AP Chemistry - Equilibrium of Acids and Bases
Arrhenius Model of Acids and Bases
The classical, or Arrhenius, model was developed by Svante Arrhenius in the nineteenth century. He defined an acid as any substance that liberates or yields hydrogen ions (H+) or protons in water. An example would be hydrogen chloride, HCl, gas, which when put in water ionizes to yield hydrogen ions, H+, and chloride ions. The resulting water solution of ionized H+ and Cl- is known as hydrochloric acid.
          HCl(g) + H2O  <------------->   H3O+(aq) + Cl-(aq)
This process involving the breakdown of a substance into ions is known as ionization.
An Arrhenius base is a substance that dissociates in water to produce hydroxide ions, OH-. Two examples of strong, or almost completely dissociated bases are potassium hydroxide, KOH, and sodium hydroxide, NaOH or lye.

Brønsted-Lowry Acid-Base Model
The Arrhenius theory applies only when water is used as the solvent. It restricts the term acid to substances yielding hydronium ions and the term base to those yielding hydroxide ions.
In 1923 J.N. Brønsted and T.M. Lowry independently proposed a much broader and more useful concept of acids and bases. According to their model, a Brønsted-Lowry acid is any substance capable of donating a hydrogen ion or proton to another substance, and a Brønsted-Lowry base is any substance capable of accepting a proton or hydrogen from another substance. In other words, acids are proton donors, and bases are proton acceptors.
According to this concept, any reaction involving the transfer of a proton or H+ from one substance to another is an acid-base reaction.
A base is a proton acceptor and an acid is a proton donor.
For ex.            NH3 + H2O ---->      NH4+    +     OH-
                       base     acid           conjugate        conjugate
                                                        acid                base
An aqueous solution of ammonia is sometimes called ammonium hydroxide, NH4OH. This usage is a matter of convenience since such a compound has never been isolated. A better notation for this is NH3(aq).
 ex.                 2 HCl + H2O ------>      H3O+    +        Cl-
                          acid    base              conjugate        conjugate
                                                             acid                base
Notice that water can act as either an acid or as a base. For this reason it is called amphoteric.

Hydronium ions
Consider HCl, a gas composed of polar covalent molecules. When HCl gas is passed through water we achieve the classic substance, hydrochloric acid, HCl(aq). The original gas does not have any of the properties of the resulting solution. It is reasonable to assume that molecules of HCl react with the water to produce ions. It is these ions that ultimately give the water and HCl solution it's acidic properties.
                    H2O + HCl ----> H3O+(aq) + Cl-(aq)
The reaction above consists of a breaking away of a proton, H+, from the HCl molecules. A stable co-ordinate bond is formed when a proton, H+, shares a pair of electrons with an oxygen atom of the highly polar water molecule. A hydrated proton, called the hydronium ion, H3O+, is formed.
Also, to some extent are the very exotic ions, H5O2+, H7O3+, and H9O4+. It is these ions along with the hydronium ion that impart the acidic properties to a solution.
Most solutions formed by the reaction of polar molecular compounds with water are observed to have either acidic or basic properties.

Concentrated vs. Dilute; Strong vs. Weak
These terms are often the most misused in chemistry.  Concentrated and dilute refer to the concentration of  an acidic or basic substance in a solvent.  eg. 16 M HCl is more concentrated than a 0.5 M solution of the same acid.
Strong and weak refer to the ability of an acid or base to dissociate. A strong acid will dissociate completely in water to form hydronium ions. i.e. 100% of it will form H3O+.   A weak acid or base will only dissociate to a certain percentage. Often a very small percentage only.

Strong and Weak Bases
NH3(aq) is a poor conductor of electricity when compared with NaOH(aq). This means that the degree of dissociation of NH3  in water is relatively small when compared with that of the NaOH.  A base which is only slightly dissociated in aqueous solution is called a weak base; one which is highly dissociated is called a strong base.  All the strong bases happen to be inorganic, that is, the NaOH, KOH, RbOH group. Even Ca(OH)2 and Ba(OH)2 are considered to be strong bases. All of the rest are too insoluble to provide a significant [OH-] in water.  The double arrow convention should be used when dealing with a weak base. A single arrow is to be used when showing the dissociation of a strong base since for all practical purposes, dissociation is 100% complete. i.e., a water solution made from NaOH(s) will have no molecules of NaOH in it.  The NaOH will be completely ionized into Na+ and OH-.

