ACIDS AND BASES. CHEMISTRY CHAPTER 12. 12.1 Acids and Bases: Introduction A. All aqueous solutions contain hydrogen ions (H + ) and hydroxide (OH-). 1. An acidic solution contains more (H + ) ions than (OH - ) ions.
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ACIDS AND BASES
12.1 Acids and Bases: Introduction
A. All aqueous solutions contain hydrogen ions (H+) and hydroxide (OH-).
1. An acidic solution contains more (H+) ions than (OH-) ions.
2. A basic solution contains more (OH-) ions than (H+) ions.
3. A neutral solution will contain equal concentrations of (H+) and (OH-) ions.
4. A hydronium ion (H3O+) is a hydrated hydrogen ion.
5. The symbols H+ and OH- can be used interchangebly in chemical reactions.
B. Arrhenius model – of acids and bases
1. An acid is a substance that contains hydrogen and ionizes in aqueous solutions to produce hydrogen ions.
2. A base is a substance that contains a hydroxide group and dissociates in aqueous solution to produce hydrogen ions.
3. Example: HCl H2+ + Cl-
4. Sulfuric acid, a polyprotic compound, ionizes in two steps.
H2SO4 H2+ + HSO4-
HSO4 H2+ + SO4-2
5. a base ionizes or dissociates in solution to produce hydroxide ions, OH-
NaOH Na+ + OH-
C. Bronstead-Lowery model
1. An acid is a hydrogen donor and a base is a hydrogen-ion acceptor.
a) When a acid donates a hydrogen-ion, a conjugatebase is formed.
b) When a base accepts a hydrogen-ion, a conjugate acid formed.
c) Two substances related to each other by donating and accepting of a hydrogen ion are a conjugate acid-base pair.
2. A base includes any substance that would accept a proton, and an acid is a proton donor.
a) A product that results from an acid-base reaction are called the conjugate acid and conjugate base.
HF + HCO3 H2CO3 + F
4. The conjugate base is a particle that remains after a proton is donated by an acid. The conjugate acid is formed when a base accepts a proton from an acid. The hydrogen carbonate ion, HCO3, behaves as a base but does not contain ionizable hydroxide.
Practice Problems Assigned WS-19A
D. Monoprotic and Polyprotic Acids
1. An acid that can donate only one hydrogen ion is called a monoprotic acid.
a) Example: hydrochloric acid (HCl) and formic acid (HCOOH) are monprotic acids, because they each contain only one ionizable hydrogen.
(Note that only those hydrogen atoms that are bonded to electronegative elements are ionizable.)
2. An acid that can donate more than one hydrogen ion.
a) Example: sulfuric acid (H2SO4) contains two ionizable hydrogen atoms called a diprotic acid.
b) Example: boric acid (H3BO3) contains three ionizable hydrogen atoms, so it is a triprotic acid.
H3BO3 + H2O H3O + H2BrO3
H2BO3 + H2O H3O + HBrO3
HBrO3 + H2O H3O + BrO3
Practice Problems Assigned WS-19B
12.2 Strengths of Acids and Bases
A. Strength of Acids
1. An acid that ionizes completely in dilute aqueous solution is a strong acid.
a) Examples: hydrochloric acid (HCl), nitric acid (HNO3), sulfuric acid (H2SO4), and perchloric acid (HClO4)
2. An acid that only partially ionizes in aqueous solution is a weak acid.
a) Examples: carbonic acid (H2CO3), boric acid (H3BO3), phosphoric acid (H3PO4), and acetic acid (HC2H3O2).
3. The ionization of a weak acid reaches a state of equilibrium in which the forward and reverse reactions occur at equal rates.
a) Example: Formic acid (HCOOH); a weak organic acid
HCOOH + H2O H3O + HCOO
b) The equilibrium constant expression for the ionization of formic acid.
c) Ka is called the acid ionization constant , which is the value of the equilibrium constant expression for the ionization of a weak acid.
d) The value of Ka indicates the extent of ionization of the acid. The weakest acids have the smallest Ka value.
e) In the case of a polyprotic acid, there is a Ka value for each ionization, and the Ka values decrease for each successive ionization.
Example: the first and second ionization of phosphoric acid is represented by this equation.
H3PO4 + H2O H3O + H2PO4
Ka = (H3O) (H2PO4)
H2PO4 + H2O H3O + HPO4
Ka = (H3O) (HPO4)
Practice Problems Assigned WS-19C
B. Strength of Bases
1. Metallic hydroxides, such as potassium hydroxide, are strong bases because they dissociate entirely into metal ions and hydroxide ions in aqueous solution.
KOH K+ + OH-
2. A weak base ionizes only partially in dilute aqueous solution to form an equilibrium mixture. (Aniline equals C6H5NH2)
C6H5NH2 + H2O C6H5NH3 + OH
a) The equilibrium constant ionization expression for the ionization of aniline in water is as follows.
