Acids and Bases
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Acids taste sour (citric acid, acetic acid)
Bases taste bitter (sodium bicarbonate)
There are 3 ways to define acids and bases, you will
learn 2 of these:
Arrhenius:
- Acids form H3O+ in water (HCl + H2O  H3O+ + Cl-)
- Bases form OH- in water (NaOH  Na+ + OH-)
Brønsted-Lowry (B-L):
- Acids donate H+ and Bases accept H+
HCl + NaOH  H2O + NaCl
HCl is the acid, it donates H+ to OH- (the base)
B-L Acids and Formation of H3O+
• In an acid-base reaction, there is always an acid/base pair
(the acid donates H+ to the base)
• H+ is not stable alone, so it will be transferred from one
covalent bond to another
Example: formation of H3O+ from an acid in water
HBr + H2O  H3O+ + Br-
Identifying B-L Acids and Bases
• Compare the reactants and the products
- The reactant that loses an H+ is the acid
- The reactant that gains an H+ is the base
Examples:
HCl + H2O  H3O+ + ClAcid = HCl and Base = H2O (HCl gives H2O an H+)
NH3 + H2O  NH4+ + OHAcid = H2O and Base = NH3 (H2O gives NH3 an H+)
CH3CO2H + NH3  CH3CO2- + NH4+
Acid = CH3CO2H and Base = NH3 (CH3CO2H gives NH3
an H+)
Conjugate Acids and Bases
• When a proton is transferred from the acid to the base
(in a B-L acid/base reaction), a new acid and a new
base are formed:
HA + B  A- + HB+
acid + base  conjugate base + conjugate acid
• The acid (HA) and the conjugate base (A-) that forms
when HA gives up an H+ are a conjugate acid/base
pair
• The base (B) and the conjugate acid (HB+) that forms
when B accepts an H+ are another conjugate acid/base
pair
Identifying Conjugate Acid/Base Pairs
1. Identify the acid and base for the reactants
2. Identify the acid and base for the products
3. Identify the conjugate acid/base pairs
acid
HF
conjugate base
+ H2O
base
H3O+
+ F-
conjugate acid
Acid and Base Strength
• Strong acids give up protons easily and completely
ionize in water:
HCl + H2O  H3O+ + Cl• Weak acids give up protons less easily and only
partially ionize in water:
CH3CO2H + H2O  CH3CO2- + H3O+
• Strong bases have a strong attraction for H+ and
completely ionize in water:
KOH(s)  K+ (aq) + OH-(aq)
NaNH2 + H2O  NH3 + NaOH
• Weak bases have a weak attraction for H+ and only
partially ionize in water:
HS- + H2O  H2S + OH-
Direction of an Acid/Base Equilibrium
• In general, there’s an inverse relationship between
acid/base strength within a conjugate pair:
- strong acid  weak conjugate base
- strong base  weak conjugate acid
(and vice-versa)
• The equilibrium always favors the direction that goes
from stronger acid to weaker acid
Example 1: HBr + H2O ? H3O+ + Brstronger acid (HBr)  weaker acid (H3O+)
(equilibrium favors products)
Example 2: NH3 + H2O ? NH4+ + OHweaker acid (H2O)  stronger acid (NH4+)
(equilibrium favors reactants)
Dissociation Constants
• Since weak acids dissociate reversibly in water, we can write
an equilibrium expression:
HA + H2O  H3O+ + AKeq = [H3O+][A-]/[HA][H2O]
• But, since [H2O] remains essentially constant we can write:
Ka = Keq x [H2O] = [H3O+][A-]/[HA]
• The acid dissociation constant (Ka) is a measure of how much
the acid dissociates (A higher Ka = a stronger acid)
• Example:
CH3CO2H + H2O  CH3CO2- + H3O+
Ka = [H3O+][CH3CO2-]/[CH3CO2H] = 1.8 x 10-5
• Can also write dissociation constants for weak bases:
NH3 + H2O  NH4+ + OHKb = [NH4+][OH-]/[NH3] = 1.8 x 10-5
Ionization of Water
• Since H2O can act as either a weak acid or a weak
base, one H2O can transfer a proton to another H2O:
H2O + H2O  H3O+ + OHKeq = [H3O+][OH-]/[H2O][H2O]
• Since [H2O] is essentially constant, we can write:
Kw = Keq x [H2O]2 = [H3O+][OH-]
(where Kw = the ion-product constant for water)
• For pure water: [H3O+] = [OH-] = 1.0 x 10-7 M
So, Kw = [H3O+][OH-] = (1.0 x 10-7 M)2 = 1.0 x 10-14
(units are omitted for Kw as for Keq and Ka)
Using Kw
• If acid is added to water, [H3O+] goes up
- for an acidic solution [H3O+] > [OH-]
• If base is added to water, [OH-] goes up
- for a basic solution [OH-] > [H3O+]
• Kw is constant (1.0 x 10-14) for all aqueous solutions
• Can use Kw to calculate either [H3O+] or [OH-] if given
the other concentration
• Example: if [H3O+] = 1.0 x 10-4 M, what is the [OH-]?
