Acid Strength And Molecular Structure Acid Strength and Percent Ionization
Do not confuse the term acid strength with pH. The strength of an acid has to do with the percentage of the initial number of acid molecules that are ionized. If a higher percentage of the original acid molecules are ionized, and therefore, donated as hydrated protons (hydronium ions) then the acid will be stonger. Strong acids are Hydrochloric (HCl(aq)), Hydrobromic (HBr(aq)), Nitric (HNO3), Sulfuric (H2SO4), and
Perchloric (HClO4) acids. In each of these molecular acids the percentage of
ionization is almost 100%.
On the other hand, there are certain acids that will not ionize very easily when added to water. These acids will only give up there protons to water with difficulty. For example, acetic acid (HC2H3O2), Hydrofluoric acid (HF(aq)), HydroCyanic (HCN), Carbonic acid (H2CO3), Sulfurous acid (H2SO3), and Nitrous acid (HNO2) are only a
few of the many "weak" acids. They will only allow 1-5% ionization. In other words, if 5% ionizatiuon takes place only 5 molecules out of 100 will ionize. The other 95 molecules go into solution as "molecules" and not ions. Weak acids can exhibit low pH readings just as strong acids. Acid strength of weak acids has more to do with the value of the ionization constant of the acid.
Acid Strength and Equilibrium Constant
For the following weak acid according to the Law of Chemical Equilibrium: HA + H2O = H3O+ + A
-ionization constant = Ka = [H3O+] [A-] / [HA]
The larger the value of Ka the more Hydronium ion will be in the solution for a given
initial concentration of the acid since the value of Ka must equal the ionization
expression on the right side of the above equation and the Hydronium ion
concentration is in the top portion of that expression. The percentage of ioninzation will therefore be larger. Strong acids have Ka's that approach infinity in value.
Therefore the percentage of ionization would approach 100% for strong acids.
Relative Strengths of Acids and Bases
To understand why the Ka approaches infinity in value, let's compare the acid
strengths of the components in the solution capable of acting as an acid. For example in HCl:
HCl + H2O ---> H3O+ + Cl
-HCl is giving up a Hydrogen ion to water to produce Chloride ion as the reaction proceeds to the right. Water is acting as the base accepting the Hydrogen ion to form Hydronium ion (Hydrated proton). Let's assume that we could reverse this process and proceed to the left. If so then the acid in the reverse process would be H3O+ as it would have to donate a Hydrogen ion to produce H2O. Now if we compare
the acid strength between HCl and H3O+ we find that there is no comparison in acid
strength between HCl and H3O+. The HCl is so much more acidic that H3O+ has no
that zero in the denominator of the equilibrium expression that would make the expression approach infinity since mathematically division by zero equals infinity.
We can come to the same conclusion if we look at the bases in the process. We identified the H2O in the above reaction as the base as the reaction proceeded to the
right. If the reaction was able to reverse and proceed to the left then Cl- would have
to accept a Hydrogen ion to become HCl. We say that the Chloride ion is the
conjugate base of the acid HCl. If we compared the base strength of H2O to Chloride,
water would be a much stronger base and that is another reason that this reaction would proceed exclusively to the right. So we can compare relative strengths between the bases and come to the same conclusion. We can say that HCl is the acid of the conjugate base Chloride ion. We say that the H3O+ is the conjugate acid
of the base water. We have two conjugate pairs in any acid-base reaction.
Notice that the weaker the base the stronger is its conjugate. A conjugate pair in the above ionization of HCl is HCl/Cl-. HCl is the stronger acid (compared to H
3O+) and
Chloride is the weaker base (compared to water). The other conjugate pair is between H2O/H3O+. We found that H3O+ was the weaker acid (compared to HCl) but
that H2O was the stronger base (compared to Cl-)
In general, in any conjugate pair, a strong acid will be paired with a weak base, and a weak acid will be paired with a strong base. Also in any acid base reaction, the strong acid and strong base always proceeds to the weaker acid and base since the strong acid and base are better able to function as acid and base. The greater the difference in the relative strength of the two acids the more one sided will be the process. Strong acids and bases always proceed exclusively in one direction (toward the weaker acid and base). These last two paragraphs above are most profound and will be used in Organic systems as well.
