Acid Base Chemistry

Introduction

In discussing equilibrium chemical systems, we must consider as one of the most important of these topics reactions involving acids and bases. Most acid-base systems are equilibrium ones. Let's look at this field of chemistry, starting with some descriptive chemistry.
Acids and bases are relative terms.  A molecule cannot be called an acid in exclusion of all else.  For example, the gaseous hydrogen chloride molecule HCl(g) is a very stable molecule and does not act as acid or base unless something else is introduced to the system.  If water is introduced to the system, the HCl readily reacts with the water to form H₃O⁺(aq) and Cl^-(aq).  In other words, in the process of dissolving into the water, the HCl molecule gives up a proton to become Cl^-(aq) and a water molecule accepts the proton and becomes H₃O⁺(aq).  According to the Brønsted-Lowrey definition, an acid is a molecule or ion that can easily give up (donate) a proton in a reaction with a base.  Similarly, a base is a molecule or ion that can bond with (accept) a proton released by an acid in a reaction.  Thus, in the reaction between hydrogen chloride (HCl) and water, the HCl is the acid and the water is a base.  In the reaction between ammonia (NH₃) and water, the water donates a proton to the ammonia molecule to make NH₄⁺(aq) and OH^-(aq).  Thus, in this reaction, the ammonia is a base and the water is an acid.  We have seen water as a Brønsted-Lowrey acid and as a Brønsted-Lowrey base.  Water is amphiprotic.
We have experiences of acids and bases in our every day life, some of which we may not have recognised as being due to acid base properties.  Take the following observations for example.

Acids	Bases
Taste sour, eg., Vinegar (acetic acid and lemon (citric acid)	Taste bitter
Changes indicators' colours	Also changes indicator colours (opposite change)
Reacts with carbonates (CaCO3), bubbles and releases CO2.	Reacts with organic matter (Caustic)
Reacts with most metals to dissolve the metal and release H2 gas. Ironically, this is an oxidation/reduction (REDOX) reaction, not an acid base reaction.  M(s) + 2 HCl(aq) MCl2(aq) + H2(g)	Will react with aluminum to dissolve it, creating a solution of aluminum tetrahydroxide complex ions. 2Al(s) + 2NaOH(aq) + 6H2O(l) 2Na+ + 2[Al(OH)4]- + 3H2(g). This is actually two reactions, an acid/base and a REDOX reaction.

Acid properties are due to the hydrogen ion H⁺, more properly named the hydronium ion (H₃O⁺).
When we represent an acid in reaction, we often use a shortcut HA (or HB) where the H is the proton that is released in reaction with bases and the A (or B) is the rest of the species.
The part, A or B may be neutral or may be charged.  The part, without the H is the conjugate base of the form with the H.

HB{	HCl: the B is the chloride ion Cl-
	HNO3: B is NO3-
	NH4+ :  B is the NH3 molecule.

A reaction between an acid and water can be represented as follows:
This representation of an acid-base reaction is a proton transfer reaction. This is the definition of acid/base systems called Brønsted-Lowrey. The double arrow is here because the vast majority of acid/water reactions reach equilibrium; they do not go to completion. The acid, HB, gives up a proton (H⁺) to become B^- while the base (water) accepts the proton to become H₃O⁺. Often, we ignore the water and simply write:
It is important to remember in this short-hand system that B, the "rest of the species", is not necessarily negatively charged.  The representation here merely indicates that the B part is relatively negative, compared to the HB. Take for example the acid/base system
 NH₄⁺ H⁺ + NH₃.
 Here, B (NH₃) is neutral but HA (NH₄⁺) is +1 charged.
Our discussions in this course will be restricted to aqueous (water) solutions.  In aqueous solutions, water plays a special role in that it can react with both an acid (as a base) and with a base (as an acid).  Water is amphiprotic.  When we dissolve any acid (or base) with water, there will be a reaction.  The amount of reaction with water, depends on the strength of the acid. 
Hydrogen chloride is a strong acid in water. That means it reacts with water completely. On the other hand, Hydrogen fluoride doesn't react completely with water.  It sets up an equilibrium with water. Some HF remains in solution undissociated. HF is a weak acid in water.
NOTE:  the terms strong and weak apply only to an acid (or a base) reacting with water.  If an acid reacts with a different base (not the conjugate base and not water), it will always react 100%. 
An acid is formed when a compound has a hydrogen bonded to an atom that is moderately stable as a negative ion after the hydrogen ion is stripped off by a base.  Inorganic acids generally are formed when hydrogens are attached to electronegative species like a halogen or complex ions like sulfates, which can exist as negative ions Organic molecules can have acidic properties.  The most common type of organic acid is a carboxylic acid, as we see in the acetic acid molecule.
The carboxylic acid portion of the molecule is a carbon with two oxygens attached with a hydrogen on one of them.  This structure (see below) can form a stable anion because of resonance that occurs between the two oxygens after the hydrogen leaves. 
Acetic acid is also a weak acid. The hydrogen on the oxygen is the acidic hydrogen. It leaves behind a carboxylate ion (called the acetate ion in this case).
Actually, the vast majority of acids are weak acids. It would do you well to memorize the few acids that are strong and then you will always know that any other acid you encounter is a weak one.
The strong acids are:

• Hydrogen Halides (except HF), HCl, HBr, HI,
• Nitric acid, HNO₃
• Sulphuric Acid H₂SO₄ (in dilute solutions only)
• Perchloric acid HClO₄.
• Chloric acid HClO₃

