Chemistry 12
UNIT IV - ACIDS AND BASES ver 1.1
I. CHARATERISTICS OF ACIDS AND BASES
II. THE ARRHENIUS THOERY OF ACIDS AND BASES
Named after Svente Arrhenius who received the Nobel prize in Chemistry in 1903 for his work with electrolytes. He was one of the first chemists to develop a complete definition of acids and bases.
ARRHENIUS THEORY
ACID
BASE
SALT
To simplify his theory:
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NEUTRALIZATION REACTION REVIEW
Ex. Write the neutralization reaction between H3PO4 and calcium hydroxide.
2 H3PO4 + 3 Ca(OH)2 à Ca3(PO4)2 + 6 H2O
Try: Write the neutralization reaction between magnesium hydroxide and hydrogen iodide.
Consider the following neutralization reaction:
HF(aq) + LiOH(aq) à LiF(aq) + H\O(l)
The complete ionic equation would be:
H+(aq) + F-(aq) + Li+(aq) + OH-(aq) à Li+(aq) + F-(aq) + H\O(l)
The net ionic equation would be:
H+(aq) + OH-(aq) à H\O(l)
DESCRIPTIVE DEFINITIONS OF ACIDS AND BASES
All acids have certain properties in common, as do all bases.
ACIDS
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BASES
Litmus is a special type of paper that turns colour in the presence of acids or bases, important to know and easily memorized by:
COMMON ACIDS
1. SULPHURIC
ACID
2. HYDROCHLORIC
ACID
3. NITRIC
ACID
4. ACETIC
ACID
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COMMON BASES
1. SODIUM
HYDROXIDE
2. POTASSIUM
HYDROXIDE
3. AMMONIA
THE NATURE OF H+
Hydrogen Atom Water
1 proton 1 electron 0 neutrons Two lone pairs of electrons,
Remove an electron – all succeptible to ‘attack’ by
It is a single proton. positive charges
(highly concentrated charge)
Now, dissociation equations for Acids look slightly different:
HCl(g) à H+(aq) + Cl-(aq) can be rewritten as:
HCl(g) + H2O(l) à H3O+(aq) + Cl-(aq) make sure whatever you write is balanced!
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III. BRONSTED-LOWRY THEORY OF ACIDS AND BASES
The theory set forth by Arrhenius worked well, however it did not take into consideration the equilibrium reactions that occurred in acids and bases, so another theory was needed. The Bronsted-Lowry Theory is more general and incorporates Arrhenius’ theory into a larger scheme.
BRONSTED-LOWRY THEORY
To state it more simply,
AN ACID IS A PROTON DONOR AND A BASE IS A PROTON ACCEPTOR
Consider the following typical Bronsted-Lowry acid base equation:
NH3 + H2O « NH4+ + OH-
NH3 has an extra H (and a positive charge) when it forms NH4+, it accepted a proton
Water has lost a hydrogen (and a positive charge) when it forms OH-. It lost a proton.
The trick to determining which is the acid and which is the base is to look at the reactant side of the equation. Pick one of the chemical speices. Find a similar looking species (having a similar composition) on the product side.
Ex. Identify the Acid and the Base in the following reaction:
CH3COOH + H2O ↔ CH3COO- + H3O+
ACID BASE
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Consider the original reaction again:
NH3 + H2O ↔ NH4+ + OH-
Forward reaction: NH3 + H2O à NH4+ + OH-
BASE ACID
But look at the reverse reaction: NH3 + H2O ß NH4+ + OH-
ACID BASE
Ex. CH3COOH + H2O ↔ CH3COO- + H3O+
The substances that differ from each other by only one proton are referred to as CONJUGATE ACID BASE PAIRS. In any Bronsted-Lowry equation there are two conjugate acid base pairs.
Ex. HNO2 + H2O ↔ NO2- + H3O+
Ex. What is the conjugate base of:
a. H2O c. HN3
b. HNO2 d. HSO4-
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Ex. What is the conjugate acid of:
a. OH- c. PO43-
b.HC2O4- d. HSO4-
Ex. Identify the conjugate acid base pairs and state which is the acid and which is the base:
a. HNO3 + H2O ↔ NO3- + H3O+
b. HCO3- + SO32- ↔ CO32- + HSO3-
Ex. Write the Bronsted-Lowry equations for the reaction between:
a. HCN- and F-
b. NO2- and HSO3-
WATER
Consider the following two Bronsted-Lowry equilibrium:
HCN(g) + H2O(l) ↔ H3O+(aq) + CN-(aq)
NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
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MONOPROTIC
ACID
DIPROTIC
ACID
TRIPROTIC
ACID
Note that diprotic and triprotic acids that have donated a proton can be AMPHIPROTIC
RECOGNIZING AMPHIPROTIC SUBSTANCES
IV. STRENGTHS OF ACIDS AND BASES
STRONG ACIDS
AND BASES
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WEAK ACIDS
ANDS BASES
Don’t confuse the terms STRONG and CONCENTRATED
Ex .10.0 M HF(aq) is concentrated, but weak. 0.001 M HCl(aq) is dilute, but strong.