Kw = [H3O+][OH-]
[OH-] = Kw/ [H3O+] = 1.0 x 10-14/1.0 x 10-4= 1.0 x 10-10 M
Is this an acidic or a basic solution?
Since [H3O+] > [OH-], it’s an acidic solution
The pH Scale
• pH is a way to express [H3O+] in numbers that are easy
to work with
• [H3O+] has a large range (1.0 M to 1.0 x 10-14 M) so we
use a log scale:
pH = - log [H3O+]
• The pH scale goes from 0 - 14
• Each pH unit = a ten-fold change in [H3O+]
• pH 7 = neutral, pH < 7 = acidic, pH > 7 = basic
• Can use an indicator dye (on paper or in solution) that
changes color with changes in pH, or a pH meter, to
measure pH
Calculating pH and pOH
• Can calculate pH from [H3O+]:
If [H3O+] = 1.0 x 10-3 M, what is the pH?
pH = - log [H3O+] = - log(1.0 x 10-3) = 3.00
• Note: sig. figs. in [H3O+] = decimal places in pH
• Can also calculate [H3O+] from pH:
If pH is a whole number, [H3O+] = 1 x 10-pH
So, if pH = 2, then [H3O+] = 1 x 10-2
• Can calculate pOH from [OH-]
pOH = - log[OH-]
• Also, since Kw = [H3O+][OH-]
then pKw = - log Kw = - log (1.0 x 10-14) = 14.00
• And, pKw = pH + pOH = 14.00
• So, if pH = 3.00, then pOH = 14.00 - 3.00 = 11.00
Reactions of Acids and Bases
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Acids and bases are involved in a variety of chemical reactions
(we’ll study 3 types here)
Acids react with certain metals to produce metal salts and H2 gas,
for example:
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
Acids react with carbonates and bicarbonates to produce salts,
H2O and CO2 gas, for example:
NaHCO3(aq) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g)
Acids react with bases (neutralization reactions) to form salts and
H2O, for example:
HBr(aq) + LiOH(aq)  LiBr (aq) + H2O(l)
Neutralization reactions are balanced with respect to moles of H+
and moles of OH-, for example:
H2SO4 + 2NaOH  Na2SO4 + 2H2O
Acidity of Salt Solutions
• Salts dissolved in water can affect the pH
• When salts dissolve, they dissociate into their ions
NaCl  Na+ + Cl• If one of those ions can donate a proton to H2O, or
accept one from H2O, the pH will change:
Na2S  2Na+ + S2S2- + H2O  HS- + OH-
S2- is a weak base that can accept an H+ from H2O
Since [OH-] is increased, the solution is basic
Salts that form Neutral Solutions
• When a strong acid dissolves in water, a weak conjugate base is
formed that can’t remove a proton from water
• When a strong base dissolves in water, the metal that dissociates
can’t form H3O+
• So, salts containing ions that come from strong acids and bases
do not affect the pH of the solution
• Example:
KBr  K+ + Br- (KOH = strong base, HBr = strong acid)
(KOH + HBr  KBr + H2O)
K+ has no proton to donate, so can’t form H3O+
Br- is too weak of a base to pull a proton off of H2O, so can’t
form OHSo, the solution remains neutral
Salts that form Basic Solutions
• When a weak acid dissolves in water, the conjugate
base formed is usually strong enough to remove a
proton from H2O to form OH• So, salts that contain ions that come from a weak acid
and a strong base form basic solutions
• Example:
NaCN  Na+ + CN- (HCN is a weak acid)
CN- + H2O  HCN + OH-
(HCN + NaOH  NaCN + H2O)
(Na+ doesn’t affect the pH, it’s from a strong base)
Salts that form Acidic Solutions
• When a weak base dissolves in water, the conjugate acid
formed is usually strong enough to donate a proton to
H2O to form H3O+
• So, salts that contain ions from a weak base and a strong
acid form acidic solutions
• Example:
NH4Br  NH4+ + Br- (from NH3 and HBr)
(NH3 + HBr  NH4Br)
NH4+ + H2O  NH3 + H3O+
(Br- doesn’t affect the pH)
Buffer solutions
• A small amount of strong acid or base added to pure water will
cause a very large change in pH
• A buffer is a solution that can resist changes in pH upon addition
of small amounts of strong acid or base