Bases are proton acceptors according to the Brønsted-Lowry definition. We've seen that water can act as a base
Ammonia is also a base
Actually, since HCl is a strong acid, it is not really a good representation of reality to show the HCl reacting with another solute in water. Since HCl dissociates (according to the previous reaction) 100% in water, we never actually have HCl (aq) existing as a species. We have instead, H₃O⁺in the solution and it reacts with the ammonia
We can also have water acting as an acid.
Note that whereas the former reaction shows a 100% reaction this latter reaction is that of a weak base with water. Hence, an equilibrium is set up.
Alcohols can also act as an acid
Like acids, The vase majority of bases are weak bases. The few strong bases include the alkali-metal hydroxides (NaOH, etc. and the oxide ion O^2-).
Brønsted-Lowry bases all react with water to produce the hydroxide ion (OH^-).
Brønsted-Lowry acids all react with water to produce the hydronium ion (H₃O⁺).
Lewis Acids:
Lewis gave a different definition of acids which involved the transfer of electron pairs, rather than the transfer of protons.  The two are compatible in that all Brønsted-Lowry acids are also Lewis acids but some Lewis acids are not Brønsted-Lowry acids.  The Lewis definition is more general than that of Brønsted-Lowry.
Lewis acids are electron pair acceptors [like a proton accepts an electron pair when it bonds with a base]
Lewis bases are electron pair donors. [like the electron pair on the oxygen of a hydroxide ion is shared into the empty 1s orbital on the hydrogen cation (proton)]
Consider BF₃ + F^- BF₄^-.  the fluoride has lone-pair electrons which it shares into the empty orbital on the boron to form a new covalent bond.  Hence, the fluoride is the electron pair donor (Lewis base) and the boron tri-fluoride is the electron pair acceptor (Lewis acid).  No protons were transferred in this reaction so this is not a Brønsted-Lowry acid base reaction.

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Acid-base Neutralization reactions

Consider
Or
In all cases, the net ionic reaction is simply
The simplest type of chemical calculation involving acid-base reactions is a titration situation where the acid and base are exactly neutralized, i.e., they are both completely used up.

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Example:  A sample containing 25.00 mL of a NaOH solution was neutralized by 32.72 mL of a 0.129 MHCl solution. What was the concentration of the original sodium hydroxide solution.
First, write a balanced chemical equation
We know that for perfect neutralization, all acid and base are exactly used up. The stoichiometry of this particular reaction shows a 1:1 ratio of acid and base, Therefore, the number of moles of acid added is equal to the number of moles of base in the original solution. Or, another way of saying this is:
Finally, the question asked for concentration of the initial NaOH solution.  So we divide the number of moles of NaOH by the initial volume of the NaOH that was used.

CNaOH =	32.72 mL	0.129 mol HCl	1 NaOH	1	= 0.169 mol/L NaOH
		1 L	1 HCl	25.00 mL

Note that there were two conversion factors (1000 mL/ 1L) and (1 L/ 1000 mL) that occurred in the individual calculations but which I ignored in the extended fraction formalism because they cancelled out.  Also note that the ratio of NaOH to HCl does not need the units moles since they too cancel out.

Conjugate acid-base pairs

The shorthand for an acid dissociation is HB H⁺+ B^-. In this system, we see that an equilibrium exists. On the left-hand side, the acid HB dissociates (in water). On the right-hand side, the base B reacts with H⁺ to reproduce HB. We call this pair of acid and base a conjugate pair since they are interconverted by exchange of a proton.
Some examples of conjugate acid-base pairs are:

• HB H⁺ + B^-
• HCl H⁺ + Cl^-
• H₂O H⁺ + OH^-
• NH₄⁺ H⁺ + NH₃
• H₃O⁺ H⁺ + H₂O
• Acid H⁺ + Base (conjugate pair)

Similarly

• B + H⁺ BH⁺
• H₂O + H⁺ H₃O⁺
• CH₃COO^- +H⁺ CH₃COOH
• F^- + H⁺ HF
• Base + H⁺ Acid (Conjugate pair)

Every Acid-base reaction actually involves two acid-base conjugate pairs

HA	+	B		A-	+	HB+
HF	+	H2O		F-	+	H3O+
H2O	+	CN-		OH-	+	HCN
H2O	+	NH3		OH-	+	NH4+
Acid 1		Base 2		Base 1		Acid 2

Notice that water is both an acid and a base. It can donate a proton and it can accept a proton. It is called amphiprotic. We'll see this later when we discuss the self-ionization of water.

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Strengths of acids and bases.

We already saw that HCl is a strong acid in water. That means that it dissociates 100% when dissolved in water. This carries two implications. First of all, we never find a species called HCl (aq) actually existing in water since it is instantly dissociated to its ions. Second, If we look at the conjugate base of HCl (Cl^-). We can immediately say that in any aqueous solution that contains the chloride ion, Cl^- will never react to form HCl. Thus, even though Cl^- is termed the conjugate "base" of HCl. It doesn't in fact act as a base. We call this a very weak base. A strong acid has a very weak base as its conjugate.
Consider the weak acid HF. If we dissolve some HF in water, some of it will dissociate to form the conjugate base F^-. However, if we dissolve some F^- (say, NaF, dissolved in water) we know that some of the fluoride will set up an equilibrium to form some HF. Thus, a weak acid has a weak base as it's conjugate.
Similarly, if we have a strong base (say, O^2-) we know that if we have a solution containing OH^-, it will never react as an acid to form the oxide ion. Thus, a strong base has a very weak acid as its conjugate.