Weaker acids ionizing in water will not be as acidic and therefore the conjugate acid, H3O+, will be able to compete and the reverse process will have a chance to occur.
An equilibrium will form with any weak acid or weak base.
We can compare the relative strengths of acids or bases by comparing their
ionization constants (Ka's or Kb's). The larger the ionization constant the stronger will
be the acid or base. However, the question arises as to what causes one acid to be stronger than another or one base to be stronger than another base?
Relative Acid Strength and Bond Strengths
For acids, this has to do with the strength of the bond holding the Hydrogen to the molecule. When that bond breaks it always breaks heterolytically. That means that both bonding electrons go to the more electronegative atom which is usually an Oxygen, Halogen, Sulfur, or carbon. When the bond breaks the Hydrogen atom is left without any electrons, a naked proton. Naked protons are not very stable so must be stablized or protected by surrounding itself with solvent molecules forming a "solvent cage". For aqueous acids that would be water. The negative end of the polar water molecules are attracted to the positive charge of the proton.
Acid Strength and Atom Size
When the bond breaks between a Hydrogen atom and another atom, both bonding electrons will go to the other atom giving it a negative charge. The larger this atom the greater will be its surface area. The negative charge can be dispersed over this larger surface area and thereby be stabilized. The more stabilized this anion the less likely it will react with the Hydrogen ion to reverse the process. In general if we compare acids with the other atoms are in the same group of the periodic table, the further down in the group the atom is the stronger will be the acid. For example the relative acid strength of the Hydrogen halides is
HI > HBr > HCl > HF. In group 16 (VIA)
H2Te > H2Se > H2S > H2O
For Group 15(VA)
SbH3 > AsH3 > PH3 > NH3
Relative Acid Strength and Electronegativity
If we compared acid strength with acids where the atoms bonded to Hydrogen were in the same period, the difference in the size of the atoms would not be significant. In that case the electronegativity becomes the deciding factor. The greater the electronegativity of the atom the stronger will be the acid. The larger difference in electronegativity between the Hydrogen and the other atom shifts the bonding pair of electrons toward the other atom thus weakens the bond between the Hydrogen and the other atom. Therefore, the bond between the atom and Hydrogen will be more easily broken.
For example:
HF > H2O > NH3 > CH4
HCl > H2S > PH3 > SiH4
Relative Acid Strengths of Oxyacids and Electronegativity
Oxyacids have the Hydrogen atoms in the acid bonded to an Oxygen. The more Oxygens there are in the acid the stronger the acid will be. This can be understood by noting that Oxygen atoms are extremely electronegative. The greater number of Oxygens in the molecule the greater the shift of electrons toward the Oxygens and the weaker the bond between the Hydrogens bonded to Oxygens becomes.
The relative acid strength of the following Oxyacids verifys this trend: H2SO4 is a stronger acid than H2SO3
HNO3 is a stronger acid than HNO2
HClO4 is a stronger acid than HClO3 is a stronger acid than HClO2 is a stronger acid
than HClO
Carboxylic Acid Strength
Acid strength is determined by comparing pKa values of acids.
pKa = -log Ka
Stronger acids have lower pKa's.
Carboxylic acids are stronger acids than other protic substances like alcohols, R-OH. This can be explained by considering the way the negative charge on the conjugate base is distributed.
R-COOH + H2O = RCOO- + H+
If we look at the carboxylate anion, we can see that the two Oxygen atoms and the carbon atom between them have "p" orbitals with the Pi electrons from the double bond between the carbon and one of the Oxygen atoms.(See Fig 1 below)
Carboxylic Acid Strength and Inductive Effect
Increasing the number of halogen atoms will spread the charge out that must more and increase the acid strength of the acid. (See Figure below)
In the three acids in the Figure above, the first acid is less acidic compared to the second acid which in turn is less acidic than the third acid. The equilibrium would be shifted more toward the right side and the Ka value would be larger as more halogen
atoms are attached to the carbon. Consequently, the pKa will decrease in value.
Here is an exercise for you to consider:
Identify each of the compounds below as an acid or base. Which are conjugate pairs?