Weak Acids
Weak acids, like weak bases, are equilibrium mixtures, consisting of a higher concentration of reactant molecules than product ions.    A weak acid like acetic acid is written as:
                     H2O + CH3COOH   <------->  C2H3O2- + H3O-
                                                 ^
                                                  |
                                                the acid hydrogen
The H at the start of the formula is termed the acid hydrogen since it is this H atom which reacts with the water to form the acidic properties of the solution. You know it as the H on the -COOH portion of the carboxylic acid in organic acids.
In solution, weak acids have a very low degree of dissociation. eg. a 0.10 mol/L solution of HCl is 100% dissociated, whereas a 0.10 mol/L solution of acetic acid is only about 1.3% dissociated.
eg.                            HCl      +     H2O -----> H3O+    +    Cl-
              before    0.10 M                                 0 M          0 M

                after         0 M                              0.1 M         0.1 M
 

For acetic acid we need some equations:
                      CH3COOH + H2O <------->  H3O+ + CH3COO-
You can see from the equation that the moles of acetic acid dissociated = the moles of H3O+ formed which is also equal to the moles of CH3COO- formed.
Since only 1.3% of the 0.10 mol/L dissociate then the [] of each ion is:
1.3% * 0.10 mol CH3COOH = 0.0013 mol H3O+= 0.0013 mol CH3COO-
100              1 L solvent                       1 L                                   1 L
Follow-Up Problem
A 0.60 mol/L solution of formic acid, HCHO2, is 1.8% dissociated. Calculate the [] of each species (excluding the water) in the equilibrium system.

The pH Scale
Every aqueous solution is either acidic, basic or neutral. There is a quantitative relationship between the concentration of hydronium and hydroxide ions in the solution.
                                       Neutral solution [H3O+] = [OH-]
                                           Acid solution [H3O+] > [OH-]
                                          Basic solution [H3O+] < [OH-]
The brackets as usual denote molar concentrations.
The pH scale is a numerical scale which, for most applications extends from 0 through to 14. The numbers on the scale represent the relative acidity of solutions and can be converted into actual hydronium ion concentrations.
The pH scale is based on the self-ionization of pure water.
Two water molecules will sometimes combine into hydronium and hydroxide ions.
                      H2O + H2O <------>  H3O+ + OH-
Pure water is considered to neutral and the hydronium ion concentration is 1.0 x 10-7 mol/L which is equal to the hydroxide ion concentration.
 ie.               [H3O+] = [OH-] = 1.0 x 10-7 mol/L
The equilibrium law for this reaction should be
Keq = [H3O+][OH-] = (1.0 x 10-7)(1.0 x 10-7) = 1.0 x 10-14
You will please note that at neutrality the molarity of the hydronium ion is 10-7. The 7 plays a part in the pH scale by indicating neutrality. The scale reaches a maximum at 14. Please note again that the hydronium and hydroxide concentrations multiply out to 10-14 M. The pH scale was derived around this relationship:
ie.     [H3O+] = 10-pH mol/L
So the pH is the -log of the [hydronium ion].
    pH      [H3O+     [OH-     pOH 
1 10-1  10-13  13
2 10-2 10-12 12
3 10-3 10-11 11
4 10-4 10-10 10
5 10-5 10-9 9
6 10-6 10-8 8
7 10-7 10-7 7
8 10-8 10-6 6
9 10-9 10-5 5
10 10-10 10-4 4
11 10-11 10-3 3
12 10-12 10-2 2
13 10-13 10-1 1

 
Sample Problems
What is the pH of an HCl solution which has a [H3O+] = 1.0 x 10-3?
             pH = -log[H3O+] = -log[1.0 x 10-3] = -(-3) = 3
What is the pH of an acetic acid solution whose [H3O+]=2.5 x 10-4?
What is the hydronium concentration of nitric acid if the pH=4.0?
         [H3O+] = 10-pH = 10-(4) = 1.0 x 10-4 mol/L
What is the [H3O+] of HCl if the pH = 2.57?
What is the pH of 0.010 mol/L hydrochloric acid?
The pH of a solution may be determined by the use of an electronic instrument known as a pH meter, or through the use of chemical indicators. Acid-base indicators are dyes which undergo slight changes in molecular structure and colour when the pH value of the solution changes.
Specific colours correspond to specific pH values. Some examples are: litmus, phenolphthalein, bromothymol blue, etc. There is a list of acid-base indicators in the databook.