Acid	Base
Strong	Very weak
Weak	Weak
Very weak	strong

Let's reconsider the strong acid HCl. It cannot exist in aqueous solution. Neither can any strong acid. They all react with water to form H₃O⁺. H₃O⁺ is the strongest acid that can actually exist in aqueous solution. All stronger acids are levelled to H₃O⁺.
Similarly, Strong bases are levelled to the strongest base that can exist in water, OH^-. Thus, any intrinsic differences in the strengths of strong acids (or bases) are not detectable in water since they can't exist in aqueous solution.  We would need another solvent system to be able to distinguish the differences experimentally.  For example, liquid sulfuric acid can be used as a solvent in which hydrochloric acid and all other 'strong' acids do not fully ionize.  In other words, HCl is a weak acid when dissolved in H₂SO₄(l).  To keep our discussions as simple as possible, our discussions of strong and weak acids and bases will generally be restricted to aqueous solutions, unless otherwise specified.

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Equilibria in acid-base systems

Consider the following reaction.
If we were to write an equilibrium constant expression for this reaction, we might at first attempt to create the following expression.
Recall that we write concentration here but actually an equilibrium constant is properly written using relative activities. Relative activities are written as:

In the case of the solute chemicals, standard concentration is 1 mol L^-1. Hence, we can use Molar concentrations directly in the equilibrium equations by simply dropping the units. However, the standard concentration of a pure liquid is it's normal molar concentration (in the case of water at 25ºC, 55.5 mol L^-1). If we're considering dilute aqueous solutions, we can easily assume that the concentration of the water (solvent) is almost identical to the pure water concentration. Hence, it is a reasonably good approximation to assume that the relative activity of the solvent in a dilute solution is 1.
Thus, we don't write the activities of solvents in the equilibrium expressions since they have no effect on the final value of K.
Our equilibrium for the reaction would thus be
The subscript a refers to the reaction type.  Specifically, this K value refers to a reaction where one mole of acid reacts with water.

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Example,
What is the concentration of the species in a solution made up of 0.010 mol of HCl in 250 mL of water?  Since at these concentrations, HCL is a strong acid, i.e., all the HCL is used up, we do a limiting reagent calculation calculation. ("IF" table). Initial concentration of HCL is

Balanced equation	HCl	+      H2O		H3O+	+   Cl-
Initial concentrations (I)	0.040 M	solvent, activity=1		0	0
Limiting Reagent=HCl:	all HCL used up, use storchiometry to get final concentrations.
Final Concentrations (F)	0			0.040 M	0.040 M

Thus, [H⁺] = [Cl^-] = 0.04 M.

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Obviously, we did extra work here in this example. We could have simply said that the reaction was complete and that the stoichiometry gives the final answers because of the one-to-one molar ratio.  The IF table just helps some people visualize the process better.  It is also useful as a precourser step in some ICE table calculations.
It's not quite that simple if the acid (or base) is a weak acid.
For example, Consider the following.
The subscript a merely indicates that the equilibrium is that of an acid reacting with water. Please don't think that there is anything else special about this relationship. In general, we can write an expression for any acid HB.

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Example: What is the concentration of H₃O⁺ in a 0.20 Msolution of acetic acid?
[K_a (CH₃COOH) = 1.8 ×10^-5]

	CH3COOH	+	H2O		H3O+	+	CH3COO-
I	0.20				0		0
C	-x				+x		+x
E	0.20-x				x		x

Obviously, this is a quadratic equation (it has a term in x²and in x). In this case, there are two options we can use in order to solve this problem
1) Use the Quadratic formula
First, convert the K_a expression into the familiar quadratic form

Note here that x is [H+], b is Ka and c is Ca×Ka. where Ca is the initial concentration of acid. We'll use these notations in the "Summary of Calculation Types" section later.

after some algebra:
2) Since the molar concentration of the acid is more than 100 times it's equilibrium constant we can be fairly sure in making the assumption that the amount of reaction (x) is very small.
Assume 0.20 -x ~ 0.20. We can now rewrite the equilibrium constant expression by replacing the denominator as follows

Note here that this equation is easily generalizes as  [H+] =(Ca×Ka)½  where Ca is the initial concentration of acid. We'll use these notations in the "Summary of Calculation Types" section later.

We should always check our assumptions.
0.20 × 0.0019 = 0.1982 (rounds to two sig figs) ~ 0.20. (no change) Actually, we might see a small change in the last digit. The limit for this type of assumption to be considered valid is 5%.
Thus, our assumption was good. We can normally consider that the assumption will be valid if
or, in a problem where we add a base to water,
For concentrations we meet regularly here in first-year chemistry, that means if the K constant is about 10^-5 or less, then the assumption will probably be good.

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Example: What is [H₃O⁺] in 0.20 M HCN? K_a= 4.9×10^-10.

	HCN	+	H2O		H3O+	+	CN-
I	0.20				0		0
C	-x				+x		+x
E	0.20-x				x		x

Thus, [H₃O⁺] = [CN^-] = 9.9×10^-6M.
To convert from activities back to concentrations in mol/L, just put the units on at the end since we dropped them when we converted to activities so we could insert the numbers into the equilibrium expression.
The percent ionization is
(This means that out of 100,000 molecules, only 5 are ionized when they dissolve in water at the specified overall concentration)

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We can do all the same things for a weak base.
Subscript b merely means that this equilibrium constant expression is for one mole of base (B) reacting with water.

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Example: What is [OH^-] in 0.10 M NH₃? K_b(NH₃)= 1.8×10^-5.

	NH3	+	H2O		NH4+	+	OH-
I	0.10				0		0
C	-x				+x		+x
E	0.10-x				x		x

Thus, [NH₄⁺] = [OH^-] = 1.3×10^-3M.

Note here that this equation is easily generalizes as  [OH-] =(Cb×Kb)½  where Cb is the initial concentration of base. We'll use these notations in the "Summary of Calculation Types" section later.