The relative strength of an acid or base can be determined by looking at the appropriate chart (Relative Strengths of Bronsted-Lowry Acids and Bases)
STRONG ACIDS
STRONG BASES
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WEAK ACIDS
Ex. Rank the following acids in strength, from strongest to weakest:
H2O HI HF H2CO3 H3PO4
WEAK BASES
Ex. Rank the following bases in order from strongest to weakest.
C2O4- OH- NH3 SO42- CO32-
Note – the relationship between conjugate pairs and relative strength.
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IMPORTANT
LEVELLING EFFECT
Consider the following example. 1.0 M aqueous solutions were prepared of four different acids.
1.0 M HClO4 produced: 1.0 M H3O+ and 1.0 M ClO4-
1.0 M HCl produced: 1.0 M H3O+ and 1.0 M Cl-
1.0 M HF produced: 0.97 M HF, 0.03 M H3O+ and 0.03 M F-
1.0 M CH3COOH produced: 0.996M CH3COOH, 0.004 M H3O+ and 0.004 M CH3COO-
The strong acids: dissociate completely. Ion concentrations are directly related to the
stoichoimetry of the dissociation. (Effectively replaced by 1.0 M H3O+ and
1.0 M of its’ conjugate base
The weak acids: Don’t dissociate completely. Since HF is a stronger acid than acetic acid,
it has dissociated more.
Strong acids are 100 % dissociated in water to form H3O+ and an anion. Water is said to have LEVELLED all the strong acids to the same strength.
Note- recall the electrolytes conduct electricity due to ions being present in solution. The greater the
concentration of ions, the more conductive the solution. So,
Since strong acids and bases dissociate completely, they will have a higher conductivity than weak acids and bases which only dissociate somewhat. Conductivity can then be used to distinguish between strong and weak (can’t be used to tell the difference between acid/base)
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V. EQUILIBRIUM CONSTANT FOR THE IONIZATION OF WATER
A solution can be classified as acidic, basic, or neutral based on the relative concentrations of H3O+ and OH-.
IONIZATION OF WATER
The reaction between a Strong acid and Strong base
HCl(aq) + NaOH(aq) à NaCl(aq) + H2O(l) + 59kJ
Net Ionic Equation: H+(aq) + OH-(aq) à H2O(l) + 59 kJ
Even if no acid or base is present, pure water will always contain small amounts of H3O+ and OH-.
This is a result of collisions between water molecules. THIS IS CALLED THE SELF-IONIZATION OF WATER
Sometimes written as 2H2O(l) + 59 kJ « H3O+(aq) + OH-(aq)
An equilibrium constant for this reaction can be written as:
Keq = [H3O+][OH-]
[H2O]2
Since water is a pure liquid, the concentration of H2O is a constant at a given temperature, and so it is incorporated into the Keq value.
The value of Kw varies only with temperature. Unless otherwise stated, assume that:
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Ex. What is the [H3O+] in pure water? 2H2O(l) + 59 kJ « H3O+(aq) + OH-(aq)
Kw = [H3O+] [OH-]
1x10-14=x2
x = 1.00x 10-7 M
Ex. What is the [OH-] in 0.25 M HCl?
Try: What is the [H3O+] in 0.0075 M NaOH?
VI. pH and pOH
When working with dilute solutions of strong or weak acids or bases, the [H3O+] or [OH-] can be very small (often less than 1.00 x 10-6). With such small concentrations, it can ofte nbe difficult to compare them, so new units were developed to make it easier for comparisons to be made. The new units were pH and pOH.
Quick math lesson – logarithms
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CONVERTING FROM [H3O+] and [OH-] to pH and pOH.
Ex. If the hydronium ion concentration is 4.67 x 10-5 M, what is the pH?
pH = -log[H3O+]
= -log (4.67x 10-5)
= 4.431
Ex. If the hydroxide ion concentration is 2.83 x 10-6 M, what is the pOH?
Consider the following:
pH = - log[H3O+]
If you want to solve for [H3O+]
To perform the calculation on a calculator:
CONVERTING FROM pH and pOH to [H3O+] and [OH-].
Ex. If the pH is 3.17, what is the hydronium ion concentration?
