5.3 Acids, Bases and Buffers Flashcards
Brønsted–Lowry acid
a species that can donate a proton
HCl(aq) → H+(aq) + Cl-(aq)
When an acid is added to water, it releases H+ ions (protons) into solution
Brønsted–Lowry base
a species that can accept a proton
NH3 (aq) + H+(aq) → NH4+(aq)
common base
Common bases are metal oxides and metal hydroxides
e.g. MgO NaOH
Ammonia is also a base
monoprotic acids
Many acids are called monoprotic acids. This means that they only donate one mole of protons per mole of acid;
e.g. HCl, HNO3, CH3COOH
diprotic acids
Some acids are diprotic acids. This means that they can donate two moles of protons per mole of acid; e.g. H2SO4, HOOCCOOH.
H2SO4(aq) + 2H2O(l) -> 2H3O+(aq) + SO42-(aq
triprotic acids
Triprotic acids donate three moles of protons per mole of acid; e.g. H3PO4.
ammonia
Ammonia is a gas that dissolves in water to form a weak alkaline solution.
Dissolved NH3 reacts with water
NH3(aq) + H2O ⇌ NH4+(aq) + OH-(aq)
acts as a base because N in NH3 is accepting a proton
alkali
An alkali is any chemical compound that gives a solution of pH greater than 7.0 when dissolved in water.
Sodium hydroxide NaOH
Potassium hydroxide KOH
Sodium hydroxide Mg(OH)2
is a soluble base that releases OH- ions in aqueous solutions
conjugate acid-base pairs
Two species differing by H
conjugate acid
is a species formed from a Brønsted-Lowry base by the addition of a proton.
conjugate base
is a species formed from a Brønsted-Lowry acid by the loss of a proton
amphoteric compounds
is a molecule or ion that can react both as an acid as well as a base.
e.g. amino acids, which have amine and carboxylic acid groups, and self-ionisable compounds such as water
HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)
base 1 acid 1
Kw
constant called the ionic product for water, Kw, is defined as:
Kw = [H+][OH–]
units= mol2 dm–6
what is meant by the strength of an acid?
the extent of dissociation
strong acids
A strong acid has a low pH (usually 0 or 1). This means that the concentration of H+ is high. This is because the acid is fully dissociated into its ions.
weak acid
A weak acid has a higher pH (but still less than 7). This means that the concentration of H+ is lower than for a strong acid. This is because the acid is not fully dissociated into its ions.
find pH of 250 cm3 of an aqueous solution containing 3.65 g hydrochloric acid
n(HCl)= 3.65g / (1 + 35.5) = 0.1 mols
conc(HCl)= 0.1 / (250/1000) = 0.4 mols dm-3
pH = –log10 [0.4]
pH=0.398
weak acid and Ka
HA(aq) + H2O(l) H3O+(aq) + A–(aq)
HA(aq) ⇌ H+(aq) + A–(aq)
Ka= [H+][A-]
[HA]
unit= mol dm–3
approximation 1
[H+(aq)] = [A−(aq)] [H+] = square root of Ka X [HA]
approximation 2
The equilibrium [HA] is smaller than the undissociated [HA]
The dissociation of weak acids is small you can assume [HA]start »_space; [H+] and you can neglect decrease in concentration of the HA dissociation
assume [HA]start»_space; [H+] and equilibrium [HA] = undissociated [HA]
limitations to approximation 1
At 25 C [H+] from dissociation of water = 10-7. If the pH > 6 then [H+] from the dissociation of water will be significant compared with the dissociation of HA.
Approximation breaks down for very weak acids or diluted solutions
limitations to approximation 2
This approximation holds for weak acids with small Ka values. It breaks down when [H+] becomes significant and there is a real difference between [HA]eqm and [HA]start – [H+]eqm
This approximation is not justified for stronger weak acids with Ka> 10-2 mol dm-3 and for very dilute solutions
Ka and pKa
The extent to which an acid dissociates is shown by the value of Ka. The larger the value of Ka, the stronger the acid.
if Ka is less than 1, the acid is weak
if Ka is greater than 1, the acid is strong
The values of Ka span a wide range. To make them easier to interpret, a new term pKa is used
pKa = –log10 Ka
calculating Ka from pKa
pKa = –log10 Ka
so
Ka = 10–pka
buffer
A system that minimises pH changes on addition of small amounts of an acid or a base.
It will remove any H+ or OH- that you might add to it, otherwise the pH will change
composition of buffers
They are composed of weak acids and their conjugate bases:
HA ⇌ H+ + A-
HA and A- act as two independent reservoirs that are capable of reacting with added alkalis or acids, respectively:
The buffer is most effective when [HA] = [A-].
The buffer is no longer effective when one of the components is completely used up.
The pH of a buffer does vary slightly when reacting with any added species, so shouldn’t be considered totally constant.
how buffers work
As buffers have two independent reservoirs, a reaction with one will cause a shift in the equilibrium to negate the change
-> adding alkali
HA ⇌ A- + H+
addition of an acid to a buffer
[H+] increases
Conjugate base (A-) reacts
Equilibrium shifts left
[H+] removed
addition of an alkali to a buffer
[OH-] increases
H+ reacts (H+ + OH- → H2O)
Equilibrium shifts right
HA dissociates restoring [H+]
forming buffers- addition
Adding solutions of a weak acid and its conjugate salt.
As the dissociation of weak acids is very low, the concentration of the acid can assume to remain the same.
As salts are ionic they completely dissociate in solution so provide a concentration for the conjugate base.
forming buffers- neutralisation
Partial neutralisation of a weak acid.
Add an alkali, e.g. NaOH(aq) to excess weak acid.
This will neutralise some of the acid to form the conjugate base and leaving some acid left over.
making a buffer from an acid and a base
When a small quantity of a strong base is added to a weak acid, a buffer solution is formed.
Some of the weak acid reacts with the strong base to form the salt of the weak acid. For example:
CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)
If the acid is in excess, the solution contains some of the salt of the weak acid, along with some of the weak acid, so the solution is a buffer.
method 1 of calculating pH of buffers
weak acid and a salt
-calculate the concentration of weak acid, HA
-calculate the concentration of the salt, A-
use:
[H+] = Ka X [HA]
[A-]
then use:
pH=-log10[H+]method 2 of calculating pH of buffer
n weak acid
n base
n acid in excess
n salt formed
total volume
conc of acid [HA]
conc of salt [A-]
use rearrangement of formula for Ka
[H+] = ka X [HA]
[A-]
then use:
pH=-log10[H+]how does a buffer act when added to acid
Equilibrium shifts to the left
CH3COOH ⇌ CH3COO– + H+
[CH3COOH] increases and [CH3COO¯] decreases
how does a buffer act when added to an alkali
Equilibrium shifts to the right
CH3COOH ⇌ CH3COO– + H+
[CH3COOH] decreases and [CH3COO¯] increases