The percent ionization is
In this case, our assumption is good. The amount of reaction x is less than 5% of 0.10.

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Self-ionization of Water

We've seen that water can act as both an acid and a base. In fact, water can be simultaneously the acid and the base in a reaction called the self-ionization of water.
One water molecule gives up a proton to another water forming one hydronium and one hydroxide ion. This is also called the autoprotolysis of water. The equilibrium constant for this reaction would be:
For pure water, the concentrations of the two ions are the same and are both 1.0 × 10^-7. Hence,
This equilibrium exists even in solutions of acids and bases and is the reason that there is always some H₃O⁺ and some OH^-ions present in solution even in strong acid and strong base solutions. Take for example, a solution of 0.010 M HCl.
In solution, 0.010 M HCl will break up immediately and completely to 0.010 M H⁺ and 0.010 M Cl^-.We' ignore the Cl^- since it doesn't contribute to the acid-base system at all.
Now, in the self-ionization of water, we set up initial and final conditions.

	H2O	+	H2O		H3O+	+	OH-
I	1				0,010*		0	*Due to the HCL
C					+x		+x
E	1				0.010+x		x

Now, we can fill this into the equation for K_W.
To simplify this calculation and avoid the quadratic formula, we'll assume that the amount of reaction x was very small. Hence, we can replace 0.010+x by ~0.010.
Obviously, x is indeed very much smaller than 0.01 so the assumption is valid.

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pH Scale

In water solution we can easily find H₃O⁺ and OH^-concentrations in the range between 1 and 10^-15. For reasons of convenience and other scientific reasons, concentrations are often expressed on a logarithmic scale. For this, the 'p' function is used. It was first proposed by Sørenson. (S1909) You will see this function used in numerous ways in Science. Don't get confused into thinking that each time you see it it is something new. In general
Whatever X is. In this case, we use pH and pOH to measure concentrations. Hence,
We already know that in aqueous solution K_W = [H⁺][OH^-] = 1.0×10^-14. Because of a law in mathematics that says
we can write
We can also convert pH and pOH values back to [H⁺] and [OH^-] values by using the inverse of the p function
Let's set up a table of pH and pOH values

pH	[H+]	[OH-]	pOH	Examples
-1	10	10-15	15	Conc. HCl {10 M, ca. 37% by wt}
0	1	10-14	14
1	10-1	10-13	13	Stomach acid
2	10-2	10-12	12	"
...
7	10-7	10-7	7	Neutral (Pure water)
...
12	10-12	10-2	2	Household ammonia
13	10-13	10-1	1
14	10-14	1	0	1 M NaOH solution
15	10-15	10	-1	Drain Cleaner (extremely caustic)

  NOTES on pH (Theoretical definition):

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The Experimental Definition:
Now let's try some calculations,
What is the pH of a solution where:
a. [H₃O⁺] = 1×10^-3M,               b. [H₃O⁺] = 5×10^-3M,                  c. [H₃O⁺] = 1 M.

1. pH = -log(1×10^-3) = 3
2. pH = -log(5×10^-3) = 2.3
3. pH = -log(1) = 0

That's relatively simple, now let's try some equilibrium calculations.

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Example: What is the pH of a 0.20 M solution of acetic acid?

                     CH3COOH + H2O     H3O+ + CH3COO�
Initial0.20 M                 0      0
x reacts: CHANGE =>    -x                  +x     +x
   Equil             0.20-x                 x      x

Assume 0.20 -x ~ 0.20. We can now rewrite the equilibrium constant expression by replacing the denominator as follows
We should always check our assumptions.
0.20 - 0.0019 = 0.1981 (rounds to two sig figs) ~ 0.20. (no change) Actually, we might see a small change in the last digit. Since the limit for an assumption to be considered valid is 5%.
Thus, [H⁺] = x = 1.9×10^-3. Hence, pH = -log(1.9×10^-3) = 2.72

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Example: What is the pH of a solution of 25 mL of 0.20 M NaOH added to 40.0 mL of 0.15 M HCl?
Whenever we add an acid to a base, even a weak acid and a weak base, we react them with each other 100% (limiting reagent calculation) and then allow the excess react with water to find the final pH.

              HCl   +   NH3   $\rightarrow$   NaCl   +   H2O
Initial      6.0×10-3    5.0×10-3      0
ammonia is limiting reagent
Final        1.0×10-3    0          5.0×10-3

So Now, we see that the HCl is excess, we allow it to react with water.

              HCl + H2O $\rightarrow$ H3O+ + Cl- (100 % reaction) 
Initial      1.0 mmoles    0      0
Change (100%, i.e., x=1 mmole)
Final          0          1.0     1.0  (mmoles)

#mol H⁺ = 1.0×10^-3 mol.  Now, divide this final number of moles by the total volume of solution to get the concentration [H⁺].
Hence, pH = -log(0.015) = 1.82.

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Examples:
Calculate the pH of the following solutions: a) 0.0100 M NaOH, b) 0.134 M NaOH

1. Since NaOH is a strong base, [OH^-] = 0.0100 M.
   K_W = [H₃O⁺][OH^-] = 1.0×10^-14
   [H₃O⁺] = 1.0×10^-14 / 0.0100 = 1.00×10^-12.
   pH = -log(1.00×10^-12) = 12.
   