1. H3O+ (aq) + NH3(aq) ----> H2O(l) + NH4+
2. HCN(aq) + H2O(l) = H3O+ + CN
-3. H2SO4 + H2O(l) ---> H3O+(aq) + HSO4-(aq)
Which of the following bases in the given base pairs will be the strongest?
1. Cl- and HCO 3
-2. HSO3- and NO3
-3. C2H3O2- and ClO4
-Which acid in the following pairs would be the strongest?
1. H2SO3 or H2SO4
2. HBr or HCl 3. H2O or NH3
When you have finished supplying answers to these problems, check your results with the Revised: 4/10/97
Identify each of the compounds below as an acid or base. Which are conjugate pairs?
1. H3O+ (aq) + NH3(aq) ----> H2O(l) + NH4+
Answer: H3O+ (aq) is the acid
NH3(aq) is the base
H2O(l) is the conjugate of the acid H3O+
NH4+ is the conjugate acid of the base NH3
2. HCN(aq) + H2O(l) = H3O+ + CN
H3O+ is the conjugate acid of the base H2O
CN- is the conjugate base of the acid HCN
3. H2SO4 + H2O(l) ---> H3O+(aq) + HSO4-(aq)
Answers:H2SO4 is the acid
H2O is the base
H3O+ is the conjugate acid of the base water
HSO4- is the conjugate base of the acid H2SO4
Which of the following bases in the given base pairs will be the strongest?
1. Cl- and HCO 3
-Answer:HCO3- is the strongest base because its conjugate is H2CO3 which is a
weaker acid than HCl the conjugate acid to Cl
-2. HSO3-2 and NO3
-Answer: HSO3- is the stronger base because its conjugate acid is H2SO3 is
weaker than the conjugate acid of NO3- which is HNO
3. Remember the
weaker the acid the stronger is its conjugate base pair. 3. C2H3O2- and ClO4
-Answer: C2H3O2- is the stronger base because its conjugate acid HC2H3O2 is
weaker that the conjugate acid HNO3 which is paired up with NO3
-Which acid in the following pairs would be the strongest?
1. H2SO3 or H2SO4
Answer:H2SO4 is the stronger acid because in Oxyacids the more Oxygens the
the weaker the bond holding the Hydrogen and the stronger the acid will be. 2. HBr or HCl
Answer:HBr is stronger. If the atom bonded to the Hydrogen is larger in size than the acid will be stronger. Bromine is lower in its group and therefore larger
3. H2O or NH3
Answer:H2O would be stronger acid than NH3.If the atoms connected to the
Hydrogens are in the same period on the periodic Table then the atom with the highest electronegativity will be the strongest acid.
How do you determine the pH of a solution?
The pH Scale
The pH scale is a logrithmic scale. The p" factor" is defined as the log of the molar concentration of whatever follows the letter p and then multiplied by a negative So the pH = -log[H+]
pCl = -log[Cl-]
pKa = -log Ka
pKw = -log Kw
pAg = -log[Ag+]
For strong acid molar concentrations equal to or less than 1, the pH value would have a value from 0-14.
In a water solvent based solution: [H+] [OH-] = K
If the Hydrogen ion concentration is 0.1 moles/liter Then the [OH-] could be found by the equation above:
[OH-] = 1 X 10-14 / 1 X 10-1 = 1 X 10-13
The pOH = -log[OH-] = -log(1 X 10-13) = -(log 1 + log 10-13) = -(0 + -13) = -(-13) =
13
For a [H+] = 0.1 = 1 X 10-1
Then pH = -log 1 X 10-1 = -(0 + -1) = 1
Therefore the pH + pOH = 14
One can have a pH that is a negative value in for example strong acid solutions greater than 1 mole/liter.
For a 10 M H+
pH =-log(1 X 101) = -(0 + 1) = -1
Most acidic and basis substances whose pH lie between the 0-14 range. Do not confuse the term acid strength with pH. The strength of an acid has to do with the percentage of the initial protons that are ionized. If a higher percentage of the original protons are ionized and therefore donated as hydrated protons (hydronium ions) then the acid will be stonger. Strong acids are Hydrochloric, Hydrobromic, Nitric, Sulfuric, and Perchloric acids. In each of these molecular acids the percentage of ionization is almost 100%.