The pOH Scale
The pOH scale is the corollary of the pH scale.
ie. pH + pOH = 14 You'll remember from math class that when you multiply two numbers you only add their logs.
[H3O+] = (1.0 x 10-7 )*(1.0 x 10-7) = 1.0 x 10-14
or
-log[H3O+] = 7 + 7 = 14
Thus a solution that has a pH = 7 must also have a pOH = 7.
What is the pOH of a 0.010 mol/L NaOH solution?
What is it's pH?
What are the hydronium ion and hydroxide ion concentration of a solution prepared by adding 1 mL of 1.0 mol/L HCl to 9 mL of water? Assume that volumes are additive and that the 1 mol/L HCl dissociates completely.
Follow-up problems
What are the hydronium ion and hydroxide ion concentrations of a solution made by adding 1 mL of 0.1 mol/L NaOH to 9 mL of water?
Find the pH, pOH, [OH-] of a 0.00010 mol/L HCl solution.
Find the pH of a 0.00325 mol/L NaOH solution.
What is the hydronium ion concentration of a solution that has a pH = 2.6?
Go to the Acid-Base Water,  pH,  and pOH Worksheet

Equilibrium in Bronsted Acid-Base Systems
In acid-base reactions, equilibrium favours the production of the weaker acid and base. ie; the stronger the reacting acid and base, the more complete the reaction.
eg.                    HClO4(l) + H2O(l) -----> H3O+(aq) + ClO4-(aq)
Think of a Bronsted acid-base reaction as a competition between the 2 bases in the system for protons. The stronger base "wins" and forces the equilibrium in the direction of the weaker acid and base. Above, the H2O and ClO4- ions, both bases, are competing for protons. H2O, the stronger of the two, "wins" and the equilibrium is shifted to the right in the direction of the weaker base, ClO4- ions. HClO4, the stronger acid, has a greater tendency to give it's protons to H2O than H3O+ ions have to give their protons to ClO4- ions. The equilibrium is displaced so far to the right that the reaction is essentially complete.
In contrast to HClO4, HCN is a weakly dissociated acid.
         HCN(g) + H2O(l) <--------=====> H3O+(aq) + CN-(aq)
The reaction above indicates that the equilibrium that is reached has a majority of the HCN molecules unreacted. The H3O+ ions are a much stronger acid than the HCN, and the CN- ions are a much stronger base than H2O. This means that the CN- ions "win" in the competition for protons and force the equilibrium left.
The stronger the acid, the weaker is its conjugate base.
Please note that HClO4, a very strong acid, is the conjugate acid of ClO4- ions. ClO4- is itself a very weak base. The perchlorate ions, are so weak that any base below it in the acid-base table in your databook can take H+ ions away from the ClO4- ions. Thus H2O completely removes the acid hydrogen from the HClO4 molecules and forms hydronium ions, H3O+.
In general, the strong acids in the upper part of the left hand column have the greatest tendency to react with the strong bases in the lower of the table.

Ion Concentration and pH in Detail
Strong Acids      HClO4, HI, HBr, HCl, HNO3 & H2SO4
If we use the general symbol, X, to represent the anion of these acids, then the general equation becomes:
                      HX + H2O ---> H3O+(aq) + X-(aq)
Dissociation is 100% complete.

Strong Bases Any alkali metal with hydroxide is considered to be a strong base and capable of undergoing complete dissociation.
                O2- and NH2- also react with water to give OH- to completion.
It is easy to calculate the [H+] and the pH of a strong acid or base. Since a strong acid is completely dissociated in dilute solution the [H3O+] essentially equals the original concentration of the solute.
eg.         a 1.0 x 10-3 mol/L (0.0010 mol/L) HCl solution
               HCl      +    H2O --------> H3O+(aq) + Cl-(aq)
  start  0.0010 M                               -------         -------

  after        0.0 M                            0.0010 M     0.0010 M
                                                   (1.0 x 10-3 M)

The pH will be 3, the pOH will be 11.
 

Weak Monoprotic Acids
The general equation for a weak acid is:
                        HX + H2O ------> H3O+ + X-
Applying the equilibrium law to this equation we get:
In dilute solutions of acids and bases the [H2O] is essentially a constant at 55.4 mol/L. (1000 mL of H2O = 1000 g of  H2O/18.02 g/mol = 55.4 moles). This, when multiplied with Keq gives us a new constant value which we call Ka.
     OR 
The Ka values for acids and bases stronger than H3O+ are not listed because of their high degree of dissociation, the denominator would approach zero and result in an infinitely large Ka.
Calculate
(a) the [H3O+],
(b) the pH and
(c) the % dissociation for a 0.100 mol/L solution of acetic acid at 25oC, Ka for CH3COOH is 1.8 x 10-5.
The equation for the dissociation is:

                 CH3COOH + H2O -------> H3O+(aq) + CH3 COO-(aq)
 

Let 'x' be the concentration of acetic acid that dissociates:

