kb is very small that we can assume that
0.10>>x. We then have:
1.8*10-5 = x2 / 0.1
x = 1.3x10-3 M
Relative error = (1.3x10-3 /0.1) x 100 = 1.3%
The assumption is valid, therefore:
[OH-] = 1.3x10-3 M
pOH = 2.89
pH = 11.11
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Lecture 23
Acid-Base Equilibria, Cont…
Buffer Solutions
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Buffer SolutionsA buffer is a solution that resists changes in
pH upon addition of small quantities of acids or bases. In other words, a buffer is a solution that keeps its pH almost constant. A buffer is a solution containing a weak acid and its conjugate base or a weak base and its conjugate acid. Below is an explanation of how buffers work:
Let us look at a buffer formed from a weak acid (like acetic acid) and its conjugate base (e.g. sodium acetate); we have the following equilibria:
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HOAc H+ + OAc-
OAc- + H2O HOAc + OH-
In the first equilibrium, acetate is produced from the dissociation of acetic acid. However, there is a lot of acetate added to solution from the sodium acetate. Therefore, the first equilibrium does not occur to any significant degree. We can fairly assume that HOAc will practically not dissociate. The same applies for the second equilibrium where acetic acid is produced. Since a great excess of acetic acid is present in solution from the first equilibrium, one can say that acetate will not practically associate in water.
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Therefore, neither acetic acid nor acetate equilibria will proceed to any significant extent. The equilibrium constant for dissociation of acetic acid can be written as:
Ka = [OAc-][H+]/[HOAc]
Where the [OAc-] = COAc- and [HOAc] = CHOAc
When an acid is added to the buffer above, H+ from the acid will combine with OAc- to form HOAc which is reflected by an increase in
HOAc and a decrease in OAc-.
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In case the amount of H+ added is not very large, the ratio [OAc-]/[HOAc] will only change slightly. A similar argument can be presented when a small amount of base is added to the buffer solution. However, we will treat the buffer problem as we described for the common ion effect
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Buffer CapacityThe degree of buffer resistance to changes in
pH is referred to as buffer capacity. The buffer capacity is defined as the concentration of acid or base ( in moles ) needed to cause a pH change equal to dpH where:
= dCbase/dpH = - dCacid/dpHThe minus sign is because addition of an acid
causes a decrease in pH. A more practical relation to use for calculation of buffer capacity is:
= 2.303 Cacid Cconjugate base / (Cacid + Cconjugate base)
61
Calculate the buffer capacity of a solution containing 0.10 M acetic acid and 0.10 M acetate. Find the pH change when you add NaOH so that
the solution becomes 0.005 M in NaOH
Solution
= 2.303 Cacid Cconjugate base / (Cacid + Cconjugate base)
= 2.303 * 0.10 * 0.10/( 0.10 + 0.10) = 0.115 M
= dCbase/dpH
0.115 = 0.005/dpH
dpH = 0.005/0.115 = 0.043
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Hasselbalch-Henderson Equation
In the acetic acid/acetate buffer described above, we have:
Ka = [OAc-][H+]/[HOAc]
Pka = pH – log [OAc-]/[HOAc]
pH = pka + log [OAc-]/[HOAc]
The above equation is referred to as Hasselbalch-Henderson equation.
63
This equation can be useful to describe buffer limits where the maximum pH limit for a
buffer is when the salt to acid ratio is 10 and the minimum pH limit of the buffer is when
the acid to salt ratio is 10. Inserting these values, one in a time, in the Hasselbalch
equation gives
pH = pka + 1Therefore, the buffer acts well within two pH
units and the midpoint pH value is equal to pka. One should first look at the pka or pkb to choose the correct buffer system which can
be used within a specific pH range.
64
65
Calculate the pH of the buffer containing 0.50 M formic acid (HA, ka = 1.8x10-4) and 0.25 M sodium formate
(NaA).
Solution
First, write the acid equilibrium equation
HA H+ + A-
66
Ka = x (0.25 + x)/(0.50 – x)
Assume that 0.25 >> x
1.8x10-4 = 0.25 x /0.50
x = 3.6x10-4
Relative error = (3.6x10-4/0.25) x 100 = 0.14%
The assumption is therefore valid and we have
[H+] = 3.6x10-4 M
pH = 3.44
67
One can simply work the problem without using x, since the solution is a buffer:
Ka = x * 0.25/0.50
x = 3.6x10-4
[H+] = 3.6x10-4 M
pH = 3.44
68
Calculate the pH of the buffer solution prepared by mixing 10 mL 0.10 M HOAc (ka =
1.75x10-5) with 20 mL of 0.20 M sodium acetate.