Determination of pH
pH = -log[H3O+] The negative sign is to be multiplied by the log answer you get.
Example: What is the pH of a solution whose [H3O+] = 1 X 10-4 M
pH = - log[1 X 10-4] = - [ log 1 + log 10-4 ]
Note: When you multiply numbers you always ADD their log forms log 1 is always zero
log 10x = x so log 10-4 = -4
therefore:
pH = - [ 0 + (-4) ] = - [-4 ] = +4
Here is a little harder Example:
Calculate the pH of a solution that has a [H3O+] = 2.5 X 10-9
pH = - log [H3O+] = - [ log 2.5 + log 10-9 ]
log 2.5 can be determined using a calculator having the log function key:
1. Enter the number in this case 2.5 2. depress the log key
3. Read the display which in this example would be .3979
log 10-9 = -9
so pH = - [ .3979 + -9 ] = - [ - 8.602 ] = + 8.602 Another example involving a base:
Calculate the pH of a solution that has a [OH-] = 1 X 10-5 M
Determine pOH
pOH = - log [OH- ] = - [log 1 + log 10-5 ] = - [ 0 + -5 ] = - [-5] = +5
pH = 14 - pOH = 14 - 5 = 9
Determining the [H3O+] when given the pH
Example: What would be the [H3O+] of a solution that has a pH = 5.4
[H3O+] = Antilog (-pH) = Antilog (-5.4) = 3.98 X 10-6
Note: Determining Antilogs is done in the following manner using a calculator:
1. Enter the number which you wish to take the Antilog of in this case -5.4 (press the change sign key (+/-) to change the sign to negative)
2. Depress the second function key sometimes called the inverse (inv) key 3. Depress the log key
4. Read the display
Determining the [OH-] when given the pOH
Example: Calculate the [OH-] of a solution that has a pOH = 8.2
[OH-] = Antilog (-pOH) = Antilog ( - 8.2 ) = 6.31 X 10-9
Determining the [H3O+] when given the [OH-]
Example: Calculate the [H3O+] when the [OH-] = 3.2 X 10-3
Water ionizes only slightly giving the following equilibrium:
H2O (l) + H2O(l) = H3O+(aq) + OH-(aq)
According to the Law of Chemical Equilibrium:
Kc = [H3O+] [OH-] / [H2O(l)]2
Note: concentration of pure water is a constant 55.5
Since the molar concentration of water is constant we multiply both sides of the above expression by [H2O]2 that will result in another constant:
Kc [H2O]2 = Kw = [H3O+] [OH-] = 1.0 X 10-14
Kw = 1 X 10-14 = [H3O+] [OH-]
Note: Kw is the ionization equilibrium constant of pure water which always has a
value of 1 X 10-14 at 25 degrees Celsius.
Kw = 1 X 10-14 = [H3O+] [ 3.2 X 10-3]
[H3O+] = 1 X 10-14 / 3.2 X 10-3 = 3.125 X 10-12
Yes, if the acid or base is a strong one, then the dissociation will be 100%
For example, HCl is classified as a strong acid so:
HCl + H2O ----> H3O+(aq) + Cl- (aq)
If I start with a 0.1 M HCl, then I will have 0.1 M H3O+ because for every HCl that
breaks apart then one H3O+ is formed and one Cl- is formed. Since strong acids (and
bases) ionize 100% then all of the original concentration will be converted to H3O+
Typical strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4
Strong Bases: All the Hydroxide compounds of Group 1 and Group 2 metals
LiOH, NaOH, KOH, etc and Be(OH)2, Ca(OH)2, Mg(OH)2, etc
What would be the [OH-] of a .2 M NaOH solution?
NaOH + H2O ----> Na+(aq) + OH- (aq)
.2 M NaOH will produce .2 M OH- since the breakdown is 100%
What would be the [OH-] of a .3 M Ca(OH) 2
Ca(OH)2 + H2O ---> Ca+2 (aq) + 2 OH- (aq)
.3 M Ca(OH)2 will produce .6M OH- because for every one Ca(OH)2 that breaks apart
TWO OH- ions are produced (twice as much) (look at the equation and note the
coefficients)
Now here is an example for you to work out.