                CH3COOH + H2O -------> H3O+(aq) + CH3 COO-(aq)
   start          0.100 M                            -------            -------
  finish          0.100 - x                               x                      x
 

1.8 x 10-3    (x) (x)
                    0.100 - x

1.8 x 10-4 - 1.8 x 10-3 x = x2

Upon rearrangement x2 +1.8 x 10-3x -1.8 x 10-4 = 0
 

Use the quadratic equation to solve for 'x'

a = 1    b = 1.8 x 10-3    c = -1.8 x 10-4

Therefore the [H3O+] is 1.25 x 10-2 mol/L

b) The pH is 1.9 based upon -log[H3O+]

c) The % dissociation is 1.25 x 10-2 M * 100 = 12.5%
                                      0.100 M
 

Go to the Acid-Base Ka, Kb, pKa, pKb Worksheet

Weak Bases
Similar calculations are used for weak bases.
eg.      NH3(g) + H2O(l) <---------> NH4+(aq) + OH-(aq)

OR 
 

This tells you that the dissociation constant for any of the conjugate bases can be obtained from the value of Kw divided by the appropriate value of Ka. (Ka can also be obtained by dividing Kw by Kb).
eg. NH3   Kw = 1.0 x 10-14               Ka for NH4+ = 5.6 x 10-10

        Kb = 1.0 x 10-14  =  1.8 x 10-5
                 5.6 x 10-10
 

Follow-Up Problem
What is the [OH-] of a 0.10 mol/L solution of NaCN?
When NaCN dissolves in water it dissolves completely because it is a sodium salt.
                  NaCN ------->    Na+    +    CN-
                   0.1 M             0.1 M      0.1 M
However the CN- ions that are produced then react with water and set up an equilibrium.
         CN-(aq) + H2O <-------->  HCN(aq) + OH-
         0.1 M -x                                x                x
When we set up the equilibrium equation we must set it us as a Kb equation. Ka would be used if HCN was reacting with water. This time the salt of HCN, the CN- is reacting and it is acting as a base. Therefore we must use the Kb value.
2.04 x 10-5  (x)(x)
                       0.1 - x

x2 +2.04 x 10-5 x -2.04 x 10-6 = 0

Apply the quadratic formula and solve for 'x'.

x = 1.42 x 10-3 mol/L      Therefore the [OH-] = 1.42 x 10-3 mol/L.
 

Follow-Up Problems
Calculate the pH of a 0.10 mol/L NaCH3COO solution?

Solutions Containing Ions Which Do Not Hydrolyse Appreciably
Cl-, NO3-, Br-, I-, ClO4- and SO42- are all the salts of the very strong acids. They are therefore extremely weak bases.(ie. Kb=0) They do not appreciably affect the acidity or alkalinity of an aqueous solution. Na+, K+ and Rb+ and any other alkali metal ion also do not affect the water's pH either. The same is true for the charged ions in Group IIA. Using these generalizations we can predict that solutions of NaCl, KI, BaCl2 and RbNO3 which are made from the above cations and anions are essentially neutral.
All other ions tend to cause pH shifts. Cations tend to make solutions more basic whereas anions tend to bring a solution closer to acidity. If both weakly dissociated anions and cations are present then the pH will have to be determined mathematically.
Follow-Up Problem
1. Without making any calculations, predict the relative acidity (ie, pH is >, < or =7) of these aqueous solutions.

a) 0.1 mol/L KCl

b) 1.0 mol/L K2CO3

c) 0.1 M NH4NO3

d) 0.1 M NaF

e) 1.1 mol/L Na2SO3
 

2. For question (e) above do the calculations and prove that the pH is what you predicted.
Go to the Acid-Base Hydrolysis of Ions Worksheet

Buffer Solutions
Buffers are solutions with the ability to resist the addition of strong acids or strong bases, within limits.
They play an important role in chemical processes where it is essential that a fairly constant pH is maintained. In many industrial and physiological processes, specific reactions occur at some optimum pH value. When the pH varies to any extent from the optimum value, undesirable reactions and effects may occur. For example, the pH of your blood lies at about 7.35. If this value drops below 7.0 (acidosis) the results are fatal. Also if it rises above 7.7 (alkalosis) the results are as well fatal. Fortunately our blood contains a buffering system which maintains the acidity at the proper level. If it were not for the protection of the buffering system, we could not eat and adsorb many of the acidic fruit juices and foods in our diet.
A typical lab buffer is CH3COOH and its salt NaCH3COO. Most buffer solutions are made up using a weak acid and its sodium salt! When a strong base such as NaOH is added to the buffer, the acetic acid reacts with and consumes the excess OH- ion. The OH- reacts with the H3O+ ion from the acid in the following reaction:
                H2O + CH3COOH   <------->   H3O+ + CH3COO-