SolutionLet us first calculate the concentrations after
mixing (final concentrations of the acid and its conjugate base)
mmol HOAc = 0.10 x 10 = 1.0 mmol[HOAc] = 1.0/30
mmol OAc- = 0.20 x 20 = 4.0 mmol
69
[OAc- = ]4.0/30
The equilibrium equation is
HOAc H+ + OAc-
70
Ka = x (4.0/30 + x)/ (1.0/30 – x)
Assume 1.0/30 >> x
1.75x10-5 = x (4.0/30)/1.0/30
x = 1.75x10-5 /4.0
x = 4.38x10-6
Relative error = {4.38x10-6/(1.0/30)} x 100 = 0.013%
The assumption is valid therefore:
[H+] = 4.38x10-6 M
pH = 5.36
71
Calculate the pH of the solution resulting from adding 25 mL of 0.10 M NaOH to 30 mL of
0.20 M acetic acid.
SolutionLet us find what happens when we mix the two
solutions. Definitely the hydroxide will react with the acid to form acetate which also
results in a decrease in the acid concentration.
mmol OH- = 0.10 x 25 = 2.5 mmolmmol HOAc = 0.20 x 30 = 6.0 mmol
72
Now the mmol base added will react in a 1:1 mole ratio with the acid . Therefore we have
mmol HOAc left = 6.0 – 2.5 = 3.5 mmol
[HOAc] = 3.5/55 M
mmol OAc- formed = 2.5 mmol
[OAc-] = 2.5/55 M
73
Ka = x (2.5/55 + x)/ (3.5/55 – x)
Assume 2.5/55 >> x
1.75x10-5 = x (2.5/55)/3.5/55
x = 1.75x10-5 x 2.5/3.5
x = 2.45x10-5
Relative error = {2.45x10-5/(2.5/55)} x 100 = 0.054%
The assumption is valid
pH = 4.61
74
A buffer solution is 0.20 M in HOAc and in NaOAc. Find the change in pH after addition
of 0.10 mmol of HCl to 10 mL of the buffer without change in volume.
Solution
We should find the initial pH (before addition of HCl) and then calculate the pH after addition.
HOAc H+ + OAc-
75
Initial pH
Ka = x (0.20 + x)/ (0.20 – x)Assume 0.20 >> x
1.75x10-5 = 0.20 x/ 0.20x = 1.75x10-5
Relative error = (1.75x10-5/0.20) x 100 = 8.8x10-3%[H+] = 1.75x10-5 M
pH = 4.76
76
After addition of HCl, the acetate concentration will decrease while the acetic acid concentration will
increase.
mmol HOAc = 0.20 x 10 + 0.10 = 2.1 mmol
[HOAc] = 2.1/10 = 0.21 M
mmol OAc- left = 0.20 x 10 – 0.10 = 1.9 mmol
[OAc-] = 1.9/10 = 0.19 M
77
Ka = x (0.19 + x)/ (0.21 – x)
Assume 0.19 >> x
1.75x10-5 = 0.19 x/ 0.21
x = 1.93x10-5
Relative error = (1.93x10-5/0.19) x 100 = 0.01%
78
[H+] = 1.93x10-5 M
pH = 4.71
pH = 4.71 – 4.76 = - 0.05
Try the same problem replacing NaOH for the HCl and get the change in pH using
the same procedure. You should realize that in this case the acetic acid will
decrease while acetate will increase.
79
Lecture 24
Acid-Base Equilibria, Cont…
Polyprotic Acids
80
Calculate the volume of 14.8 M NH3 and the
weight of NH4Cl (FW = 53.5) you would have
to take to prepare 100 mL of a buffer at pH 10.00 if the final salt concentration is to be 0.200 M. kb = 1.75x10-5
Solution
The key to solving any problem is the equilibrium of substances in solution.
