# Edexcel Chemistry - Topic 12: Acid-Base Equilibria

• Created by: Ryan C-S
• Created on: 13-04-18 17:29

## Acid-Base Definitions

• A Lewis Acid is a substance that accepts a pair of electrons
• A Lewis Base is a substance that donates a pair of electrons
• A Bronsted-Lowry Acid is a substance that donates a proton
• A Bronsted-Lowry Base is a substance that accepts a proton
• Each acid is linked to a conjugate base on the other side of the equation
HCl(g) + H2O(l) --> H3O+(aq) + Cl-(aq)
• Strong Acids are acids that completely dissociate
e.g. HCl(aq) --> H+(aq) + Cl-(aq)
• Weak Acids are acids that don't completely dissociate
e.g. CH3COOH(aq) --> CH3COO-(aq) + H+(aq)
• Monoprotic acids only donate one proton to an aqueous solution
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## Calculating pHs

• The acidity of a solution is based off of the concentration of protons in an aqueous solution [H+]

pH = -log[H+]

• e.g. For a 0.1M solution of HCl, the pH will be -log[0.1] = 1.00

[H+] = 1x10^-pH

• e.g. For a solution of pH 2.00 the concentration of protons present will be 1X10^-2.00 = 0.01M
• For a monoprotic strong acid, the concentration of H+ ions present will be the same as the concentration of the acid
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## Acid-Base Equilibria and Kw

• In all aqueous solutions and pure water, the following equilibrium occurs:
H2O(l) ⇌ H+(aq) + OH-(aq)
• Therefore the following equilibrium expression occurs:

Kc = [H+(aq)][OH-(aq)] => Kc x [H2O(l)] = [H+(aq)][OH-(aq)]
[H2O(l)]

• The Kw expression is equal to the Kc of the above equilibrium multiplied by the concentration of water so Kw = [H+(aq)][OH-(aq)]
• At 25C the value of Kw for all aqueous solutions is 1x10^-14 mol2 dm-6
• The Kw expression can be used to calculate the value of [H+(aq)] if the value of [OH-(aq)] is known
• pKw = -log(Kw) so Kw = 10^-pKw
• If Kw =1x10^-14 then pKw = 14
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## Finding pHs of Pure Water and Strong Bases

pH of Pure Water

• Pure water is neutral because the concentration of H+ ions is equal to the concentration of OH- ions
• Using Kw = [H+][OH-] then we can say Kw = [H+]^2 so the square root of Kw = [H+]
• At 25C [H+(aq)] = 1x10^-14 = 1x10^-7 so pH = 7.00

pH of a Strong Base

• For strong bases, the concentration of OH- ions is usually given so the concentration of H+ ions is required in order to work out the pH
• For a solution of 0.1M NaOH: [H+] = Kw/[OH-] = 1x10^-14 / 0.1 = 1x10^-13 mol dm-3
pH = -log(1x10^-13) = 13.00
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## Finding pHs of Weak Acids and Ka

• Weak Acids only slightly dissociate when in water so form the equilibrium:
HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq) or HA(aq) ⇌ H+(aq) + A-(aq)
• This equilibrium can give an expression for Ka:
Ka = [H+][A-]
[HA]
• The Ka value of an acid can be used to determine the concentration of H+ ions and thus the pH
• The half equivalence point is the position when the concentration of added base is equal to half the original concentration of acid
• Assuming that [HA] = [A-] then pH = pKa
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## Diluting an Acid or Alkali

• Dilutions of an acid or alkali can be worked out by the following equations:

[H+]new = [H+]old x (old volume / new volume)

[OH-]new = [OH-]old x (old volume / new volume)

• These new values can be used in the pH equation to work out the new pH of a solution

pH = -log[H+]

• pH is a logarithmic scale so diluting a strong acid 10 times will increase its pH by one unit and 100 times will increase its pH by two units
• Weak acids will increase by less than one unit for being diluted 10 times
• Diluting a weak acid pushes the equilibrium to the right
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## Buffer Solutions

• A buffer solution is one where the pH does not change significantly if small amounts of acid or alkali are added to it
• An acidic buffer solution is made from a weak acid and a salt of that weak acid (made by reacting the weak acid with a strong base) e.g. CH3COOH and CH3COO-Na+
• A basic buffer solution is made from a weak base anf a salt of that weak base (made by reacting the weak base with a strong acid) e.g. NH3 and NH4+Cl-
• An ethanoic acid buffer works by creating an equilibrium between the acid and conjugate base:
CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)
• In a buffer solution there is a higher concentration of the salt CH3COO- ion than in the pure acid
• If small amounts of acid is added to the buffer, the above equilibrium will shift to the left removing nearly all the H+ ions
• As there is a large salt ion concentration, the ratio of [CH3COOH]:[CH3COO-] stays relatively constant so therefore the pH remains similar
• If small amounts of alkali is added to the buffer, OH- ions react with the H+ ions to form water and the equilibrium will shift to the right to produce more H+ ions
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## Calculating pH of Buffers

• The Ka equation is used in order to determine the pH by finding the concentration of H+ ions present in the solution first

Ka = [H+][A-]
[HA]

pH = -log[H+]

• The equation is assuming that the [A-] concentration is due to the added salt only and that the initial concentration of acid has remained constant
• The salt content can be added as a solution or a solid
• A buffer can also be made by partially neutralising a weak acid with alkali therefore making a salt
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## Titration Curves

• Titration curves show the change in pH when a volume of an acid or a base is added

Constructing Titration Curves

• Measure the initial pH of the acid
• Stir the mixture to equalise the pH
• Measure and record the pH to 1 decimal place
• When approaching the end point add in smaller volumes of alkali
• Add alkali until in excess

The equivalence point lies at the midpoint of the extrapolated vertical position of the curve
The meter must be calibrated first by measuring a known pH of buffer solution as pH meters can lose accuracy on storage
Accuracy can be improved by maintaining a constant temperature

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## Indicators

• Indicators can be considerred as a weak acid. The acid must have a different colour to its conjugate base
• An indicator changes colour from HIn to In- over a narrow range
• HIn(aq) ⇌ H+(aq) + In-(aq)
• Le Chateliers Principle can be applied to give the colour change
• In an acid solution, the H+ ions present will push the equilibrium towards the reactants and therefore the colour of HIn will be displayed
• In an alkaline solution, the OH- ions present will react and remove H+ ions causing a shift to the products and therefore the colour of In- will be displayed
• The end point in a titration is reached when [HIn] = [In-] so an indicator whose end point coincides with the equivalence point should be chosen
• An indicator will work if the pH range of the indicator lies on the vertical part of a titration curve

Phenolphthalein - Strong Bases: Colourless (acid) --> Pink (alkali)
Methyl Orange - Strong Acids: Red (acid) --> Yellow (alkali)

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