The Nature of Acids and Bases
The properties of acids and bases were well known long before there was a workable theory that explained and predicted their behaviour. Early theories of acids and bases were proposed by men like:
· Antoine Lavoisier: Proposed that oxygen created the acidic properties and that acids were combinations of oxides and water.
o Explains why carbonated water (CO2 (g) in water ) is acidic
o Could not explain why MgO is a base.
· Sir Humphrey Davy: Proposed that hydrogen was responsible for the acidic properties of acids.
o Explains why HCl is acidic or HNO3 is acidic.
o Could not explain why NH3 is base
· Svante Arrhenius: Presented the first modern theory of acids and bases was presented by Svante Arrhenius
Arrhenius Theory
1) According to the Arrhenius theory the following is the general form for writing the dissolving of an acid, using the general form of the acid as HX.( X could represent any number of anions)
HX (aq) ------> H + (aq) + X - (aq)
2) According to the Arrhenius theory what is the general form for writing the dissolving of a base? The general equation of the base would be XOH. (X represents any number of cations)
3) Write the general nonionic or molecular equation for the reaction of an acid with a base according to Arrhenius theory. ( for example: HCl and NaOH )
4) Disappearing Ink is a interesting application of how neutralization works according to Arrhenius .
5) Arrhenius model had several shortcoming (problems). Two such problems were:
- he predicted that substances that did not produce H + ions or OH- ions in solution would be neutral
- he predicted that acids would produce H+ ions, which are free protons ( hydrogen without an electron) in solution.
Bronsted Lowry Theory
The shortcomings (problems) of the Arrhenius theory lead scientists to seek other explanations for the nature of acids and bases. Two groups of scientists in the same year (1923) independently proposed a new theory . Their theory explained the behaviour of all of the acids and bases that Arrhenius did but also was able to resolve some of the problems with his theory. That is they were able to explain why some salts are acidic and basic and why no free protons are found in solutions of some acids.
To understand their theory examine the following reaction between vinegar and ammonia solutions. If we test each solution for conductivity we see that neither is a conductor in dilute solution. They therefore can be represented as molecules. When equal volumes are mixed the resulting solution conducts electricity readily. The reaction that occurs can be written as seen below.
CH3COOH ( aq) + NH3 (aq) NH4 + (aq) + CH3COO - (aq)
When Bronsted and Lowry examined this reaction, they defined vinegar as an acid because it donated a hydrogen ion to the ammonia. They defined ammonia as a base because it accepted this hydrogen ion.
Acid | substance that donates hydrogen ions (protons) in solution. |
Base | substance that accepts hydrogen ions (protons) in solution. |
Bronsted- Lowry solutions to Arrhenius problems:
· Bronsted and Lowry stated that when acids dissolve in water they donate a proton to water to form H3O + ions. H3O + ions were later proved to exist through independent experimentation.
HCl (g) + H2O (l) H3O + (aq) + Cl - (aq)
· Bronsted and Lowry could also explain why some salts were acidic, while other were basic by saying that the ions formed when they dissolve react with water . When the solution were acidic they formed H3O + ions, when basic they formed OH- ions.
o Why sodium bicarbonate is basic in solution. The bicarbonate ion reacts with water to form OH- ions. HCO3 - is behaving as a base because they are accepted protons.
HCO3 - (aq) + H2O (l) H2CO3 (aq) + OH - (aq)
o Why sodium bisulfate is acidic in solution. The bisulfate ion reacts with water to form H3O + ions. HSO4 - is an acid because it is donating a proton to the water.
HSO4 - (aq) + H2O (l) SO4 2- ( aq) + H3O + (aq)
Amphoteric Substances
Bronsted and Lowry not only solved the problems with Arrhenius theory, they also explained how some substances can behave as acid under one set of conditions and bases under a different set.
Examine water in the above reactions . In the bicarbonate reaction water is behaving like an acid and is donated a proton. In the bisulfate reaction water is behaving as a base and is accepting a proton. Substances that can behave in this way are said to be Amphoteric.
Demonstration of Amphoteric substances Amphoteric substance may behave as either acids or bases. Certain metal hydroxide will behave as amphoteric substances. Click on the test tube on the right and complete the following demonstration to illustrate this concept |
Unfortunately there is no way to tell by looking at the formula if the substance will behave as acid or base.