81
Here, we have ammonia and ammonium which are combined in the equation:
NH3 + H2O NH4+ + OH-
Kb = [NH4+][OH-]/[NH3]
We are aware of the pH which means that we can find [OH-] and we are given the concentration of NH4
+ as
0.200 M. Therefore:
pOH = 14 – 10 = 4
[OH-] = 10-4 M
Now we can solve the equilibrium relation to find [NH3]
1.75x10-5 = (0.200x 10-4)/[NH3]
[NH3] = 1.14 M
82
We need 100 mL of 1.14 M to be prepared from 14.8 M so we have
mmol ammonia needed = 1.14x100 = 114 mmol
mL ammonia = mmol/molarity = 114/14.8 = 7.7 mL
Or simply, MiVi = MfVf
14.8* VmL = 1.14 * 100
VmL = 7.7 mL
The weight of NH4Cl can also be found as the volume
and molarity are given in the problem
Mmol NH4Cl = 0.200 x 100 = 20.0 mmol
Mg NH4Cl = 20.0 x 53.5 = 1070 mg
83
How many g ammonium chloride (FW = 53.5) and how many mL of 3.0 M NaOH should be added to 200 mL water and diluted to 500 mL to prepare a buffer at pH 9.5 and a salt concentration of 0.10 M.
84
Here, we have ammonia and ammonium which are combined in the equation
NH3 + H2O NH4+ + OH-
Kb = [NH4+][OH-]/[NH3]
We are aware of the pH which means that we can find [OH-] and we are given the concentration of
NH4+ as 0.10 M. Therefore:
pOH = 14 – 9.5 = 4.5
[OH-] = 10-4.5 = 3.2x10-5 M
85
Now we can solve the equilibrium relation to find [NH3]
1.75x10-5 = 0.10 x 3.2x10-5/[NH3]
[NH3] = 0.18 M
mmol NH3 = 0.18 x 500 = 90 mmol
mmol NaOH = mmol ammonia
VmL NaOH = mmol/molarity = 90/3.0 = 30 mL
mmol NH4+ = 0.10 x 500 = 50 mmol
Total mmol salt = mmol ammonia + mmol ammonium = 90 + 50 = 140 mmol
mg NH4Cl = 140 x 53.5 = 7.49x103 mg = 7.49 g
86
5 .Solutions of Polyprotic AcidsPolyprotic acids are weak acids, except sulfuric acid where the dissociation of the first proton is complete, which partially dissociate in water in a multi step equilibria where hydrogen ions are produced in each step. Examples include carbonic, oxalic maleic, phosphoric, etc. A general simplification in the calculation of pH of such acids is to compare ka1 and ka2 where ,usually ka1/ka2 is a large value (>102) and thus equilibria other than the first dissociation step can be ignored.
87
Let us look at the following example:
H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2
H2PO4- H+ + HPO4
2- ka2 = 7.5 x 10-8
HPO42- H+ + PO4
3- ka3 = 4.8 x 10-13
[H+] = [H+]H3PO4 + [H+]H2PO4- + [H+]HPO4
2-
Looking at the values of the acid dissociation constants for the three steps, it is obvious that the first step occurs about 106 times greater than the second and thus the amount of protons in the second step is negligible ( [H+]H2PO4
- )
compared with the first.
88
In addition, the third step comes from the second step and since the second step contributes a negligible amount of H+ we can also neglect the third step ([H+]HPO4
2-) or any
other consecutive steps. Therefore, only the first equilibrium contributes to the H+ concentration.
89
Example
Find the pH of a 0.10 M H2SO4 (ka2 = 1.7*10-2)
H2SO4 H+ + HSO4-
HSO4- H+ + SO4
2-
The first dissociation is 100% complete, therefore, we have:
[H+] = [HSO4-] = 0.10 M from first dissociation.
The second dissociation is as follows:
HSO4- H+ + SO4
2-
90
1.7*10-2 = x(0.10 + x)/(0.10 – x)
assume that 0.10 >> x
x = 1.7*10-2
Relative Error = (1.7x10-2/0.10) * 100% = 17%
Therefore, the assumption is invalid and the equation must be solved by the quadratic equation.
91
Find the pH of a 0.10 M H3PO4 solution.
Solution
H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2
H2PO4- H+ + HPO4
2- ka2 = 7.5 x 10-8
HPO42- H+ + PO4
3- ka3 = 4.8 x 10-13
Since ka1 >> ka2 (ka1/ka2 > 102) the amount of H+ from the second and consecutive equilibria is negligible if compared to that coming from the first equilibrium.
92
Therefore, we can say that we only have:
H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2
93
Ka1 = x * x/(0.10 – x)
Assume 0.10>>x since ka1 is small (!!!)