   Alternatively, we could have calculated pOH first and converted to pH
   
   pOH = -log(0.01) = 2
   pH = 14- 2 = 12
2. [H₃O⁺] = 1.0×10^-14 / 0.134 = 7.46×10^-14
   pH = -log(7.46×10^-14) = 13.13
   
   or
   
   pOH = -log(0.134) = 0.87
   pH = 14 × 0.87 = 13.13

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Distribution diagrams

It is useful to look at a graphical technique of viewing acid-base equilibria. In this technique, we plot the relative amount of the components as a function of pH. Using this type of diagram, we can tell at a glance what the major constituents of a solution will be for any given pH (or pOH). The equations for creating these plots are found in the text book. We will not be expecting you to memorize them. However, those equations can be used to program an excel spreadsheet so that the plots can be created easily. A sample spread-sheet file is available here for your use.
Let's look at a distribution diagram for Acetic acid (K_a = 1.8×10^-5).
Here, we see two curves. These represent the concentrations of the acetic acid (red) and the acetate ion (blue). The K_a value of 1.8×10^-5 gives a pK_a = 4.74. You can see that in a solution of exactly pH=4.74, half of the original acetic acid would be ionized to acetate ion (and H⁺). At pH lower than 4.74, the proportion of acetic acid is larger and at pH higher than 4.74, more acetate ion is present.
Recall the example in the pH discussion above where we calculated the pH of a 0.20 M solution of acetic acid. We assumed that the amount of dissociation of acetic acid (x) was small compared to the initial amount. Once we had calculated the value for [H⁺] (and hence, the pH) we found that our assumption was true. The calculated pH was 2.72. Checking our chart, at a pH of 2.72 we can easily see that almost all of the original acetic acid is still present (relative amount ~ 1). Hence, by using this distribution chart, we can see visually that our assumption is good.
It is important to think of distribution diagrams in the correct terms. Remember, pH is the x axis. That means, this is a plot of how the acid-base system will behave in a solution of given pH. It is not usually used to tell us what pH a solution will attain by addition of some amount of acetic acid. For that, you need to do the calculations as we did above.
In yet another example, we looked at the equilibrium reached in 0.20 MHCN? K_a = 4.9×10^-10. We calculated the value of [H⁺] to be 9.9×10^�6M. This gives a pH of 5. Let's look at the distribution curve for HCN to see if pH 5 really does have mostly HCN present.
Since pKa = 9.3, we see that the cross-over point (where [HCN]=[CH^-]) occurs there. The value pH =5 is well below the region where any significant dissociation of HCN occurs. Hence, it is safe to approximate the amount of HCN in solution as equal to the initial amount (to an accuracy of a few digits, anyway).

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Relationship between K_a and K_b for conjugate acid/base pairs

Consider the reaction between the weak acid HF and water.
HF and F^- are conjugate acid and base respectively. We could say HF is the conjugate acid of F^- or equally well, we could say that F^-is the conjugate base of HF.
The K_a (HF) can be written as:
We could also have written the equilibrium as that of the weak base F^-reacting with water along with the expression for K_b(F^-).
Now, if we add the two chemical equations together, we get the reaction for the self-ionization of water.
Thus, the equilibrium constant K_W can be represented as
Or
This is a general relation, good for any conjugate weak acid/base pair.
It is possible to use the p function on the K values just as we did for pH and pOH. Thus, we can write
where the K_a and the K_b refer to a conjugate acid/base pair.

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Example: What is the base dissociation constant for the F^- ion? Given K_a = 3.5×10^-4.
What if we had been given a value for pKa rather than K_a? Say, pK_a = 3.46.
Notice that the reverse of function p (-log X = Y) is written according to mathematical principals as p^-1
[X = antilog(-Y) or X = 10^-Y].

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Buffer Solutions

In many chemical processes, both industrial and biological reactions can only take place in a narrow range of [H₃O⁺] concentrations (narrow pH range). To ensure that the desired reactions take place, buffers are used which hold the pH constant (within limits).
Examples of such chemical processes are:

• Sewage treatment: Bacteria only survive in a narrow pH range
• Photographic developing
• Electroplating
• Most metabolic processes

• Stomach pH ~ 1.5
• Saliva pH ~ 6.8
• Blood pH ~ 7.4

Buffers are used to maintain pH at some desired range. A Buffer is made up of approximately equimolar amounts of a weak acid and its conjugate base. To make these solutions, we could, for example add a weak acid (like Acetic acid) and the salt of the acid (like Sodium acetate) in equal proportions.
Video demonstration of the Buffer effect: (Broadband) (Dialup)
Let's consider the solution made up of CH₃COOH and CH₃COO^-in approximately equal amounts.
According to La Châtelier's principal If we add a small amount of acid, the acetate ion will react with it and reduce the effective change in pH. If we add a small amount of base, the acetic acid will react with it and again reduce the effective change in pH. The amount of acid and base added must be smaller than the amount of buffer or it will simply use up all the buffer and then proceed to change the pH.
To restate what we said in more mathematical terms:
Here, we see that the equilibrium constant for the reaction of the acetate ion with acid is very large. Hence, any acid added to the buffer solution will be immediately used up in the buffer equilibrium. Similarly, the K for the reaction of the acetic acid with a base is large and the base will be used up.

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Buffer Example: What is the pH of a solution of 0.050 mol acetic acid and 0.050 mol of acetate ion in 1 L of solution?

             CH3COOH (aq) + H2O  CH3COO- (aq) + H3O+ (aq) 
Initial:        0.05                  0.05           0
Change;           -x                    +x          +x
Equilibrium:   .05-x                 .05+x           x

Here, we assume that x is small and that we can replace 0.050+x by 0.050 and 0.050-x by 0.050.
(Yes, x is small compared to 0.050)
NOW, let's add 0.001 mol HCl to the solution (Pretend no volume change just for simplicity in this example).
The HCl will react immediately with the acetate to use 0.001 mole of acetate and produce the same amount of acetic acid.

