Edexcel A Level Chemistry: 12: Acid Base Equilibria

A proton donor

A proton acceptor

A species formed by the reception of a proton (H+) by a base in other words, it is a base with a hydrogen ion added to it.

What is left after an acid has donated a proton in a chemical reaction.

One which can donate two protons, such as H2SO4

One which can accept two protons, such as CO3 2-

One which can act either as an acid or a base, such as water

One which is almost completely dissociated in aqueous solution

With a single arrow in the dissociation equation, not reversible

One that is only partially dissociated in aqueous solution

With a reversible arrow in the dissociation equation

Directly, so HCl of 0.100 moldm-3 will have an H+ concentration of 0.100 moldm-3

pH = -log [H+]

[H+] = 10^-pH

The acid dissociation constant

[H+] x [A-]
_______
[HA]

[H+]^2
_____
[HA]a

By rearranging the equation to find a value for [H+]^2, then square rooting the value.

-log Ka

…the stronger the acid

…the weaker the acid

Ka HSO4 = 0.0100
In 0.500 moldm-3 solution of H2SO4, contribution to [H+] from H2SO4 will be 0.500moldm-3. If we assume that contribution to [H+] from HSO4- is (x)moldm-3:

Ka(HSO4-) = 0.0100 = (0.500- x) x/0.500
Solving this equation gives x = 0.0098
So total [H+] = 0.05098 with a pH of 0.293

The ionic product of water

[H+] x [OH-]
_________
[H2O]

[H+] x [OH-], as pure water is neutral

-log Kw

Because water ionises to form H+ and OH-

Find the concentration of the base
Divide the Kw by the base concentration
This will give you [H+], so you can then calculate pH

By measuring the pH of equimolar aqueous solutions of the acids/bases, at the same temperature

7

7, alkaline

<7, acidic

pH increases by one for each 10-fold dilution

pH increases by 0.5 for each 10-fold dilution

By dissolving a known mass of acid in a small volume of water and making it up to a known volume, then measuring its pH
Calculate concentration of solution
Work backwards to calculate Ka

When the acid and base are mixed in exactly the right proportions to react in equivalent amounts

When the acid and base have reacted together in the exact proportions dictated by the stoichiometric equation

Starts at high pH, sudden drop as equivalence point reached, ends at low pH

Starts at high pH, equivalence point around pH 8/9, shorter middle section, ends at pH 3/4

Starts at pH 11, slight decrease in pH before equivalence point, around pH 5/6, ends around pH 1

Starts around pH11, no steep section, short, equivalence point at pH 7, ends around pH 4/5

The point on a weak acid/base titration curve where the equivalence point lies, no steep section

Because a buffer solution is formed

K(in) = [H+] x [In-]

[HIn]

The red colour of methyl orange will first predominate when [HIn] is ten times [In-], and the yellow colour will predominate when [In-] is ten times [HIn].

When [HIn] = 10[In-]:

[H+] x [In-]
= 2.00 x 10^-4 mol dm-3 (K(In))
10[In-]
[H+] + 2.00 x 10^-3

…one drop of acid is added from the burette to accurately determine the end point of titration

A solution whose pH remains the same when small amounts of acid are base are added

Mixing a weak acid with its conjugate base

| Mixing a weak base with its conjugate acid

Ethanoic acid and sodium ethanoate

Find Ka
Then find [H+] by rearranging the equation for Ka with the acid or base and the conjugate ion
Use [H+] to calculate pH

When small amounts of acid are added, majority of H+ react with the CH3COO- ions to form CH3COOH molecules.

Majority of OH- added react with CH3COOH to form CH3COO- and H2O

[H+] = Ka x [acid]
____
[salt]

The pH of a buffer solution

A mixture of a weak acid and its conjugate base

A mixture of a weak base and its conjugate acid

At the half equivalence point, pH = pKa

Dissociation (endothermic) and hydration (exothermic)