1.1*10-2 = x2/0.10
x = 0.033
Relative error = (0.033/0.10) x 100 = 33%
The assumption is invalid and thus we have to use the quadratic equation. If we solve the
quadratic equation we get:
X = 0.028
Therefore, [H+] = 0.028 M
pH = 1.55
94
Find the pH of a 0.10 M H2CO3 solution.
Ka1 = 4.3x10-7, ka2 = 4.8x10-11
Solution
We have the following equilibria
H2CO3 H+ + HCO3- ka1 = 4.3 x 10-7
HCO3- H+ + CO3
2- ka2 = 4.8 x 10-11
Since ka1 is much greater than ka2, we can
neglect the H+ from the second step
95
H2CO3 H+ + HCO3- ka1 = 4.3 x 10-7
Ka1 = x * x/(0.10 – x)
Assume 0.10>>x since ka1 is small
4.3*10-7= x2/0.10
x = 2.1x10-4
Relative error = (2.1x10-4/0.10) x 100 = 0.21%
The assumption is valid and [H+] = 2.1x10-4 M
pH = 3.68
96
If we would like to calculate the amount of H+ coming from the second equilibrium ([H+]second
step = [CO32-]) we substitute 2.1x10-4 for [H+] as
follows:
Ka2 = [H+][CO32-]/[HCO3
-]
But from the first step we have
[H+] = [HCO3-]
ka2 = [CO32-] = 4.8x10-11 = [H+]second step
97
Therefore we are justified to omit second dissociation where the hydrogen ion
concentration obtained from the first step (2.1x10-4 M) is much greater than the [H+]
obtained from the second dissociation (4.8x10-11 M).
98
Buffer Calculations for Polyprotic Acids
A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can form a buffer when combined with its conjugate base (dihydrogen phosphate).
H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2
This buffer operates in the range:
pH = pka + 1 = 0.96 – 2.96
99
Also, another buffer which is commonly used is the dihydrogen phosphate/hydrogen
phosphate buffer.
H2PO4- H+ + HPO4
2- ka2 = 7.5 x 10-8
This buffer operates in the range from 6.1 to 8.1
A third buffer can be prepared by mixing hydrogen phosphate with orthophosphate
as the following equilibrium suggests:
HPO42- H+ + PO4
3- ka3 = 4.8 x 10-13
This buffer system operates in the pH range from 11.3 to 13.3
100
The same can be said about carbonic acid/bicarbonate where
H2CO3 H+ + HCO3- ka1 = 4.3 x 10-7
This buffer operates in the pH range from 5.4 to 7.4; while a more familiar buffer is composed of carbonate and bicarbonate according to the
equilibrium:
HCO3- H+ + CO3
2- ka2 = 4.8 x 10-11
The pH range of the buffer is 9.3 to 11.3.
Polyprotic acids and their salts are handy materials which can be used to prepare buffer solutions of
desired pH working ranges. This is true due to the wide variety of their acid dissociation constants.
101
Find the ratio of [H2PO4-]/[HPO4
2-] if the pH of the solution containing a mixture of both
substances is 7.4. ka2 = 7.5x10-8
SolutionThe equilibrium equation combining the two
species is:
H2PO4- H+ + HPO4
2- ka2 = 7.5 x 10-8
Ka2 = [H+][HPO42-]/[H2PO4
-][H+] = 10-7.4 = 4x10-8 M
7.5x10-8 = 4x10-8 [HPO42-]/[H2PO4
-]
[HPO42-]/[H2PO4
-] = 1.9
Buffers with Specific Ionic Strength
How many mL of 12.0 M acetic acid and how many grams of sodium acetate (FW = 82 g/mol) are needed to prepare a 500 mL buffer at pH 5.0, and having an ionic strength of 0.2. ka = 1.8*10-5