                             H3O+ + OH- <----------> H2O
 

The OH- reduces the H3O+ ion concentration, which causes a shift to the right, forming additional CH3COO- and H3O+ ions. For practical purposes each mole of OH- added consumes a mole of CH3COOH and produces a mole of CH3COO-.
                   OH- + CH3COOH <------------>   CH3COO- + H3O+
When a strong acid such as HCl is added to the buffer, the hydronium ions react with the CH3COO- ions of the salt and form more undissociated CH3COOH.
                  H3O+ + CH3COO-   <----------->CH3COOH + H2O
As you would expect, there is a limit to the quantity of H+ or OH- that a buffer can absorb without undergoing a significant change in pH. If a mole of HCl is added to a litre of buffer solution containing 0.5 moles of sodium acetate/acetic acid buffer the H+ completely consumes the buffer and results in a drastic change in pH.
The blood buffer is made up from the dissolved carbon dioxide in the plasma.
              CO2(g) + H2O <------->    H2CO3  <------>  HCO3- + H3O+
When a base is added it reacts with the carbonic acid.
                       OH- + H2CO3 <-------->  HCO3- + H2O
When an acid is added it reacts with the bicarbonate ion.
                     H3O+ + HCO3-    <--------->  H2CO3 + H2O
Because there are both a base-neutralizer and an acid-neutralizer then we have a buffer.

Buffer Components
A buffer has two components.
         HA                      NaA ---> Na+ + A-
    a weak acid   &     a soluble salt of the acid
Therefore any extra H3O+ will be neutralized by the A- in the buffer.
                  H3O+ + A-   <------->  HA + H2O
And any extra OH- that is added will be neutralized by the acid.
                       HA + OH-   <------>  A- + H2O
There are 3 basic types of calculations that can be done with buffer system and these will be covered next.

The pH of a Buffered Solution
Because HA is a weak acid, very little of it is dissociated at equilibrium. Even if HA were the only solute. But the solution also contains A- from the dissolved salt. The presence of A- suppresses the already slight ionization of HA by shifting the following equilibrium to the left. You should recognize this for the common ion effect which it is.
HA   <-------->    H+ + A-
ie. The value of [HA] at equilibrium = [HA] initially

and the value of [A-] at equilibrium = [A-] initially.

Therefore we can make 2 assumptions:

[HA]equilibrium = [HA]from the initial [acid] = [acid]

[A-]equilibrium = [A-]from the initial [salt] = [anion]

Therefore we can do this:

[H+] = Ka x [acid]
                   [anion]
 

If we take the -log of both sides:

-log[H+] = -log Ka + (-log  [ acid] )
                                          [anion]

pH = pKa - log   [acid]
                         [anion]
 

Since  we end up with the following:
(Henderson-Hasselbach Equation)
Two factors govern the pH in a buffered solution.
(1) pKa of the weak acid;
(2) ratio of the initial molar []'s of the acid and it's salt.
If we prepare a solution where [anion] = [acid] then the

log [anion] = log 1 = 0
      [acid]
 

Therefore the pH of the solution will turn out equal to the pKa of the weak acid.
What mostly determines where on a pH scale a buffer can work best is the pKa of the weak acid. Then by adjusting the ratio of [anion] to [acid], we can cause shifts so that the pH of the buffered solution comes out on one side or the other of this value of pH.

Calculating the pH of a Buffered Solution
To study the effect of a weakly acidic culture medium on the growth of a certain strain of bacterium, a microbiologist prepared a buffer solution by making it with 0.11 mol/L NaCH3COO, sodium acetate, and also 0.09 mol/L CH3COOH, acetic acid, What is the final pH?
Ka for acetic acid = 1.8 x 10-5
Therefore pKa = 4.74
[acid] = 0.090 mol/L      [anion] = 0.11 mol/L

pH = 4.74 + log (0.11) = 4.74 + log 1.2 = 4.74 + 0.079 = 4.82
                          (0.09)
 

You may sometimes see the Henderson-Hasselbach equation given in some chemistry sources as pH = pKa + log [salt]/[acid]. DON'T USE IT! It only works for monovalent salts. Use the formula given previously in these notes.

The Effectiveness of a Buffer
Suppose we drop 0.01 moles of strong base into our buffer from the last example. What will be the measured effects?
            HA     +   OH-<------->  A-       +    H2O
start   0.09 M      -----               0.11 M
finish -0.01 M    -0.1 M           +0.01 M
          0.08 M      0.0 M            0.12 M
pH = pKa + log [anion]
                          [acid]

      = 4.74 + log (0.12)
                          (0.08)
 

      = 4.74 + log 1.5
      = 4.74 + 0.18
      = 4.92     The change in pH is small compared to what it would have been in pure water!