                CH3COOH (aq) + H2O  CH3COO- (aq) + H3O+ (aq) 
Initial:        0.051                   0.049           0
Change;           -x                      +x           +x
Equilibrium:   .051-x                  .049+x           x

Again, we assume that x is small.
(Yes, x is small compared to 0.049)

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We see in the above example, that the pH has changed from 4.74 to 4.72, a very small change, considering that if we had added the HCl to 1 L of pure water, the resulting pH would have been 3.
What if we had added .01 mol HCl rather than .001 mol. That's ten times as much.
Our initial values would have changed by 0.01 rather than by 0.001.

                CH3COOH (aq) + H2O  CH3COO- (aq) + H3O+ (aq) 
Initial:            0.06                 0.04            0
Change;               -x                   +x           +x
Equilibrium:       .06-x                .04+x            x  

Again, we assume that x is small.
(Yes, x is small compared to 0.050)
Even adding 0.01 mol of HCl only changed the pH by about .16 units. Had we added the HCl to pure water, the pH would have changed from 7 to 2 (5 units). Thus, the buffer was effective in holding the pH steady.
Let's look at the distribution diagram for acetic acid/acetate again to see this effect visually.

The initial pH of the buffer solution, before adding the HCl was 4.74. This is exactly the cross-over point on the graph where the relative amounts (concentrations) of the acetate and the acetic acid are equal. Notice that at this point, the slope of the curves is very steep. This means that to make any change in pH, we must make large changes in the relative amounts of the acid and conjugate base. Note too that this range extends from just below pH=4 to pH=6. A pH range of about 2 or a relative concentration range of 100. Thus, we can have a buffer solution of any weak acid/conjugate base system as long as the ratio of the acid to the base is between about 10 and 0.1.

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Buffer Example:  What is the pH of a solution of 1.00 mol of NH₃ and 0.40 mol of NH₄Cl in 1.0 L of solution? Is this a buffer solution?
NOTE: NH₄Cl is a salt that dissolves completely to form ammonium ion and chloride ion.

                  NH3 (aq) + H2O  NH4+ (aq) + OH� (aq) 
Initial:          1.00                0.40         0
C                   -x                  +x        +x
Equilibrium:      1.00�x             0.04+x        x  

assume x is small:
To see if this is a buffer, we can look at the ratio of acid to base:
This ratio is well within the range 10 to .1 so it is a buffer solution.
Looking at the distribution diagram for ammonia (pK_b = 4.74 so pK_a= 9.26) we can also see that pH = 9.66 is in the buffer zone (slope of the line is steep).
We can derive a general equation to deal with calculations of pH of buffer solutions.
We are using initial concentrations because if we are truly in a buffer zone the amount of change in the relative concentrations of the acid and conjugate base is small.
This equation is called the Henderson-Hasselbalch equation. And it is a very useful shortcut to calculating the pH of a buffer solution once we have established the initial concentrations like we did in the examples above.
We see that the primary contribution to the pH of a buffer solution is the pK_aof the acid/base pair. And modifications to the pH are affected by adjusting the relative concentrations of the acid and base. Thus, by carefully choosing the correct acid/base pair, we can create a buffer of any pH we wish.

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A buffer is most effective at resisting change in pH when [base] = [acid]
A one-to-one molar ratio is not always desired. What if we need to achieve a specific pH value, not quite equal to pKa for the buffer system we are using?

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Example: Blood has a pH of 7.4. and is buffered by a carbonic acid/hydrogen carbonate buffer system. What is the ratio of acid to base required to keep the pH at 7.4, given pK_a(H₂CO₃) = 6.37
We need a ratio of 10.7 base to acid to maintain a pH of 7.4. This is right on the edge of the buffer region for the buffer system we used here. In reality, the buffer system used by our blood is much more complex system than we used here and so thankfully, we are not quite on the edge, so to speak.

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Well, we've learned how to look at distribution diagrams and how to do equilibrium calculations. In this next section, we'll review a bit of both and show more how these two ideas can be combined to help in the understanding of acid-base equilibria.
We've seen that for a monoprotic acid/base system (one proton is exchanged) the distribution diagram consists of two curves that cross at a value pH = pK_a. (or pOH = pK_b). We've also seen that the width of the curved space is about 2 pH units. This means that we can always draw a quick distribution diagram for any acid/base conjugate system knowing only the K_a (or pK_a), or the K_b (pK_b) of the weak acid or weak base respectively.
Let's take for example, the acetic acid / acetate system.
We know that K_a(HAc) = 1.8×10^-5. So pKa = -log(1.8×10^-5) = 4.74. Thus, we can quickly draw a distribution diagram by drawing a rectangle to represent the plot area. The axes are:

• Y: fraction, range = 0 to 1.
• X: pH, range= 0 to 14.

The cross-over of the two curves occurs at about 4.7 and the width of the cross-over area is 1 pH unit to either side of 4.7 (about 3.7 to 5.7). The resulting sketch of the distribution diagram is shown below.
Here, we see the cross-over at 4.74 and we know the width of the area is 2 units (NOTE). Our drawing doesn't have to be perfect. The information is still there.

Summary of calculation types.