We need to find the concentration of the salt:
= ½ CiZi2
0.2 = ½ (CNa+ * 12 + COAc
- * 12)
CNa+ = COAc
-
0.2 = ½ (2CNa+ )
CNa+ = COAc
- = 0.2 M = CNaOAc
mmol NaOAC = 0.2*500 = 100
mg NaOAc = 100*82 = 8200 mg or 8.2 g
102
HOAc H+ + OAc-
We can now find the concentration of the acid where:
1.8*10-5 = 10-5*0.2/[HOAc]
[HOAc] = 0.2/1.8 = 0.11 M
mmol HOAc = 0.11*500 = 55.6
12.0*VHOAc = 55.6
VHOAc = 4.6 mL
103
How many mL of 12.0 M acetic acid and how many grams of sodium acetate (FW = 82 g/mol) and how many grams of NaNO3 (FW = 85 g/mol) are needed to prepare a 500 mL buffer at pH 5.0, and a salt concentration of 0.1 M and an ionic strength of 0.15 . ka = 1.8*10-5
= ½ CiZi2
= {(NaNO3)+(NaOAc)}0.15 = ½ {(CNa
+ * 12 + CNO3- * 12)+ (0.1*12 + 0.1*12)}
CNa+ = CNO3
- = CNaNO3
0.15 = ½ {(2CNaNO3) + 0.2}CNaNO3 = 0.05 Mg NaNO3 = 0.05*500*85 = 2125mg = 2.125 g
104
CNaOAc = 0.1 M mmol NaOAC = 0.1*500 = 50mg NaOAc = 500*82 = 4100 mg or 4.1 g
HOAc H+ + OAc-
We can now find the concentration of the acid where:
1.8*10-5 = 10-5*0.1/[HOAc][HOAc] = 0.1/1.8 = 0.056 Mmmol HOAc = 0.056*500 = 27.85.0*VHOAc = 27.8VHOAc = 5.6 mL
105
How many grams of Na2CO3 (FW = 106 g/mol) and how many grams of NaHCO3 (FW = 84 g/mol) are needed to prepare a 1000 mL buffer at pH 10.0, and having an ionic strength of 0.2. ka2 = 4.8*10-11
HCO3- CO3
2- + H+
4.8*10-11 = 10-10 [CO32-]/[HCO3
-]
[HCO3-] = 2.1[CO3
2-]
= ½ CiZi2
106
In NaHCO3, CNa+
= CHCO3-
In Na2CO3, CNa+ = 2CCO3
2-
0.2 = ½ (CNa+ *12 + CHCO3
- *12 + CNa+ *12 + CCO3
2- *22)
0.2 = ½ (CHCO3- *12 + CHCO3
- *12 + 2 CCO32- *12 + CCO3
2- *22)
However, [HCO3-] = 2.1[CO3
2-]
0.2 = ½ (2* 2.1[CO32-] + 2 CCO3
2- *12 + CCO32- *22)
[CO32-] = 0.0392 M
[HCO3-] = 2.1[CO3
2-] = 2.1*0.0392 = 0.0824 M
g Na2CO3 = 0.0392 *1000 * 106= 4.16
g NaHCO3 = 0.0824*1000*84 = 6.92g
107
Lecture 25
Acid-Base Equilibria, Cont….
Salts of Polyprotic Acids
109
Fractions of Dissociating Species at a Given pH
Consider the situation where, for example, 0.1 mol of H3PO4 is dissolved in 1 L of solution.
H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2
H2PO4- H+ + HPO4
2- ka2 = 7.5 x 10-8
HPO42- H+ + PO4
3- ka3 = 4.8 x 10-13
Some of the acid will remain undissociated (H3PO4), some will be converted to H2PO4
-,
HPO42- and PO4
3- where we have, from mass
balance:
CH3PO4 = [H3PO4] + [H2PO4-] + [HPO4
2-] + [PO43-]
111
We can write the fractions of each species in solution as
0 = [H3PO4]/CH3PO4
1 = [H2PO4-]/CH3PO4
2 = [HPO42-]/CH3PO4
3 = [PO43-]/CH3PO4
0 +1 + 2 + 3 = 1 ( total value of all fractions sum up to unity).
112
The value of each fraction depends on pH of solution. At low pH dissociation is suppressed and most species will be in the form of H3PO4 while high pH values will result
in greater amounts converted to PO43-.
Setting up a relation of these species as a function of [H+] is straightforward using the equilibrium constant relations. Let us try finding 0 where 0 is a function of
undissociated acid. The point is to substitute all fractions by their equivalent as a function of undissociated acid.