Kb = (C6H5NH3) (OH)
3. Kb is called the base ionization constant, which is the value of the equilibrium constant expression for the ionization of a weak base.
Practice Problems Assigned WS-19D
12.3 What is pH
A. Pure water self-ionizes slightly to form H3O and OH ions as shown in the equation.
H2O + H2O H3O + OH
B. The equation for the equilibrium can be simplified by removing one water molecule from each side.
H2O H + OH
C. A special equilibrium expression for the self-ionization of water is defined as:
Kw = (H+) (OH-)
D. Kw is called the ion product constant for water, which is the value of the equilibrium content expression for the self-ionization of water. In pure water at 298K, the concentrations of H+ ions and OH- ions both equal 1.0 x10-7, so the value of Kw can be calculated.
Kw = (H+) = (1.0 x 10-7) & (OH-) = 1.0 x 10-7
Kw = 1.0 x 10-14
E. Using Kw to calcualte (H+) and (OH-)
At 298K, the OH- ion concentrations of an aqueous solution is 1.0 x 10-11 M. Find the H+ ion concentration in the solution and determine whether the solution is acidic, basic or neutral.
First: Write the ion product constant expression.
Kw = (H+) (OH-) = 1.0 x 10-14
Second: Divide both sides of the equation by (OH-)
(H+) = Kw / (OH-)
Third: Substitute the value for Kw and (OH-) and solve.
(H+) = 1.0 x 10-14 / 1.0 x 10-11 = 1.0 x 10-3 M
(H+) is greater than (OH-),so the solution is acidic
Practice Problems Assigned WS-19E
F. pH and pOH
1. The pH scale has a value of 0 to 14.
a) Acidic solutions have pH values between 0-7, with the value of 0 being the most acidic.
b) The pH if a basic solution is between 7 and14, with 14 representing the most basic solution.
c) A neutral solution has a pH of 7.
2. A pOH scale expresses the basicity of a solution. The pOH of a solution is the negative log rhythm of the hydroxide ion concentration.
pOH = -log (OH-)
a) If either pH or pOH is known, the other may be determined by using the following relationship.
pH + pOH = 14.00
b) The pH and pOH values for a solution may be determined if either (H+) or (OH-) is known.
3. Sample Problem: Calculating pH and pOH from (H+)
If a certain carbonated soft drink has a hydrogen ion concentration of 7.3 x 10-4 M, what are the pH and pOH of the soft drink?
Because (H+) is given, it is easier to calculate pH frist.
pH = -log (H+)
pH = -log (7.3 x 10-4)
pH = -(log 7.3 + log 10-4)
pH = -(0.86 + (-4)) = 3.14
The pH of the soft drink is 3.14. Note that the number of the decimal places retained in the pH value equals the number of significant figures in the H+ ion concentration.
To find the pOH, recall that pH + pOH = 14.00. Isolate pOH by subtracting pH from both sides of the equation.
pOH = 14.00 – 3.14 = 10.86 (Therefore the soft drink is acidic)
Practice Problems Assigned WS-19F
G. Calculating ion concentration from pH
1. When the pH of a solution is known, you can determine the concentration of H+ and OH-.
pH = -log (H+)
Multiply both sides of the equation by -1
-pH = log (H+)
Now take the antilog of both sides of the eqaution
antilog (-pH) = (H+)
(H+) = antilog (-pH)
Similar for (OH+)
(OH-) = antilog (-pOH)
3. Example Problem- Calculating (H+) and (OH-) from pH
What are (H+) and (OH-) in an antacid solution with a pH of 9.70?
Use pH to find (H+)
(H+) = antilog (-pH)
(H+) = antilog (-9.70)
(H+) = 2.0 x 10-10 M
To determine (OH-), first use the pH value to calculate pOH
pOH = 14.00 – pH
pOH = 14.00 – 9.70 = 4.30
(OH-) = antilog (-pOH)
(OH-) = antilog (-4.30)
(OH-) = 5.0 x 10-5 M
As expected (OH-) is greater than (H+) , therefore a basic solution
Practice Problems Assigned WS-19G
H. Using pH to calculate Ka
1. If you know the pH and the concentration of a solution of a weak acid, you can calculate Ka for the acid. The following example problem illustrates this types of calculation.
2. Example Problem – Calculating Ka from pH
The pH of a 0.200M solution of acetic acid (CH3COOH) is 2.72. What is Ka for acetic acid?