Buffer Capacity
No buffer has an unlimited capacity. ie; buffers can only absorb so much abuse before they are destroyed. The capacity of a buffer is the amount of acid or base it can handle before the pH of the solution changes drastically.
If you add enough strong acid to neutralize all of the buffers basic component, then additional strong acid will make the pH drop rapidly. The same applies for a strong base and the buffers acidic component. The buffer's pH is a function of it's pKa and the ratio of concentrations of anion and acid, but the buffer's capacity depends upon actual concentrations.

Preparation of a Buffer
To prepare a buffer first choose an acid with a pKa within a ±1 pH unit of the desired value. Then manipulate the ratios to get the desired pH. A solution buffered at pH 5.00 is needed in a chemistry experiment. Can we use acetic acid and sodium acetate to make it? If so, what ratio of acetate ion to acetic acid is needed?
Ka of acetic acid = 1.8 x 10-5 pKa = 4.74
the pKa is within a range of 5 ± 1      ie. between the values of  4 - 6.
therefore, acetic acid is okay to use.
next use  pH = pKa + log [anion]
                                        [acid]

            5.00 = 4.74 + log [anion]
                                         [acid]

            0.26 = log [anion]
                             [acid]

          100.26 = [anion]
                        [acid]

                    = 1.8
 

Therefore the ratio of anion to acid is 1.8 to 1. Therefore, use a ratio of 1.8 moles of acetate ion to 1.0 mole of acetic acid. The solution will then be buffered at a pH of 5.00

OR
 

We could use a ratio of  0.18 moles to 0.100 moles if we wanted a smaller buffering capacity.
Go to the Acid-Base Buffers Worksheet
Go to the Acid-Base More Buffers Worksheet

A Basic Buffer
NH3 + H2O <------->  NH4+1 + OH-1
Take the -log of both sides
A generalized version of this basic Henderson-Hasselbach equation is
A chemistry student needs 250 mL of a solution buffered at a pH of 9.00. How many grams of ammonium chloride have to be added to 250 mL of 0.2 mol/L NH3 to make such a buffer? (Volume is assumed not to change.)
pH = 9.0      pOH = 5.0

[base] = 0.2         pKb of ammonia = 4.74 (look up from datatable)

pOH = pKb +log [cation]
                             [base]

5.00 = 4.74 + log [cation]
                              (0.2)

0.26 = log [cation]
                   (0.2)

[cation] = 100.26 X 0.2

            = 1.8 * 0.2

            = 0.36 mol/L of the NH4+1.

But we only need enough for 250 mL  so   0.36 mol  = __x___
                                                                 1000 mL     250 mL

                                                                 x = 0.09 moles of NH4+1 ions needed

Since the NH4+1 comes from NH4Cl then we also need 0.09 moles of NH4Cl.
g = n * mm
   = 0.09 moles * 53.5 grams/mole
   = 4.8 grams of the salt are required.

Volumetric Analysis
In many acid-base reactions, the equilibrium is displaced almost completely toward the product side. These reactions may be considered quantitative and can be used as the basis for the analysis of the amount of acid or base in a given sample. The process is termed volumetric analysis.
The requirements are:
1) Only a single, specific reaction must take place between the unknown substance and the known substance used for the analysis.
2) The unknown substance must react completely and rapidly with the added standard reagent. ie; it must be a quantitative reaction.
3) An indicator or method must be available to signal when all the unknown substance has reacted with the added standard reagent.
The usual objective is to determine the mass or percentage of a qualitatively identified component in a sample whose quantitative amount is unknown. If the sample is a solution, the objective may be to determine its molar concentration.

Acid-Base Titration
An acid-base titration is just a method by which we can perform a volumetric analysis. The concentration of an acid-base solution may be determined by measuring the volume of the base of known concentration needed to react completely with a specific volume of an acid solution.
Definitions
Standard solution: A solution of known concentration.

Titration: the process of adding the standard solution from a graduated tube in controlled amounts.

Buret: a graduated tube with a dispensor control at the bottom.

Stoichiometric Point: the point at which equal molar quantities of reactants are present.

Chemical Indicators: a chemical that will change colour at or very near the ph of the stoichiometric point.

Endpoint: The place in the titration when the indicator changes colour.
 