113
Ka1 = [H2PO4-][H+]/[H3PO4]
Therefore we have
[H2PO4-] = ka1 [H3PO4]/ [H
+]
ka2 = [HPO42-][H+]/[H2PO4
-]
Multiplying ka2 time ka1 and rearranging we get:
[HPO42-] = ka1ka2 [H3PO4]/[H
+]2
ka3 = [PO43-][H+]/[HPO4
2-]
Multiplying ka1 times ka2 times ka3 and rearranging we get:
[PO43-] = ka1ka2ka3 [H3PO4]/[H
+]3
But we have:
CH3PO4 = [H3PO4] + [H2PO4-] + [HPO4
2-] + [PO43-]
114
Substitution for all species from above gives:
CH3PO4 = [H3PO4] + ka1 [H3PO4]/ [H+] + ka1ka2 [H3PO4]/[H
+]2 +
ka1ka2ka3 [H3PO4]/[H+]3
CH3PO4 = [H3PO4] {1 + ka1 / [H+] + ka1ka2 /[H
+]2 + ka1ka2ka3
/[H+]3}
[H3PO4]/CH3PO4 = 1/ {1 + ka1 / [H+] + ka1ka2 /[H
+]2 + ka1ka2ka3
/[H+]3}
o = [H+]3 / ([H+]3 + ka1[H+]2 + ka1ka2[H
+] + ka1ka2ka3)
Similar derivations for other fractions results in:
1 = ka1 / ([H+]3 + ka1[H+]2 + ka1ka2[H
+] + ka1ka2ka3)
2 = ka1ka2 [H+] / ([H+]3 + ka1[H
+]2 + ka1ka2[H+] + ka1ka2ka3)
3 = ka1ka2ka3 / ([H+]3 + ka1[H
+]2 + ka1ka2[H+] + ka1ka2ka3)
115
Calculate the equilibrium concentrations of the different species in a 0.10 M phosphoric acid
solution at pH 3.00.
Solution
The [H+] = 10-3.00 = 1.0x10-3 M
Substitution in the relation for o gives
o = [H+]3 / ([H+]3 + ka1[H+]2 + ka1ka2[H
+] + ka1ka2ka3)
o = (1.0x10-3)3/{(1.0x10-3)3 + 1.1x10-2 (1.0x10-3)2 +
1.1x10-2 * 7.5x10-8 (1.0x10-3) + 1.1x10-2 * 7.5x10-
8 * 4.8 * 10-13}
116
o = 8.2x10-2
0 = [H3PO4]/CH3PO4
8.2x10-2 = [H3PO4]/0.10
[H3PO4] = 8.2x10-3 M
Similarly, 1 = 0.92,
1 = [H2PO4-]/CH3PO4
0.92 = [H2PO4-]/0.10
[H2PO4-] = 9.2x10-2 M
Other fractions are calculated in the same manner.
117
6 .pH Calculations for Salts of Polyprotic AcidsTwo types of salts exist for polyprotic acids.
These include:
1. Unprotonated saltsThese are salts which are proton free which
means they are not associated with any protons. Examples are: Na3PO4 and Na2CO3. Calculation of pH for solutions of such salts is straightforward and follows the same scheme described earlier for salts of monoprotic acids.
118
Find the pH of a 0.10 M Na3PO4 solution.
Solution
We have the following equilibrium in water
PO43- + H2O HPO4
2- + OH-
The equilibrium constant which corresponds to this equilibrium is kb where:
Kb = kw/ka3
119
We used ka3 since it is the equilibrium constant
describing relation between PO43- and HPO4
2.
However, in any equilibrium involving salts look at the highest charge on any anion to
find which ka to use.
Kb = 10-14/4.8x10-13
Kb = 0.020
120
Kb = x * x/0.10 – x
Assume 0.10 >> x
0.02 = x2/0.10
x = 0.045
Relative error = (0.045/0.10) x 100 = 45%
Therefore, assumption is invalid and we have to use the quadratic equation. If we solve the
quadratic equation we get:
X = 0.036
Therefore, [OH-] = 0.036 M
pOH = 1.44 and pH = 14 – 1.44 = 12.56
121
2. Protonated Salts
These are usually amphoteric salts which react as acids and bases. For example,
NaH2PO4 in water would show the following
equilibria:
H2PO4- H+ + HPO4
2-
H2PO4- + H2O OH- + H3PO4
H2O H+ + OH-
[H+]solution = [H+]H2PO4- + [H+]H2O
– [OH-]H2PO4-
[H+]solution = [HPO42-] + [OH-] – [H3PO4]
122
Now make all terms as functions in either H+ or H2PO4
-, then we have:
[H+] = {ka2[H2PO4-]/[H+]} + kw/[H+] – {[H2PO4
-][H+]/ka1}Rearrangement gives
[H+] = {(ka1kw + ka1ka2[H2PO4-])/(ka1 + [H2PO4
‑]}1/2
At high salt concentration and low ka1 this relation may be approximated to:
[H+] = {ka1ka2}1/2
Where; the pH will be independent on salt concentration but only on the equilibrium
constants.