CH3COOH H+ + CH3COO-
Ka = (H+) (CH3COO-)
pH = -log (H+)
(H+) = antilog (-pH)
(H+) = antilog (-2.72)
(H+) = 1.9 x10-7 M
(CH3COO-) = (H+) = 1.9 x 10-3 M
(CH3COOH) = 0.200 M – (H+)
(CH3COOH) = 0,200 M – 1.9 x 10-3 M = 0.198M
Ka = (1.9 x 10-3) (1.9 x 10-3) = 1.8 x 10-5
Practice Problems Assigned WS-19H
A. The reaction of an acid and a base in aqueous solution is called a neutralization reaction.
1. Products of a neutralization reaction are a salt and water.
a) Salt – is an ionic compound of a positive ion from a base and a negative ion from an acid.
b) An example of an acid-base neutralization is the reaction of nitric acid and calcium hydroxide to form calcium nitrate and water.
2HNO3 + Ca(OH)2Ca(NO3)2 + 2H2O
2. Acid-base neutralizations are used in the procedure called titration, which is a method of determining the concentration of a solution by reacting it with another solutions known concentration.
a) to find the concentration of an acid solution, you would slowly add a basic solution of known concentration.
b) the neutralization reaction would proceed until it reaches the equivalence point, where the moles of H+ ion from the acid equal the moles of OH- ion form the base.
c) At the equivalence point, a large change in pH occurs that can be detected by a pH meter or an acid-base indicator, which is a chemical dye whose color is affected by pH changes.
B. Example Problem – Calculating concentration from Titration Data
In a titration, 53.7-mL 0.100M HCl solution is needed to neutralize 80.0-mL of KOH solution. What is the molarity of the KOH solution?
1. Write the balanced equation for the reaction.
HCl + KOH KCl + H2O
2. Convert millileters if HCl solution to liters.
53.7-mL HCl x 1-L HCl = 0.0537-L HCl
3. Determine the moles of HCl used to multiplying the volume of the solution by its molarity, or mol/L.
0.0537-L HCl x 0.100 molHCl = 5.37 x 10-3
4. Use the mole ratio in the balanced equation to calculate the moles of KOH in the unknown solution.
5.37 x 10-3 x 1 mol KOH = 5.37 x 10-3mol KOH
5. Convert millimeters of KOH solution to liters
80-mL KOH x 1-L KOH = 0.0800-L KOH
6. Determine molarity of the KOH solution
M = 5.37 x 10-3mol KOH = 6.71 x 10-2 M
Practice Problems Assigned WS-12I
12.5 Percent of Ionization
A. When weak acids or bases dissolve in water, they ionize slightly. Most of the acid or base remains in the molecular form. The amount of original acid or base that ionizes is expressed as the percent of ionization.
% of ionization = amount ionized x 100
original acid or base
B. The percent of ionization is used as an indicator of acid or base strength. The stronger acids usually have a higher percent of ionization. If the percent ionization of the weak electrolyte is known, the Kaor Kb can be calculated if the concentration of the solution is given.
C. Example Problem A:
If an acetic acid solution has an initial concentration of 0.0800M and is 1.50% ionized, determine the experimental Ka of the CH3COOH.
If the initial CH3COOH concentration is 0.0800, the H3O+ and the CH3COO can be obtained by the multiplying the concentration by the percent ionization.
(H3O+) = (CH3COO-) = (0.0800) (0.0150) = 0.00120M
At equilibrium (CH3COOH) = 0.0800 – 0.00120 = 0.0788M
From the ionization equation CH3COOH + H2O H3O+ + CH3COO-
Ka = (H3O+) (CH3COO-) = (1.20 x 10-3)2 = 1.83 x 10-5
(CH3COOH) 7.88 x 10-2
D. Example Problem B:
If the Ka of hydrochloric acid is 3.53 x 10-4 and the HF solution has an initial concentration of 0.150M, calculate the percentage of ionization.
From the ionization equation determine the amount ionized by solving the Ka expression for x.
HF + H20 H3O+ + F-
Ka = (H3O+) (F-) .
The (HF) is assumed to be 0.150 instead if 0.150 –x, as x is very small number. Substitute the Ka value and solve for x.
The (HF) is assumed to be 0.150 instead if 0.150 –x, as x is very small number. Substitute the Ka value and solve for x. (Repeat)
3.53 x 10-4 = (x) (x) .
x2 = 5.30 x 10-5
x = 7.28 x 10-3M
Check your answer using iteration.
x2 = (3.53 x 10-4) (0.150 – 7.28 x10-3)
x2 = 5.04 x 10-5
x = 7.10 x 10-3
Since there is a difference between your two values for x, continue to interate until your value for x does not change.
x = (3.53 x 10-4) (0.150 – 7.10 x 10-3)
x = 7.10 x 10-3
Now that you have corrected value for x, continue to solve for percent of ionization.
% ionization = amount ionized x 100
= 7.10 x 10-3 M x 100 = 4.73%
Practice Problems Assigned WS-12J