When we have a problem involving adding a weak acid or base to water, we can sometimes make simplifying assumptions in our calculations.  The ICE table process for acid/base calculations will always lead us to a quadratic equation and so we can always simply plow through the work of using the full quadratic solution to get to our desired answer.  In many circumstances, however, we can simplify the calculations by assuming certian variables are insignificant and therefore can be ignored.  When can we make such assumptions?  It turns out that the decision as to when we can make the assumptions are based both on the concentration of the acid (or base) involved and on the strength (K_a or K_b value) of the starting material.
We'll do the math here for acid added to water but the same can be done for base added to water by substituting the appropriate variables C_b, K_b and pOH, as needed.
Now, lets look again at the acetic acid example we did above. In that example, we asked for the pH of a 0.20 M solution of acetic acid. After some work and the assumption that x was small, we worked out that [H⁺] = (C_aK_a)^1/2= 1.9×10^-3. And therefore that pH = 2.72.
So, by making the asumption x << Ca, we were able to avoid using the quadratic equation and shorten the amount of work we needed to do.  We can use the equation we get out of the ICE table in two forms, depending on what final answer we need. 
In words, the pH is the average of the pC_a and pK_a.
This is easily visualized in the diagram where the pC_a and pH are indicated. pH is half-way between pK_a and pC_a.  Obviously, if the halfway point between the pC_a and pK_a lies within the buffer zone then all bets are off as far as simplifying assumptions.
We could do the same in the case of a solution of weak base where the pOH is half-way between pK_b and pC_b.
Example: We can determine the pOH of a solution of 0.30 M ammonia as follows
In both of these cases, the concentration C is larger than Kby a factor of at least 100. C > 100 * K. This is called the concentrated limit. In the case where the concentration is not so high (100K> C_a> .1K), the assumption that the reaction is small is not valid and we must use the quadratic formula.
Take acetic acid again, if C_a were to be, say, .0002 we would find that the amount of dissociation is no longer negligible (x is not small) and we would need to use the quadratic formula.  After rearranging the resulting K_a expression at equilibrium to fit the quadratic expression 0 = ax² + bx + c, we determine that x = [H⁺] , a = 1, b = K_a and c = -K_aC_a. Thus,
Notice that in this last example, the amount of dissociation (0.00007) is a sizable fraction of the initial amount (0.0002)
0.00007/0.0002×100% = 35%, definitely not negligible.  If we look at the distribution curve diagram for acetic acid, we see that at a pH of -log(7.0×10^-5)= 4.15 is in the buffer region; the slope of the distribution curve in the buffer region (Near pK_a) is very steep and hence, we cannot make any assumptions.
In an extremely dilute weak acid, we can always assume that the acid dissociates 100% just like a strong acid. The amount of dilution necessary to allow for 100% ionization is called the strong acid limit.  The strong acid limit occurrs at the edge of the buffer region (~K_a/10). Thus, any acid whose concentration C_a is less than about K_a/10 is in the strong acid region.
In the strong acid region, even weak acids behave like strong acids so the assumption, C_H⁺ = C_a and pH = pC_a is a reasonable approximation unless C_a is less than or equal to about 1×10^-7. In this case, we must include considerations of self-ionization of water in any calculations.
For example, a concentration of acetic acid of C_a =1×10^-6 is in the strong acid region for acetic acid.  So, all the acetic acid will dissociate, resulting in a pH = 6.  If we use an acetic acid solution with C_a = 1×10^-8M , the final solution would still be in the strong-acid region for acetic acid but the pH would not go higher than about 7, since water alone will contribute 1×10^-7M of H⁺ and hence the additional small amount of H⁺ from the acetic acid will only slightly change that.  Another way of thinking of it... we cannot get a pH of 8 (basic solution) by adding a small amount of acid to water.
We summarize the shortcuts we've just discussed in the following table. The table is given for weak acid added to water and so the subscripts are all a and we calculate [H⁺] and pH.

Conc.	Range of conc.	p-function equivalent	Equations
High	Ca> 100Ka	pCa< pKa - 2	[H+] = (KaCa)1/2 pH = (pKa + pCa)/2
Medium	100K> Ca> 0.1K	pKa - 2 < pCa< pKa + 1	\[\mathrm{[H^+]}=\frac{-K_a + \sqrt{K_a^2 + 4K_a C_a}}{2}\]
Low	0.1K > Ca	pKa + 1 < pCa	[H+] ~ Ca pH ~ pCa

 The equations for weak base added to water can all be derived by changing all references for acid into references for base, i.e.,
 [H⁺] ==> [OH^-], pH ==> pOH, C_a==> C_b and K_a ==> K_b.  Additionally, when looking at the distribution diagrams, we swap left and right references, i.e., the weak base region is right of the buffer region and the strong base region is left of the buffer region.

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Acid-base properties of salts.

In an earlier part of this section on acid-base reactions, we said that the reaction between an acid and a base results in a salt and water. The questions arises, Is this solution neutral? In other words, are the products neutral?  Is the Salt Neutral?
To know the answer, we can look at the acid/base properties of the two ions that make up the salt. If we don't know their properties offhand, we can sometimes figure out the answer by checking the conjugates of the ions. There are four possible cases:

Salt of strong acid and strong base

Consider an example, hydrochloric acid and sodium hydroxide. If they are added together in aqueous solution such that they exactly neutralize each other, the result will be a NaCl solution in water.
In this case, the solution is the result of the neutralization of a strong acid with a strong base and the pH is exactly 7. In other words, the salt NaCl is neither acidic nor basic. The Na⁺ is a very weak acid (the conjugate acid of a strong base, NaOH), in other words, it is not an acid. The Cl^- ion is a very weak base (the conjugate base of a strong acid, HCl), and is therefore not a base. So NaCl is a salt made up of a cation that is not an acid and an anion that is not a base. It is neutral.
Such is not the case if the original acid and base are not both strong..

Salt of strong acid and weak base

Salt of weak acid and strong base

What about a solution of sodium acetate, NaCH₃COO? Is it acidic or basic?