Given 0.02M Ba(OH)2 solution:
1. Determine the Hydroxide ion molar concentration 2. Determine the Hydrogen ion concentration 3. Determine the pH
4. Determine to pOH
Given 0.02M Ba(OH)2 solution:
1. Determine the Hydroxide ion molar concentration 2. Determine the Hydrogen ion concentration 3. Determine the pH
4. Determine to pOH
Here is the solution:
1. Write the dissociation Equation
-2. Determine the Hydroxide Concentration
Since all Hydroxides of Group 2 metals are strong bases dissociating 100% [Ba(OH)2] = .02 M
According to the balanced equation for every Ba(OH)2 that dissociates twice as many Hydroxide ions forms
[OH-] = 2(.02) = .04 M
3. Determine the Hydrogen ion concentration
Kw = 1 X 10-14 = [H3O+] [OH-]
[OH-] = 0.04 M (step 2)
1 X 10-14 = [H
3O+] [0.04]
[H3O+] = 1 X 10-14/ .04 = 2.5 X 10-13
4. Determine the pH
pH = -log[H3O+] = -log[ 2.5 X 10-13]
pH = -[-12.6] = 12.6
5. Determine pOH
pH + pOH = 14
12.6 + pOH = 14
pOH = 1.4
Determining Ka of a Weak Acid
Percent of Ionization
The degree of ionization of a weak acid or base is defined as the amount of acid or base that undergoes ionization divided by the initial concentration. It is often expressed as a percentage by multiplying the degree of ionization by 100
The ionization constant of a weak acid may be determined if you know the pH or Hydrogen ion concentration and you have the initial concentration of the acid. The equation showing the ionization of the acid must be known and the equation must be balanced.
When a 0.125M aqueous solution of benzoic acid (C6H5CO2-H) reaches equilibrium,
the pH of the solution is 2.55.
1. What is the Ka for this acid?
2. What is the % of ionization?
Question 1 Solution:
The given pH allows us to determine the Hydrogen ion molar concentration:
[H3O+] = Antilog(-2.55) = 2.8 X 10-3
initial 0.125 0 0
C6H5COOH + H2O = H3O+(aq) + C6H5COO - (aq)
change eq conc 2.8 X 10-3 M
You can solve such a problem by applying the principles of Stoicheometry:
1. Since the final eq. concentration of H3O+ was given to be 2.8 X 10-3 M then
the amount of H3O+ produced had to be 2.8 X 10-3 M
2. If this is so, then the amount of C6H5COO- produced should be the same since
according to the balanced equation for every H3O+ that is produced a
C6H5COO- is also produced (refer to the balanced equation above). So 2.8 X
10-3 M was produced
3. That means that the final eq. conc of C6H5COO- will be 2.8 X 10-3 M
4. If 2.8 X 10-3 M H
3O+ has been produced then the same amount of C6H5COOH
was ionized (ie consumed) 5. If 2.8 X 10-3 M C
6H5COOH was consumed then the final equilibrium conc left
would be the amount ionized subtracted from the initial conc. or
.125 - .0028 = .1222 M = eq conc C6H5COOH
6. State the equilibrium expression according to the Law of Chemical Equilibrium:
Ka = [H3O+] [C6H5COO-] / [C6H5COOH]
These concentration terms are EQUILIBRIUM concentrations found in steps 1,3,and 5 so they may be inserted into the right side of the above expression to get the value of Ka
Ka = [H3O+] [C6H5COO-] / [C6H5COOH]
Ka = [2.8 X10-3] [2.8 X 10-3] / 1.222 X 10-1
Ka = 6.41 X 10-5
Question 2 Solution
Since the amount of Benzoic acid that ionized was 2.8 X 10-3 M the the
degree of ionization = amount of acid ionized / initial concentration of the acid
degree of ionization = 2.8 X 10-3 / 1.25 X 10-1
degree of ionization = 3.5 X 10-2
The percentage of ionization is: P 3.5 X 10-2 ( 100) = 3.5 %
Here is an example that you can do:
Nicotinic Acid is a monoprotic weak acid with the formula HC6H4NO2. A
solution that is 0.012 M in Nicotinic Acid has a pH of 3.39.