The indicator used, should in theory coincide with the stoichiometric point of the reaction.
A 0.660 mol/L NaOH solution is used to determine the molar concentration of H2SO4 solution. What is the molarity of the acid, 20.0 mL of which is just neutralized by 36.0 mL of the standard base?
This is a standard problem which can be solved using:
MaVa = MbVb where M is the molarity, V is the volume, a is 'of the acid', b is 'of the base'.
Therefore Ma = Mb * Vb = 0.660 mol/L * 36.0 mL = 1.19 mol/L
                                Va                                 20.0 mL
Follow-Up Problems
1. What is the molarity of a hydrochloric acid solution, 30.0 mL of which is just neutralized by 48.0 mL of 0.100 mol/L NaOH?

2. How many mL of 0.100 mol/L HCl are required to neutralize 25.0 mL of 0.100 mol/L Ba(OH)2

Standard Solutions
To make up a standard solution you usually add a liquid solution of unknown concentration to a solid that has been accurately weighted.
eg. An HCl solution is standardized using pure Na2CO3(mm=106.00) as a primary standard. What is the molarity of the acid if 30.00 mL of the acid solution is required to react completely with a 0.500 gram sample of Na2CO3?
Na2CO3(s) + 2 HCl(aq) ---------> 2 NaCl(aq) + H2O(l) + CO2(g)
moles of Na2CO3 = 0.500 grams = 4.72 x 10-3 moles
                               106.00 g/mol
Reaction works on a 1:2 basis 1 Na2CO3 = 2 HCl
                                                   4.72 x 10-3        x

                                            x = 9.44 x 10-3 moles of HCl

                                  M = moles = 9.44 x 10-3 moles = 0.315 M
                                              L       0.030 L (30 mL)
 

The molarity of the HCl is not standardized at 0.315 mole/L
Follow-Up Problem
A sodium hydroxide solution is standardized by reaction with benzoic acid, HC7H5O2, (mm=122.00 grams/mole). A 2.00 gram sample of benzoic acid required 35.00 mL of NaOH to reach the endpoint. What is the molarity of the base? Benzoic acid is monoprotic.

Percentage Purity Of A Sample
We can use the HCl standardized solution from above to analyze an impure sample of Na2CO3. In a calculation of this type of analysis, convert the moles of unknown into mass units and then apply the general formula:
% of component = grams of component * 100
                                     grams of sample
What is the percentage of Na2CO3 in an impure sample if 25.00 mL of 0.315 mol/L HCl is required to react completely with a 0.600 gram sample of the impure salt? The impurities do not react with HCl.
2 HCl(aq) + Na2CO3 ------> 2 NaCl(aq) + CO2(g) + H2O(l)
moles of HCl  =  0.315 mol/L * 0.025 L = 0.007875 moles of HCl
0.007875 mol of HCl * 1 mole of Na2CO3 = 0.00394 moles Na2CO3
                                            2 moles HCl
0.00394 moles of Na2CO3 * 106.00 grams/moles = 0.418 g Na2CO3
% purity = 0.418 g * 100% = 69.7%
                   0.600 g
Follow-Up Problem
What is the percentage of acetic acid, CH3COOH, in a sample of vinegar if 35.00 mL of 0.468 M NaOH solution is required to neutralize a 25.00 mL sample of the vinegar which has a density of 1.06 g/mL. Acetic acid is monoprotic and has a mm = 60.00 g/mole.

Indicators
Acid-base indicators are weak acids or bases themselves which establish an equilibrium between their molecular and ionic forms. The molecular form has a different colour than the ionic form. Changes in the pH cause a shift in the equilibrium which favours one species over the other. For example if we look at phenolphthalein indicator:
                HPh         +      B <-------->       HB      +     Ph-
             colourless          base                  conjugate       magenta
        phenolphthalein                                   acid
           protonated
Test your understanding of indicator behaviour: The acid colour of a certain indictor, HIn, is yellow and the basic colour is blue. When this indicator is added to a certain acid HB, the solution turns yellow. The equation for the reaction may be written as:
HB + In  <--------->  HIn + B-
 
Which is the stronger acid, HB or HIn?
Take a look at the Indicator Table in the Databook.
Methyl Orange is red in a pH of 3.2 or less and yellow in a solution of 4.4 or greater. The transition colour of orange is seen between these two points.
At the transition point [HIn]=[In-]. that is, the concentration of the red-coloured species is equal to that of the yellow-coloured species so that the solution appears orange.
Thus methyl orange can be used for a titration reaction in which the pH of the solution at the stoichiometric point is between 3.2 and 4.4. The Ka for methyl orange at 25oC is 4.0 x 10-4. Thus ideally, the solution at the stoichiometric point should have a [H3O+] = 4.0 x 10-4 mol/L and a pH of 3.4