123
Protonated Salts with multiple charges
HPO42- is a protonated salt which behaves as an amphoteric
substance where the following equilibria takes place:
HPO42- H+ + PO4
3-
HPO42- + H2O H2PO4
- + OH-
H2O H+ + OH-
[H+] = [H+]HPO4- + [H+]water – [OH-]HPO4
-
[H+] = [PO43-] + [OH-] – [H2PO4
-]
[H+] = ka3 [HPO42-]/[H+] + kw/[H+] – [H+][HPO4
2-]/ka2
Rearrangement of this relation gives
[H+] = {(ka2kw + ka2ka3 [HPO42-])/(ka2 + [HPO4
2-])}1/2
Approximation, if valid, gives:
[H+] = (ka2ka3)1/2
Lecture 26
Acid-Base Equilibria, Cont…
Mixtures of Acids
Acid-Base Titrations
125
Find the pH of a 0.10 M NaHCO3 solution. ka1 = 4.3 x 10-7, ka2 = 4.8 x 10-1
HCO3- H+ + CO3
2-
HCO3- + H2O OH- + H2CO3
H2O H+ + OH-
[H+] = {(ka1kw + ka1ka2[HCO3-])/(ka1 + [HCO3
-]}1/2
[H+] = {(4.3x10-7 * 10-14 + 4.3x10-7 * 4.8x10-11 * 0.10)/(4.3x10-7 + 0.10)}1/2
[H+] = 4.5x10-9 MpH = 8.34
126
The same result can be obtained if we use
[H+] = {ka1ka2}1/2
[H+] = {4.3x10-7 * 4.8x10-11}1/2 = 4.5x10-9 M
This is since the salt concentration is high enough. Now look at the following example
and compare:
127
Find the pH of a 1.0x10-4 M NaHCO3 solution. ka1
= 4.3 x 10-7, ka2 = 4.8 x 10-11
[H+] = {(4.3x10-7 * 10-14 + 4.3x10-7 * 4.8x10-11 * 1.0x10-4)/(4.3x10-7 + 1.0x10-4)}1/2
[H+] = 7.97x10-9 MpH = 8.10
Substitution in the relation [H+] = {ka1ka2}1/2 will
give[H+] = {4.3x10-7 * 4.8x10-11}1/2 = 4.5x10-9 M, which
is incorrectYou can see the difference between the two
results.
128
Find the pH of a 0.20 M Na2HPO4 solution. Ka1 =
1.1x10-2, ka2 = 7.5x10-8, ka3 = 4.8x10-13.
HPO42- is doubly charged so we use ka2 and ka3
as the relation above
[H+] = {(ka2kw + ka2ka3 [HPO42-])/(ka2 + [HPO4
2-])}1/2
[H+] = {(7.5x10-8 * 10-14 + 7.5x10-8 * 4.8x10-13 * 0.20)/(7.5x10-8 + 0.20)}1/2 = 2.0x10-10 M
pH = 9.70
129
Using the approximated expression we get:
[H+] = (7.5x10-8 * 4.8x10-13)1/2 = 1.9x10-10 M
pH = 9.72
This small difference is because ka2kw is not
very small as compared to the second term and thus should be retained.
130
7. pH Calculations for Mixtures of Acids
The key to solving such type of problems is to consider the equilibrium of the weak acid and consider the strong acid as 100% dissociated as a common ion.
131
Find the pH of a solution containing 0.10 M HCl and 0.10 M HNO3.
Solution
[H+] = [H+]HCl + [H+]HNO3
Both are strong acids which are 100% dissociated. Therefore, we have
[H+] = 0.10 + 0.10 = 0.2
pH = 0.70
132
Example
Find the pH of a solution containing 0.10 M HCl and 0.10 M HOAc. ka = 1.8x10-5
Solution
HOAc H+ + OAc-
133
Ka = (0.10 + x) x/(0.10 – x)
Assume 0.1 >> x since ka is small
1.8x10-5 = 0.10 x/0.10
x = 1.8x10-5
Relative error = (1.8x10-5/0.10) x100 = 1.8x10-2%
Therefore [H+] = 0.10 + 1.8x10-5 = 0.10
It is clear that all H+ comes from the strong acid since dissociation of the weak acid is limited
and in presence of strong acid the dissociation of the weak acid is further
suppressed.