1. Determine the Ka of the weak acid
2. Determine the Percent ionization
Nicotinic Acid is a monoprotic weak acid with the formula HC6H4NO2. A
solution that is 0.012 M in Nicotinic Acid has a pH of 3.39.
3. Determine the Ka of the weak acid
4. Determine the Percent ionization
Question 1 Solution
For the sake of convenience lets abbreviate the formula for Nicotinic Acid as HNic
5. Determine the equilibrium concentration of Hydrogen ion from the given pH
[H3O+] = Antilog(-3.39) = 4.07 X 10-4
6. Write the ionization equation indicating the initial concentration of the Nicotinic Acid above the formula and the Hydrogen ion equilibrium concentration below the Hydrogen ion:
7. initial .012 0 0
8. HNic + H2O = H3O+ + Nic
-9. Change
10. Eq conc 4.07 X 10-4
11. Determine the Equilibrium Concentration of Nic
-Since we started with zero Molar of both the Hydrogen ion and the Nicotinate anion and we have 4.07 X 10-4 M of Hydrogen ion then we
must have produced 4.07 X 10-4 M of the Hydrogen ion. Since
according to the balanced equation for every Hydrogen ion is
produced the same amount of Nicotinate ion was produced, then the equilibrium concentration of Nic- would be the same 4.07 X 10-4 M
According to the balanced equation for every Nicotinic Acid molecule that ionizes one Hydrogen ion is formed. It was determined in Step 3 that we had to produce 4.07 X 10-4 M Hydrogen ion so 4.07 X 10-4 M
Nicotinic Acid must have ionized
13. Determine the Equilibrium concentration of Nicotinic Acid
We started with .012 M Nicotinic Acid and according to the previous step we ionized or consumed 4.07 X 10-4 M so what is left is the
equilibrium concentration:
.012 - .000407 = 0.011593 M = 1.1593 X 10-2
14. Write the Ka expression
Ka = [H3O+] [Nic-] / [HNic]
15. Plug in the equilibrium concentrations from steps 1,3,and 5 into the expression.
Ka = (4.07 X 10-4) (4.07 X 10-4) / 1.1593 X 10-2
Ka = 16.6 X 10-8 / 1.1593 X 10-2 = 1.429 X 10-5 Question 2 Solution
% ionization = [HNic] ionized / [HNic]initial(100)
% ionization = 4.07 X 10-4 (100) / 1.2 X 10-2 = 3.39 % Return To the Lesson
Determining Equilibrium concentrations
There are certain acids that will not ionize very easily when added to water. These acids will only give up there protons to water with difficulty. For example, acetic acid, citric acid, Hydrofluoric acid, Hydrogen Cyanide, Carbonic acid, Sulfurous acid, and Nitrous acid are only a few of the many "weak" acids. They will only allow 1-5% ionization. In other words, if 5% ionizatiuon takes place only 5 molecules out of 100 will ionize. the other 95 molecules go into solution as "molecules" and not ions. Weak acids can exhibit low pH readings just as strong acids. Acid strength of weak acids has more to do with the ionization constant of the acid.
Weak Acids and Bases
Here the determination of H3O+ and OH- is more difficult since these substances are
not 100% ionized. Consequently, they form an equilibrium with their ions.
HC2H3O2 + H2O = H3O+(aq) + C2H3O2- (aq)
Note: the = means double arrow
Ka = [H3O+] [C2H3O2-] / [HC2H3O2]
To determine the H3O+ concentration we need to know two pieces of data:
1. The Ka value for this acid
2. The initial concentration of this acid
Example: Determine the [H3O+], the pH, and the % ionization of a 0.1 M HC2H3O2.
the Ka = 1.8 X 10-5
1. Write the equilibrium equation for the weak acid
HC2H3O2 + H2O = H3O+ + C2H3O2
-2. Define algebraically the amount of Acetic Acid ionized.