The acid - base reactions that Bronsted and Lowry described are equilibrium. The proton when transferred from acid to base can be transferred back in a reverse reaction. The substances formed when the proton is transferred are termed conjugate acids and bases.
Conjugate Acid: chemical species formed when a reactant base accepts a proton. Chemical species has one more proton than base.
Base | Conjugate Acid | |
OH - (aq) | H2O (l) |
Conjugate Base: chemical species formed when a reactant acid donates a proton. Chemical species has one less proton than acid.
Acid | Conjugate Base | |
H2SO4 (l) | HSO4 - (aq) |
An overall reaction could be :
HCO3- (aq) | + | H2O (l) | H2CO3 (aq) | + | OH - (aq) | |
Base (acceptor) | Acid (donator) | Conjugate Acid | Conjugate Base |
Comparing Strength to Concentration
Why is it that acid solutions of the same concentration have different pH's?
Chemistry explain this fact by stating that acids (and bases) can have different strengths.
Strength in chemistry is very different than concentration. We define strength as follows
STRENGHT | Strength is defined as the degree of ionization or dissociation of the acid or base in aqueous solution. It can be measured using the equilibrium constant expression for the dissolving of the substance in water. The degree of ionization is the extent to which the acid, or ionic solids become ions. The equilibrium constant expression (Keq) for acid are termed Ka and for bases termed Kb. Tables of acid and base strength are listed in many textbooks. |
A. Comparing strong to weak acids
Strong acids are completely dissociated (100%). If the acid is less than 100% dissociated it is termed a weak acid. The kA of these strong acids is very high. (> 20). Table of Relative Strengths of acids in aqueous solution at room temperature, 25C
When chemists write the dissolving of strong acids in water they only show a one way arrow.
Example: | HCl (l) |
|
Weak acids are partially ionized or dissociated and therefore form few ions. The kA values of these acids are small in comparison to strong acids. Chemists use a double arrow to indicate their dissolving in water.
Example: | CH3COOH (aq) |
|
Concentration: | Concentration is a ratio of the amount of solute to the amount of solvent in a solution. Measures of concentration can include PPM, ppb, % , and molarity. |
B. Comparing concentrated to dilute solutions
Concentrated and dilute are relative terms but chemists tend to agree that:
Concentrated acids or bases have high molarities usually in excess or 6 mol/l.
Dilute acids and bases have smaller molarities usually less than 6 mol/l.
Check your Understanding
Using the table of acid and base strength provided, complete the following table:
Acid | Strong or Weak | Concentrated or Dilute |
12 M HF | ||
0.02 M HBr | ||
12 M HTe- | ||
0.03 M HClO |
for the answers...Click Here
How do we calculate the acidity of an acid or base in an aqueous solution? In other words how do we calculate the [H+] and
[OH-] of an acid or base?
To answer this question we have to consider what happens to the water when an acid or base is added to it.
When acids and bases are dissolved into water to form solutions they disturb an equilibrium of ions that already exists in the water. The water equilibrium forms because water ionizes to a small degree to form H30+ and OH- ions, according to the following equilibrium.
2H2O (l) |
Only about 1 molecule in a billion ionizes in this way and therefore there are so few ions in water that it does not conduct a measurable current (unless you use very sensitive instruments).
Because the reactions are reversible and an equilibrium exists we can write an equilibrium constant expression (Kc ) for the above equation. It is:
K = [H3O + (aq)] * [OH- (aq)] [H2O]2 |
Since water is a liquid the [H20] does not change. Since the water can't change concentration this term is not included in the above expression and the equation becomes:
Kw = [H3O + (aq)] * [OH- (aq)] |
Rearranging the above formula we can produce two others, one to calculate [H3O + ] and one to calculate [OH- (aq)].
[H3O + ] | = | Kw | [OH- (aq)] | = | Kw | |
[OH- ] | [H3O + ] |
This equilibrium constant is labeled Kw and its value for pure water at 25 0C has been calculated. As with other equilibrium constants the units may be dropped. The value of Kw has been calculated as;
Kw = 1.0 x 10 - 14 |
Calculating the [H3O +] in neutral water
Let's examine the ionization of water.