Inverse log (-3.17) = 6.8 x 10-4 M
Ex. If the pOH = 5.32, what is the hydroxide ion concentration?
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There is a very important relationship between pH, pOH, and Kw.
Start with the Kw expression.
Using the following chart (showing the relationships between H3O+, OH-, pH, and pOH, you can work back and forth between any of the values.
[H3O+] [OH-]
pH pOH
PH FACTORS TO REMEMBER
1. It is an important fact to realize that since pH and pOH are logarithmic scales, a difference in one
pH or pOH unit is equivalent to a 10 fold difference in concentration of the ion.
pH scale
1 2 3 4
0.1 M 0.01M 0.001M
[H3O+] 0.0001 M
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2. pH and significant figures are slightly different than significant figures in other calculations.
Ex. How many significant figures in pH = 4.14?
Take the following example:
[H3O+] = 5.28 x 10-5 M
3. Since pH and pOH are the negative of the exponent,
Low pH and pOH values INDICATE HIGH [ ] pH scale
1______14
14 1
pOH scale
High pH and pOH values INDICATE LOW [ ]
4. At 25oC (AND ONLY AT 25oC)
5. It is possible to have negative pH values, although this will only occur in concentrated strong acids.
Recall that the pH scale was developed to express and compare SMALL [H3O+], so negative pH
values are of little use.
Ex. 2.00M HCl has a [H3O+] of 2.00 M
pH = -log (2.00) = -0.301
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VII. MIXING STRONG ACIDS AND BASES
When a strong acid and base are mixed, a neutralization reaction occurs. The resulting solution may be acidic, basic, or neutral depending on the relative amounts of each of the reactants that were used.
(recall limiting reactant calculations from Chemistry 11)
Ex. What is the pH that results when 25.0 mL of 0.250 M HCl is mixed with 35.0 mL of
0.200 M NaOH?
1. write rxn HCl + NaOH à NaCl + H2O
HCl / + NaOH / à NaCl / + H2OInitial
Change
End
2. setup ICE table (modified)
3. calculate moles of each
4. compare values (LR)
5. determine [OH-]
6. answer
Try: What is the resulting pH of a solution made by mixing 45.0 mL of 0.450 M KOH with 75.0 mL
of 0.275 M HClO4?
CALCULATING MOLES OF ACID NEEDED TO REACH A DESIRED pH
Ex. How many moles of HCl(g) must be added to 40.0 mL of 0.180 M NaOH to produce a solution
that has a pH of 12.500? Assume that there is no change in volume when the HCl is added.
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VIII. ACID AND BASE EQUILIBRIUM CONSTANTS
Weak acids and bases are represented as equilibrium systems as they do not completely dissociate.
ACID
IONIZATION
CONSTANT
To find the Ka values, look down the LEFT side (the acid side) of the Strengths of Acids table and find the species that you are looking for that is acting as an acid.
Ka values for a STRONG acid are not listed in the table of Strengths of Acids. Since they are 100 % ionized, the concentration of the un-ionized acid in the denominator of the Ka expression would be zero.
BASE
IONIZATION
CONSTANT
The table of Strengths of acids does not list the values of Kb, but these can be calculated using the Ka values given in the table. Since here is a very important relationship existing between Ka and Kb for CONJUGATE ACID BASE PAIRS.
Consider the following(ammoinium acting as an acid): NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
Ka expression =
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Conjugate Base Ionization Equation: NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
Kb expression =
CALCULATING Kb from Ka
Ex. Calculate the Kb for C2O42-.
To calculate Kb, you need to find the Ka of the conjugate acid of C2O42-. (Look for C2O42-
on the base side, and you will find the Ka for its conjugate)
Try: Calculate the Kb for NO2-.
Note – because of the relationship between the acid and base ionization constants:
IX. RELATIVE STRENGTHS OF ACIDS AND BASES
If solutions containing H2CO3 and SO32- are mixed, the SO32- can only act as a base since it has no protons that it can donate.
H2CO3(aq) + SO32-(aq) ↔ HCO3-(aq) + HSO3-(aq)
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Note that all Bronsted-Lowry equilibrium reactions that are studied in Chemistry 12 will only involve the transfer of a single proton.
(There won’t be two or three proton transfers such as
H2SO4 + Mg(OH)2 ↔ MgSO4 + 2H2O)
If solutions containing amphiprotic ions such as HCO3- and H2PO4- are mixed:
The STRONGER of the two acids will donate a proton while the other accepts.
Compare Ka values.
HCO3- Ka = 5.6 x 10-11
H2PO4- Ka = 6.2 x 10-8 since this is larger, it will act as the acid
HCO3- + H2PO4- « H2CO3 + and HPO42-