Let x = [HC2H3O2] ionized (ie used up or converted)
initial 0.1 0 0 HC2H3O2 + H2O = H3O+(aq) + C2H3O2- (aq)
Change -x x x
eq conc 0.1 - x 0 + x 0 +x
3. Write the equilibrium expression.
Ka = 1.8 X 10-5 = [H3O+] [C2H3O2^-] / [HC2H3O2]
4. Express the equilibrium concentrations of acetate ion, Hydrogen ion, and Acetic acid in terms of x
Since x = amount of Acetic Acid that ionizes, according to the balanced equation in step 1 for every acetic acid molecule that ionizes one Hydrogen ion and one acetate ion forms so
x = [H3O+] produced = [C2H3O2^-]produced
Since the initial concentrations of Hydrogen ion and Acetate ion was zero then:
[H3O+]eq = [C2H3O2^-]= 0+x = x
Since x amount of Acetic Acid ionizes and we started with 0.1 M then:
5. Plug in the equilibrium concentrations and the Ka into the expression formed
in step 3 above
1.8 X 10-5 = (x) (x) / 0.1 - x
6. In order to simplify the quadratic equation make the assumption that :
If x << 0.1
Then 0.1 - x = 0.1
and
1.8 X 10-5 = x2 / 0.1
7. Solve for x
1.8 X 10-5 (0.1) = x2
1.8 X 10-5 (1 X 10-1) = x2
1.8 X 10-6 = x2
Take the square root of both sides
1.34 X 10-3 = x
8. Check to see if the assumption above is valid by comparing its value with 5% od the initial concentration
If x < .05(initial conc of acid)
then the approximated x is valid
x = 1.34 X 10-3
.05(0.1) = 5 X 10-3
Comparing the value of x with the 5% figure shows that x is valid
9. Determine the Hydrogen ion concentration
x = [H3O+] = x = 1.34 X 10-3
10. Determine the pH
pH = -log[H3O+] = -log[1.34 X 10-3
pH = -[-2.87] = 2.87
% ionization = amount ionized (100) / initial amount
% ionization = x (100) / 0.1 = 1.34 X 10-3 (100) / 1 X 10-1
% ionization = 1.34
Here is an example that you can do:
The ionization constant of a monoprotic acid, HA, is 2 X 10-10. What is the pH
of a 0.20 molar solution of this acid? What is the % ionization?
The ionization constant of a monoprotic acid, HA, is 2 X 10-10. What is the pH
of a 0.20 molar solution of this acid? What is the % ionization?
1. Write the ionization equation for the acid and balance it
HA + H2O = H3O+ (aq) + A- (aq)
2. Write the Ka expression for the weak mono-protic acid
Ka = [H3O+] [A-] / [HA]
3. Let x = [HA] that undergoes ionization
4. Identify the moles / liter H3O+ and A- that is produced as a result of
the ionization from step 3
According to the balanced equation in step 1 for every HA that ionizes one H3O+ and one A- is produced:
If x = [HA] that ionizes = [H3O+] that is produced = [A-] that is
produced
5. Determine the equilibrium concentrations of HA, H3O+ and A
-If the initial concentration of HA is 0.20 M and x M undergoes ionization then how much HA is left in the equilibrium mixture?
initial - ionized = eq conc= .20 - x = [HA]eq
If there is 0 M of H3O+ and 0 M A- initially and x M of each is produced
(added to system) then what will be the equilibrium conc of H3O+ and
A-?
[H3O+]eq = [A-]eq = 0+x = x
6. Plug in the equilibrium conc of HA , H3O+, and A- from step 5 into the
Ka expression in step 2 and include the given Ka value:
1 X 10-10 = [x] [x] / 0.20 - x
As long as Ka is less than 10-2 in value and the initial concentration of
the acid is large then an approximation can be made which will prevent the need to use the quadratic equation:
Assume that x << 0.20 if so
then 0.20 - x = .20 (approximately)
and
1 X 10-10 = [x] [x] / 0.20
0.20 X 10-10 = x2
4.47 X 10-6 = x = [H 3O+]
8. Check using the "5% Rule" if assumption in step 7 was valid
If x < .05(initial concentration)
then assumption is valid and x value is correct
.05(.2) = .010 = 10 X 10-3
x = 4.47 X 10-6 < 1.0 X 10-5
so x value is valid as calculated in step 7
9. Calculate the pH using the pH definition
pH = -log [H3O+] = - log[4.47 X 10-6] = -[ log 4.47 + log 10-6] =
