Transcript Chapter 14 Acids and Bases

Bronsted-Lowry acids and
bases
The difference between
dissociation and ionisation
• Dissociation refers to a reaction where a
molecule or substance breaks apart into
smaller units.
• The units are not necessarily ions,
although this is often the case.
• Ionization generally refers to a reaction
which forms ions from an uncharged
species.
Defining acids and bases
• In chemistry, the Brønsted-Lowry theory
is an acid-base theory, proposed
independently by Johannes Nicolau
Brønsted and Thomas Martin Lowry in
1923
Bronsted-Lowry definition of acid
• A substance behaves as an acid when it:
1.donates a proton (H+) to a base.
2.Acids are proton donors.
3.When acids react with water, hydronium
(H3O+) ions are produced. H+ ions cannot
exist by themselves
• HCl + H2O  H3O+ + Clacid
base
H+ is attracted to the negative end of H2O to
become H3O+
Bronsted-Lowry definition of a base
• A substance behaves as a base when it:
1. accepts a proton from an acid (Bases are
proton acceptors.
2. When bases react with water, hydroxide
(OH−) ions are produced.
Acids and bases
• HCl is an acid because it donates H+
• NH3 accepts H+ and therefore is the base
• (NH4+ and Cl– then form an ionic
compound)
• Lewis acid: electron pair acceptor
Lewis base: electron pair donor
H
H
N + H
H
+
H
Cl
H
N
H
-
H
+
Cl
Acid/base conjugate pairs
• Conjugate means joined together
• When an acid and a base react together a
conjugate acid and base are formed
• HNO3 + H2O  H3O+ + NO3acid base
acid 2
base 2
• The conjugate pairs are (HNO3 /NO3-)
(H2O / H3O+)
• They differ by a H+
Acid/base conjugate pairs
HCN(l) + H2O  CN–(aq) + H3O+(aq)
•
•
•
•
HCN is acid, H2O is base
H3O+ is acid, CN– is base
A conjugate acid-base pair are two
substances that differ from each other by
just one proton (H+)
HCN and CN– and H2O and H3O+ are
conjugate acid-base pairs
Questions
• Pg 245 Q 1,2,3
Hydrolysis
• Hydrolysis is a chemical reaction during
which an anion reacts with water to
produce OH– or a cation reacts with water
to produce H3O+
• H2O + Cl-  OH- + HCl
• H2O + H+  H3O+
Acid and base strength
• Acids and bases have different strengths
• Some acids donate protons more readily
than others
• The strength of an acid is its ability to
donate hydrogen ions to a base.
• The strength of a base is its ability to
accept hydrogen ions from an acid.
• Strength is different from concentration (pg
248 Figure 14.9)
Strong acids and bases
• A strong acid will completely ionise in
solution (producing many ions)
• A strong base will accept protons (H+) easily
Weak acids and bases
• A weak acid does not ionise to any great
extent and so contains a larger number of
molecules compared with the number of
ions produced in solution.
Completely
ionised
Partially
ionised
Reversible reaction
• Reversible arrows in an equation show that the
products on the right can react together and
produce the left hand side.
• A chemical equation without a double arrow
isn't reversible and can only go in one direction.
Polyprotic acids
• Polyprotic acids are acids capable of
donating more than one proton (H+).
1.Monoprotic
HCl
2. Diprotic
H2SO4
3. Triprotic
H3PO4
Amphiprotic substances
• Some substances act as acids and bases
• They can donate or accept protons and
are called amphi (meaning both) protic
(hydrogen ions)
• Example water, ammonia and amino acids
• Write the formulae of the conjugate bases
of the following acids:
• H2SO4
H 2S
HSNH4+
• Write the formulae of the conjugate acids
of the following bases:
• OHHCO3H 2O
CN-
pH and pOH
• pH stands for the potential of hydrogen or
concentration of H+
• pOH is a measure of the hydroxide ion
concentration
• The acidity of a solution is a measure of the
concentration of H+
• In a neutral solution there is the same
concentration of H+ or H3O+ and OH• Basic solutions have a lower concentration of
H3O+ than OH• Acidic solutions have a greater concentration of
H3O+ than OH-
Ionic product
• [H3O+ ] [H+ ] represents the concentration of
H3O+ or H +
• [OH-] represents the concentration of OH• Experiments show that all aqueous solutions
contain H + and OH- that the product of their
molar concentrations is 10-14M2 at 25C
• The ionic product is:
[H3O+ ]x[OH-] =
In pure water [H3O+ ]=[OH-]
10-7M = 10-7M
At 25C a solution is:
• Acidic if [H3O+ ] > 10-7M [OH-] < 10-7M
• Neutral if [H3O+ ][OH-]
• Basic if [H3O+ ] < 10-7M [OH-] > 10-7M
Calculating pH and pOH
•
pH is calculated using the following
formula:
pH = -log [H+]
• pOH is calculated using the following
formula:
pOH = -log [OH-]
• pH + pOH = 14.0
• [H+] = 10 -pH
• [OH-] = 10 -pOH
pH calculations for weak acids
• Can we use the pH calculation for weak
acids?
• No
• Why?
• Because weak acids have not fully ionized
so we do not know the H+ concentration
• You have to wait till year 12 and you do
the equilibrium constant
Calculating pH and pOH
• Find the pH of a solution of sodium
hydroxide that has a pOH of 2
• pH = 14 – pOH
• pH = 14 - 2 = 12
• Find the pOH of a solution of hydrochloric
acid that has a pH of 3.4
• pOH = 14 – pH
• pOH = 14 - 3.4 = 10.6
Questions
• Find the pH of 25.0 mL of a 0.045 M
(mol/L) solution of HCl. What is the pOH?
• Note that HCl is a strong monoprotic acid
which means that...
[HCl] = [H+]= 0.045 M
pH = -log [H+]
pH = -log 0.045
pH = 1.35
• The pOH is given by pH + pOH = 14
We substitute in the pH of 1.35 and get:
1.35 + pOH = 14
So, pOH = 12.65
Questions
• a) Find the pOH of 0.000685 M solution of NaOH.
• b) What is the pH of the solution?
a)Note that NaOH is strong and monobasic which
means that...
• [NaOH] = [OH-] = 0.000685 M
hence, pOH = -log [OH-]
pOH = -log 0.000685
pOH = 3.164
b) pOH is 3.164.
The pH can be found by using pH + pOH = 14.
Substituting in gives us pH + 3.164 = 14
So, pH = 10.836
Question
• What is the H+ concentration in the
solution of pH 3.47
• [H+] = 10-pH
= 10 -3.4
= 3.39x10-4 mol/L or M
Questions
• What is the concentration OH- ions in a
solution of pH of 10.47
• Find concentration of H+ ions
[H+] = 10-pH
= 10-10.4
= 2.51x10-11
• Find OH- concentration
[H+ ] x [OH-] = 10-14
[OH-] = 10-14
2.51x10-11
= 3.98 x 10-4 mol/L or M
Find the pH of a 0.2 mol L-1 (0.2M) solution of
H2SO4
• Write the balanced equation for the dissociation
of the acid
• H2SO4  2H+(aq) + SO42-(aq)
• Use the equation to find the [H+]:
0.2 mol L-1 H2SO4 produces 2 x 0.2 = 0.4 mol
L-1
• Calculate pH:
pH = -log[H+]
pH = -log [0.4] = 0.4
• Pg 254 Q 9 a,b,c, 10 a,b 11