134
Find the pH of a solution containing 0.10 M HCl and 0.10 M H3PO4. Ka1 = 1.1x10-2, ka2 = 7.5x10-8,
ka3 = 4.8x10-13.
Solution
It is clear from the acid dissociation constants that ka1>>ka2 and thus only the first
equilibrium contributes to the H+ concentration. Now treat the problem as the
previous example:
H3PO4 H2PO4- + H+
135
Ka = (0.10 + x) x/(0.10 – x)
Assume 0.1 >> x since ka is small (!!!)1.1x10-2 = 0.10 x/0.10
x = 1.1x10-2
Relative error = (1.1x10-2/0.10) x100 = 11 %The assumption is therefore invalid and we have to solve the
quadratic equation. Result will beX = 9.2x10-3
Therefore [H+] = 0.10 + 9.2x10-3 = 0.11pH = 0.96
136
In some situations we may have a mixture of two weak acids. The procedure for pH calculation of
such systems can be summarized in three steps:
1.For each acid, decide whether it is possible to neglect dissociations
beyond the first equilibrium if one or both are polyprotic acids.
137
2. If step 1 succeeds to eliminate equilibria other than the first for both acids, compare
ka1 values for both acids in order to check
whether you can eliminate either one. You can eliminate the dissociation of an acid if ka for the first is 100 times than ka1 for the
second (a factor of 100 is enough since the acid with the larger ka suppresses the
dissociation of the other).
3. Perform the problem as if you have one acid only if step 2 succeeds.
138
Find the pH of a solution containing 0.10 M H3PO4 (ka1 =
1.1x10-2, ka2 = 7.5x10-8, ka3 = 4.8x10-13) and 0.10 M
HOAc (ka = 1.8x10-5).
Solution
It is clear for the phosphoric acid that we can disregard the second and third equilibria since ka1>>> ka2.
Therefore we treat the problem as if we have the following two equilibria:
H3PO4 H+ + H2PO4-
HOAc H+ + OAc-
139
Now compare the ka values for both equilibria:
Ka1/ka = 1.1x10-2/1.8x10-5 = 6.1x102
Therefore the first equilibrium is about 600 times better than the second. For the moment, let us neglect H+
from the second equilibrium as compared to the first.
Solution for the H+ is thus simple
H3PO4 H2PO4- + H+
140
Ka = x * x/(0.10 – x)
Assume 0.1 >> x since ka is small (!!!)
1.1x10-2 = x2 /0.10
x = 0.033
Relative error = (0.033/0.10) x100 = 33 %
The assumption is therefore invalid and we have to solve the quadratic equation. Result
will be
X = 0.028
141
Now let us calculate the H+ coming from acetic acid which is equal to [OAc-]
HOAc H+ + OAc-
Ka = [H+][OAc-]/[HOAc]
1.8x10-5 = 0.028 * [OAc-]/0.10
[OAc-] = 6.4x10-5 = [H+]HOAc
Relative error = (6.4x10-5/0.028) x100 = 0.23%
Therefore, we are justified to disregard the dissociation of acetic acid in presence of
phosphoric acid. Never calculate the H+ concentration from each acid and add them up.
This is incorrect.
Find the pH of a mixture containing 0.10 M H3PO4 and
0.1 M H2CO3 solution.
H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2
H2PO4- H+ + HPO4
2- ka2 = 7.5 x 10-8
HPO42- H+ + PO4
3- ka3 = 4.8 x 10-13
Ka1/ka2 ~ 106, therefore, only first equilibrium is important
H2CO3 H+ + HCO3- ka1 = 4.3 x 10-7
HCO3- H+ + CO3
2- ka2 = 4.8 x 10-11
Ka1/ka2 ~ 104, therefore, only first equilibrium is important
H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2
H2CO3 H+ + HCO3- ka1 = 4.3 x 10-7
Also comparing the two ka values reveals that the first is about 105 times greater than the
second. Therefore, only the first equilibrium is important.
H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2
144
Therefore, we can say that we only have:
H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2
145
Ka1 = x * x/(0.10 – x)
Assume 0.10>>x since ka1 is small (!!!)
1.1*10-2 = x2/0.10
x = 0.033
Relative error = (0.033/0.10) x 100 = 33%
The assumption is invalid and thus we have to use the quadratic equation. If we solve the
quadratic equation we get:
X = 0.028
Therefore, [H+] = 0.028 M
pH = 1.55