Salt of weak acid and weak base

Now for the hard one. What if both the anion and the cation are non-neutral?
Take, for example a salt ammonium acetate, NH₄CH₃COO. Is it acidic or basic?
We now have a way of determining the acid-base properties of any salt solution. In each of these cases, we could have said that the stronger ion wins, in the first three cases, at least one of the ions was very weak and we could easily discard it. In the fourth case both ions were weak and we had to compare directly the two to determine the answer.
What if you were asked to calculate the pH of a salt solution. In the first three cases above, the calculations are relatively simple.
In case 1, the pH is 7. That's simple.

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	CN-	+ H2O		HCN +	OH-
I	0.1			0	0
C	-x			+x	+x
E	0.1-x			x	x

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Polyprotic acids.

So far, in our discussion of acids and bases, we have focused on acids that donate only one proton (and bases that accept only one). Acids of this type are called monoprotic acids. Some acids can have more than one proton to donate. The general class of these is called polyprotic acids.
Polyprotic acids can be categorized according to how many protons they donate:

• Diprotic acids donate two protons. Examples include H₂SO₄, H₂SO₃, H₂CO₃ and (COOH)₂
• Triprotic acids donate three protons. An example of this class of acids is H₃PO₄

In the case of strong acids, we can merely consider that the ratio of acid to base in a stoichiometric calculation is not 1:1, for example, in dilute solutions H₂SO₄ can be considered to be a strong acid. And especially in a reaction with a base, we can write,
Here we can plainly see that the ratio of acid to base is 1:2, not 1:1 as would be the case if a monoprotic acid had reacted with the same base.
If the acid is not a strong acid, then we must consider the equilibria that occur in the solution.
The generalized reaction scheme for reaction of a diprotic acid with water can be written as follows. We take the protons off the acid one at a time and set up an equilibrium for each step.

1. H₂B + H₂O H₃O⁺ + HB^-
   
   
2. HB^- + H₂O H₃O⁺ + B^2-
   
   

These two equilibria exist simultaneously in the solution.  This multiple equilibrium senario is not easy to work with calculation-wise.  Under certain circumstances, we can make some simplifying assumptions in our calculations. There are three different sets of conditions we should consider.

1. The K_a values are separated by at least 4 orders of magnitude (pK_a values are separated by at least 4). In this case, we can consider the equilibria separate from each other. One does not affect the other. It allows for certain assumptions to be made in the calculations.
2. The K_a values are less than 4 but greater than 1 order(s) of magnitude apart (1< ΔpK_a< 4) we are in an intermediate area where the equilibrium of step 1 is never completely separated from that of step 2 but where some separation is still visible.
3. The K_a values are only one order of magnitude apart (1 pH unit separation of pK_a values). In this case, it is nearly impossible to distinguish the two equilibria using experimental techniques. Calculations can still determine the amounts of all reactants.

Lets look at the distribution diagrams for cases like these. (Spreadsheet)
Here, the curves are colour coded: Red represents H₂B, Green is HB^-and Blue is B^2�. We see that the red-green crossover area is completely separated from the green-blue crossover area. This means that the acid H₂B is completely converted to HB^- before any significant amount of HB^-dissociates to form B^2�.
Here, we wee that the green curve never completely reaches mole fraction of 1. In this case, the second equilibrium takes effect and uses up the HB^-. In the intermediate region between two crossover points, significant concentrations of all three species H₂B, HB^- and B^2� are present simultaneously. Obviously, calculations in this region will not be as simple as in case 1.
Here, we see that the amount of HB^- is always quite low and experimentally, it would be very hard to distinguish the two different equilibria. We have what looks like a distorted crossing zone between the red and blue curves with the green poking through as a slight perturbation on the curves. We might think of the HB^- as an intermediate species whose concentration doesn't build up very much. In these cases, except under extremely carefully controlled pH situations, the reaction of H₂B with a base occurs in "nearly" a single step.
Let's look at some calculations involving the triprotic Phosphoric acid, H₃PO₄.
We will look at each equilibrium in turn:

1. H₃PO₄ (aq) + H₂O H₃O⁺ (aq) + H₂PO₄^- (aq)
   
   
   
2. H₂PO₄^- (aq) + H₂O H₃O⁺ (aq) + HPO₄^2- (aq)
   
   
   
3. HPO₄^2- (aq) + H₂O H₃O⁺ (aq) + PO₄^3- (aq)
   
   

We see that the K_a values are all separated by about 5 orders of magnitude. That means that the equilibria can all be treated independent of each other, as in Case 1 above.   We can see this even more clearly if we look at the distribution diagram for H₃PO₄.
Here, the colored lines represent the following species:

• RED => H₃PO₄,
• GREEN =>  H₂PO₄^-,
• BLUE =>  HPO₄^2�,
• VIOLET =>  PO₄^3�,

It is clear to see in the distribution diagram that the three crossover regions corresponding to the three K_a values are completely separate from each other.
Using the three K_a relationships, we can now calculate the concentration of all species involved in the equilibria.

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Example:  Calculate the concentration of H₃PO₄, H₂PO₄^-, HPO₄^2-, PO₄^3- and H₃O⁺ in a 0.100 M solution of H₃PO₄.

first Ka value	H3PO4		H+	+	H2PO4-
I	0.100		0		0
C	-x		+x		+x
E	0.100-x		x		x

We cannot assume that x<<0.100 since C_a is less than 100×K_a. We must use the quadratic equation. (NOTE)
recall: so, after some algebra, we find
However, some of the H₂PO₄^- will dissociate in the second equilibrium. Will it affect the concentration found here significantly? Probably not since the value of K_a2 is about 10⁵ smaller than K_a1. Let's do the calculation to prove it.
STEP 2: the second proton is removed.

Second Ka value	H2PO4-		H+	+	HPO42-
I	0.023		0.023 (NOTE)		0
C	-x		+x		+x
E	0.023-x		0.023+x		x