• Body fluids, such as blood, are buffered to maintain a fairly
constant pH
• Buffers are made from conjugate acid/base pairs (either a weak
acid and a salt of its conjugate base or a weak base and a salt of
its conjugate acid)
• Thus, they contain an acid to neutralize any added base, and a
base to neutralize any added acid
• Buffers can’t be made from strong acids or bases and the salts of
their conjugates since they completely ionize in H2O
How to Make a Buffer Solution
• An acetate buffer is made from acetic acid and a salt of its
conjugate base:
CH3CO2H
and
CH3CO2Na
• The salt is used to increase the concentration of CH3CO2- in the
buffer solution
• Recall: CH3CO2H + H2O  CH3CO2- + H3O+
(the equilibrium favors reactants, so the concentration of
CH3CO2- is low)
But, CH3CO2Na  CH3CO2- + Na+
CH3CO2H + CH3CO2Na + H2O  2 CH3CO2- + H3O+ + Na+
• If acid is added: CH3CO2- + H3O+  CH3CO2H + H2O
• If base is added: CH3CO2H + OH-  CH3CO2- + H2O
• Buffer capacity = how much acid or base can be added and still
maintain pH (depends on buffer type and concentration)
Calculating pH of a Buffer
• The pH of a buffer solution can be calculated from the
acid dissociation constant (Ka)
• Example (for acetate buffer):
CH3CO2H + H2O  CH3CO2- + H3O+
Ka = [H3O+][CH3CO2-]/[CH3CO2H] = 1.8 x 10-5
[H3O+] = Ka x [CH3CO2H]/[CH3CO2-]
• What is the pH of an acetate buffer that is 1.0 M
CH3CO2H and 0.50 M CH3CO2Na?
[H3O+] = 1.8 x 10-5 x 1.0 M/0.50 M = 3.6 x 10-5 M
pH = - log[H3O+] = - log(3.6 x 10-5) = 4.44
Dilutions
• Often solutions are obtained and stored as highly
concentrated stock solutions that are diluted for use (i.e.
cleaning products, frozen juices)
• When a solution is diluted by adding solvent, the
volume increases, but amount of solute stays the same,
so the concentration decreases:
Mol solute = concentration (mol/L) x V (L) = constant
So,
C1V1 = C2V2
• For molarity, it becomes:
M1V1 = M2V2
Dilution Calculations
• Example 1:
What volume (in mL) of 8.0 M HCl is needed to prepare
1.0 L of 0.50 M HCl?
M1V1 = M2V2
V1 = M2V2/ M1
V1 = 0.50 M x 1.0 L/ 8.0 M = 0.0625 L = 63 mL
• Example 2:
How many L of water do you need to add to dilute 0.50 L
of a 10.0 M NaOH solution to 1.0 M ?
V2 = M1V1/ M2
V2 = 10.0 M x 0.50 L/ 1.0 M = 5.0 L
volume of water needed = 5.0 L - 0.50 L = 4.5 L
Acid-Base Titration
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Molarity of an acid or base solution of unknown
concentration can be determined by titration:
1. A measured volume of the unknown acid or base is
placed in a flask and a few drops of indicator dye
(such as phenolpthalein) are added
2. A buret is filled with a measured molarity of known
base or acid (the “titrant”) and small amounts are
added until the solution changes color (neutralization
endpoint)
- At neutralization endpoint [H3O+] = [OH-]
3. Molarity of unknown is calculated from moles of
titrant added (mole ratio comes from balanced
chemical equation)
Example: Titration of H2SO4 with NaOH
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What is the molarity of a 10.0 mL sample of H2SO4 if the
neutralization endpoint is reached after adding 15.0 mL of 1.00 M
NaOH?
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Calculate moles NaOH added:
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15.0 mL x (1 L/ 1000 mL) x (1.00 mol/ 1 L) = 0.0150 mol NaOH
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Write the balanced chemical equation:
H2SO4 + 2NaOH  Na2SO4 + 2H2O
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Calculate moles H2SO4 neutralized:
0.0150 mol NaOH x 1 mol H2SO4/ 2 mol NaOH = 0.00750 mol
H2SO4
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Calculate molarity of H2SO4:
0.00750 mol H2SO4/ 0.0100 L = 0.750 M H2SO4