The equilibrium constant expression is Kw = [H3O +] [OH-] 1) The mole ratio of H3O + to OH- is 1:1 , Therefore [H3O +] = [OH-] 2) Since the concentrations are the same let us give them a value of x . 3) The Kw expression becomes K w = (x)(x) or x2 4) Placing in the value of Kw we get 1.0 x 10 -14 = x2 , or x = 1.0 x 10 -7 5) Therefore [H3O +] and [OH-] = 1.0 x 10 -7 mol/l in neutral water |
Calculating [H3O +] in strong acids solutions
Chemists use the fact that an equilibrium exists in water to calculate the [H3O +] and [OH-] in acid and base solutions
What happens when strong acids (a source of H3O + ions) or strong bases ( a source of OH- ions) are added to water?
Answer:
The addition of H3O+ or OH- ions disturbs the water's equilibrium. We can predict the changes that will occur using Le Chatelier's principle. The following presentation illustrates how the water equilibrium reacts.
Calculating [H3O +] and [OH -]
Acids and bases when added to water disturb an equilibrium that already exists.The changes in the concentration of H3O + and OH - can be determined using the equilibrium law expression stated in the previous section. The calculation is different for strong vs weak acids and bases. We will consider strong acids and base calculations first.
Strong Acids / Bases
Important Idea
When strong acids are added to water the [H30+] in the solution will become equal to the concentration of [H30 +] produced from the dissolved acid (unless the concentration of the acid is very dilute). The same is true of [OH-] and bases.
Explanation:This means that if the concentration of the [H3O +] produced by an acid is 2.0 mol/l the concentration of [H3O +] in the water solution will also be the same.
Consider adding 2 mol/l HCl to water. The equation below describes it's dissolving.
HCl (aq) | + | H2O(l) | H3O + | + | Cl- |
- The acid dissolves to form H30 + ions. Notice that the mole ratio of HCl: H30 + is 1:1.
· therefore the [HCl] = [H3O +].
· The concentration of [H3O +] added to this solution is therefore 2.0 mol/l.
· This concentration is added to pure water whose [H3O +] is 1.0 x 10 - 7 mol/l). If we add the two concentrations we get 2.0 mol/l + 1.0 x 10 - 7 mol/l = 2.0000007 mol/l.
· Considering significant figures then the concentration of the solution = 2.0 = concentration of HCl
To calculate the concentrations of strong acids and bases we need to perform the following procedure
important procedure | Calculating [H3O + ] or [OH-] for strong acids and bases | |
Write the dissolving of the strong acid or base in water. | ||
Calculate the concentration of the strong acid or base in mol/l. | ||
Using the mole ratio from the equation calculate the concentration of [H30 +] or [OH-] | ||
Using the Kw formula calculate the concentration of the other (H3O or OH-) | ||
Examples
Weak Acids and Bases
Key Idea:
When weak acids and bases are added to water the concentration of [H3O+] and [OH-] is not the same as the concentration of the dissolved acid. The concentration can be determined using the equilibrium constant values for acids and bases termed Ka and Kb. Ka values are the equilibrium constant expressions for the dissolving of various acids in water. The larger the value of Ka the stronger the acids are. The same is true of bases and Kb. The following link contains a table of acids and their Ka values
Calculating [H3O+] and [OH-] of weak acids and bases
Try to answer this! Click HERE. | |||||||||||||||||||||
Formulas for pH, pOH
A. Chemists express the wide range in concentration of [H3O+] and [OH-] in solution by expressing the concentrations in the form of pH and pOH. The following formulas are commonly used in these calculations.
pH = - log 10 [H3O+] | |
pOH = - log 10 [OH-] | |
[H3O+] = 10 -pH | |
[OH-] = 10-pOH | |
pH + pOH = 14 |
B. The diagram below describes the relationships between [H3O+], [OH-], pH, pOH and acidity of solutions.
Calculations with pH and pOH:
When using your calculators to calculate pH and pOH, we will usually use two buttons, log / 10 x button and the exponent button (EXP or EE button).
1. To calculate pH we enter the [H3O+] concentration in mol/l, which may require us to use the EXP button or EE button. For the number 6.2* 10 -7 we would use the following procedure
1) enter 6.2
2) press exponent button ; EE or EXP button
3) enter 7
4) press +/- button to remove the negative sign.
5) press log button.