Titration Curves
Below is a graph depicting the changes that occur in the pH of a solution during a titration of a typical strong acid with a typical strong base.
At the start, before any NaOH has been added, the solution is simply 25.00 mL of 0.20 M HCl. Since the HCl is a strong acid and is 100% dissociated, the initial pH is quite low.
As we add more and more NaOH solution we are of course affecting the concentration because of volume changes. Between 0 and approximately 24 mL of NaOH added, a large increase in the pH is not seen. All that is happening is that the added base is changing some of the acid to the salt, NaCl. NaCl, neither of which ions hydrolyses, so the concentration of remaining acid gradually becomes more and more dilute. By adding just enough base to neutralize the acid the pH should change drastically to a pH of 7.00 or neutrality. The addition of a little more base will cause the resulting solution to become quite caustic.
The graph above indicated that a rapid change in pH with a small change in volume of standard solution is desirable if the indicator is to give a sharp colour change at the endpoint.

Titration of a Weak Acid by a Strong Base
Below is a graph of the titration curve for a typical weak acid with a typical strong base.
At the equivalence point, the solution contains sodium acetate, which has an anion, but not a cation, that hydrolyses. The acetate ions however do ionize to give a slightly basic solution. For this reason, at the equivalence point where all the acid has neutralized all the base, there is still the acetate ions in solution which ionize the water.

Titration of a Weak Base by a Strong Acid
The graph below shows the titration curve for a typical weak base and a typical strong acid.
At the equivalence point, when all the acid has neutralized all the base there will still be the cation NH4+ in solution. This ion will cause the water to ionize and force the resulting pH of the solution downwards.

Go to the Acid-Base Titrations Worksheet
Polyprotic Acids
Most of the acids like HCl, HNO3, CH3COOH are monoprotic because they only produce 1 H+ per molecule. Sulphuric acid, H2SO4, and phthalate acid, C6H4(COOH)2 are diprotic acids. Phosphoric acid, H3PO4, and citric acid are triprotic acids. There are two commonly encountered acids that have never been isolated as pure species. These are carbonic acid, H2CO3, and sulphurous acid, H2SO3.
There is a separate ionization constant for the ionization of each hydrogen ion from a polyprotic acid.
              H2CO3(aq) <-------->  H+(aq) + HCO3-(aq)
              carbonic                                       bicarbonate
                 acid                                                 ion
Ka1 = [H+][HCO3-] = 4.5 x 10-7 pKa1 = 6.35
             [H2CO3]
The second acid ionization Ka2 for this equilibrium:
               HCO3-(aq)   <------->   H+(aq)    +    CO32-(aq)
               bicarbonate                                         carbonate
                    ion                                                       ion
Ka2 = [H+][CO32-] = 4.7 x 10-11 pKa2 = 10.33
             [HCO3-]
In general for any diprotic acid the value of Ka1 > Ka2 since it is easier to pull a H+ from a neutral molecule. Ka1 is usually 104 to 105 times larger than Ka2.
Because the first acid ionization constant, Ka1, is so much greater than the second, Ka2, the pH of a dilute solution of a weak diprotic or triprotic acid can be calculated just from the value of Ka1. The contributions to [H+] from the 2nd or 3rd dissociation are so small that they can be ignored.

Calculating [H+] and [A2-] at Equilibrium in a Solution of a Weak Diprotic Acid H2A
Suppose that for the diprotic acid, H2A,
Ka1 = 1.0 x 10-5 and Ka2 = 1.0 x 10-9.
What are the values of [H+] and [A2-] at equilibrium in 0.100 mol/L H2A?
                 H2A  <-------->  H+  +   HA-
Initial        0.1 M                   -----    -----
Changes     -x                         +x       +x
Corrections
caused by
second
equilibrium   0                        +y       -y   
The second conjugate of H+ at the expense of HA-

Final         0.1 -x                    x+y      x-y   
Since y is very very small x+yx and x-yx.

 

Ka1 = (x)(x) = 1.0 x 10-5
          0.1 - x

1.0 x 10-6 - 1.0 x 10-5x = x2

x2 + 1.0 x 10-5x -1.0 x 10-6 = 0 (use the quadratic equation on this)
 

x = 9.95 x 10-4 mol/L

Concentration of [A2-] at equilibrium

Ka2 = [H+][A2-] but [H+] = [HA-]
              [HA-]

Ka2 = [A2-] = A2-

[Ka2] = Ka2 = 1.0 x 10-9 mol/L
 

Go to the Acid-Base Polyprotic Acid Worksheet
Go to the Acid-Base Unit Review