2. To calculate the [H3O+] and [OH-] given the pH or pOH we would use the following procedure.
1) press INV log or 2nd function log
2) enter the pH or pOH
3) press the +/- button to add the negative sign
4) press enter to get the [H3O+] or [OH-]
QUIZZES: pH calculations
Buffers: An application of the Bronsted-Lowry Theory
1) Definition: Buffers are solutions in which the pH does not change a great deal when moderate amounts of acid or base are added to them. These solutions are made by:
1) combining a weak acid and it's conjugate base (one of the acid's salts) in a mixture
2) combining a weak base and it's conjugate acid (one of the base's salts) in a mixture.
Weak acid and conjugate base: Acetic acid and Sodium Acetate | |||
Examples | Weak Base and conjugate acid: Ammonia and Ammonium Chloride | ||
How Buffers work Buffers work by removing H3O+ or OH- ions from solution as they are added. How ?? LeChatelier's principle describes how this works. Take for example the acetic acid and sodium acetate buffer mentioned above. The equation for the equilibrium that exists in the solution is: CH3COO - (aq) + H3O+ (aq) H3O+ (aq) + OH- (aq) -------> 2H2O (l) | |||
Characteristics of a buffer
What determines how much acid or base a buffer can absorb without dramatic changes in pH?
The composition of the buffer determines how effectively it works. The information below describes three important characteristics that determine the effectiveness
The important characteristics of a buffer are: | |
Buffer capacity is the amount of acid and base that can be added before the pH change is significant . It is determined by the factors of the initial concentration of the relative amounts of the buffer components. | |
Acid - Base Neutralization Reactions
A key characteristic of both acids and bases is that they react in a way to neutralize or remove the properties of the other. These are double replacement reactions and the general way to write such a reaction is as follows:
Acid | + | Base | a Salt | + | water | ||
2 HCl (aq) | + | Mg(OH)2 (aq) | MgCl2 (aq) | + | 2 H2O (l) |
Definition: A salt is the substance formed when the cation of the base combines with the anion of the acid. These substance generally have a neutral pH.
When acids and bases react with each other, they can form a salt and (usually) water. This is called a neutralization reaction and takes the following form:
HA + BOH --> BA + H2O
Depending on the solubility of the salt, it may remain in ionized form in the solution or it may precipitate out of solution. Neutralization reactions usually proceed to completion.
The reverse of the neutralization reaction is called hydrolysis. In a hydrolysis reaction a salt reacts with water to yield the acid or base:
BA + H2O --> HA + BOH
More specifically, there are four combinations of strong and weak acids and bases:
· strong acid + strong base, e.g., HCl + NaOH --> NaCl + H2O
When strong acids and strong bases react, the products are salt and water. The acid and base neutralize each other, so the solution will be neutral (pH=7) and the ions that are formed will not reaction with the water.
· strong acid + weak base, e.g., HCl + NH3 ---> NH4Cl
The reaction between a strong acid and a weak base also produces a salt, but water is not usually formed because weak bases tend not to be hydroxides. In this case, the water solvent will react with the cation of the salt to reform the weak base. For example:
HCl (aq) + NH3 (aq) <--> NH4+ (aq) + Cl- while
NH4- (aq) + H2O <--> NH3 (aq) + H3O+ (aq)
· weak acid + strong base, e.g., HClO + NaOH --> NaClO + H2O
When a weak acid reacts with a strong base the resulting solution will be basic. The salt will be hydrolyzed to form the acid, together with the formation of the hydroxide ion from the hydrolyzed water molecules.
· weak acid + weak base, e.g., HClO + NH3 <--> NH4ClO
The pH of the solution formed from the reaction of a weak acid with a weak base depends on the relative strengths of the reactants. For example, if the acid HClO has a Ka of 3.4 x 10-8 and the base NH3 has a Kb = 1.6 x 10-5, then the aqueous solution of HClO and NH3 will be basic because the Ka of HClO is less than the Ka of NH3.
Now that you’ve already learned that acids react with bases to form salts and water is a chemical reaction that is called neutralization. You must be able to predict names and formulas of salts that incorporate polyatomic ions using the same methods introduced for binary ionic compounds. You must be able to write balanced equations for neutralization reactions using inspection and your knowledge of the neutralization process..
Acids are compounds that dissolve in water to produce solutions containing hydronium ions, H3O+ (aq), and bases dissolve in water to produce hydroxide ions, HO- (aq). When an acidic solution and a basic solution are mixed, hydroxide ions react with hydronium ions to produce water molecules:
H3O+ (aq) + HO- (aq) | 2 H2O | |
or H+ (aq) + HO- (aq) | H2O |
Abbreviating the hydronium ion as H+ (aq) makes it easier to count the water molecules produced.
This chemical process is called neutralization. What happens to the other ions that were present in the original solutions? Nothing! They are still separately solvated and moving independently, unaffected by the reaction going on around them. Let's look at the reaction between hydrochloric acid and sodium hydroxide:
HCl (g) | H+ (aq) + Cl- (aq) | ...(1) | |
NaOH (s) | Na+ (aq) + OH- (aq) | ...(2) | |
H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) | Na+ (aq) + Cl- (aq) + H2O | ...(3) | |
H+ (aq) + OH- (aq) | H2O | ...(4) | |
HCl (aq) + NaOH (aq) | NaCl (aq) + H2O | ...(5) |
Equations 1 and 2 represent the formation of the solutions of hydrochloric acid and sodium hydroxide, and equation 3 shows what happens when they are mixed. Notice that sodium ions and chloride ions appear on both sides of equation 3, called the total ionic equation. If we cancel out these spectator ions we are left with equation 4, the net ionic equation, which represents the chemistry of the neutralization reaction. If we represent the sodium ions and chloride ions as NaCl (aq), and write formulas for hydrochloric acid and aqueous sodium hydroxide, equation 3 becomes equation 5, which summarizes the overall reaction.
If we were to evaporate the water from the solution after this neutralization reaction, we would be left with solid sodium chloride, commonly known as table salt. In general, the term salt refers to any ionic compound that could be produced by a neutralization reaction. Salts contain metal cations, and anions that may be monatomic (like chloride) or polyatomic anions derived from oxyacids. The formula of a salt can be predicted from the known charges of the cation and anion as we have seen previously for binary ionic compounds. Sodium sulfate, for example, has the formula Na2SO4. When a formula includes more than one polyatomic ion, it is enclosed in parentheses with a subscript to indicate the relative number of ions. Magnesium nitrate is Mg(NO3)2. There are two nitrate ions (charge 1- each) for every Mg2+ ion. When necessary, we include the charge on the cation as part of the name. Mercury(II) acetate is Hg(C2H3O2)2.
Some salts form crystalline solids which include some water molecules as part of the structure. Such salts are called hydrates, and we include the water molecules in the formula using a dot and a coefficient. For example, copper(II) sulfate forms blue crystals with five water molecules per formula unit, so the formula is CuSO4.5H2O. Hydrates can be dried by strong heating, which drives the water molecules out of the crystal.
Equations for neutralization reactions are usually written in the form of equation 5 above, a summary using the formulas for the reactants and products. To be useful, a chemical equation must be balanced, i.e. for each element there must be the same number of atoms on both sides of the equation. This is because the law of conservation of mass requires that atoms are not created or destroyed during a chemical reaction.
One way to balance the equation for a neutralization reaction is to remember that the underlying chemical process is the reaction of hydrogen ions and hydroxide ions in a one to one ratio. We balance the overall equation by adding coefficients so that the numbers of hydrogen ions and hydroxide ions are equal.
Consider as an example the reaction of hydrogen bromide with barium hydroxide. Hydrogen ions from the acid will react with hydroxide ions from the base, and barium and bromide ions will be left in solution. The ionic compound made up of barium ions and bromide ions is barium bromide, which has the formula BaBr2 (remember that barium is a group IIA metal and forms Ba2+ cations). We first write an overall equation for the process using the correct formulas for all of the compounds involved:
HBr (aq) + Ba(OH)2 (aq) | ------> | BaBr2 (aq) + H2O | unbalanced |
We can then quickly balance the equation by recognizing that we have 2 moles of hydroxide ions from 1 mole of barium hydroxide, so we need 2 moles of hydrogen bromide and will get 2 moles of water:
2 HBr (aq) + Ba(OH)2 (aq) | -------> | BaBr2 (aq) + 2 H2O | balanced |
Inspection shows us that the other elements, barium and bromine are also balanced. When balancing an equation, start with the correct formula for each compound and change the coefficients not the formulas.
Most equations that we will meet can be balanced by direct inspection, but for more complicated situations an understanding of the chemical processes going on in the reaction can help.
Tutorial Videos that will help you to understand more about Acids and Bases
