pH calculationspH calculations

What is pH?What is pH?

pH = - log10 [H+(aq)]

where [H+] is the concentration of hydrogen ions in mol dm-3

to convert pH intohydrogen ion concentration [H+(aq)] = antilog (-pH)

IONIC PRODUCT OF WATER Kw = [H+(aq)] [OH¯(aq)] mol2 dm-6

= 1 x 10-14 mol2 dm-6 (at 25°C)

Calculating pH - Calculating pH - strong acids and alkalisstrong acids and alkalis

Strong acids and alkalis completely dissociate in aqueous solution

It is easy to calculate the pH; you only need to know the concentrationonly need to know the concentration.

Calculate the pH of 0.02M HCl

HCl completely dissociates in aqueous solution HCl H+ + Cl¯

One H+ is produced for each HCl dissociating so [H+] = 0.02M = 2 x 10-2 mol dm-3

pH = - log [H+] = 1.7

WORKEDEXAMPLEWORKEDEXAMPLE

Calculating pH - Calculating pH - strong acids and alkalisstrong acids and alkalis

Strong acids and alkalis completely dissociate in aqueous solution

It is easy to calculate the pH; you only need to know the concentrationonly need to know the concentration.

Calculate the pH of 0.02M HCl

HCl completely dissociates in aqueous solution HCl H+ + Cl¯

One H+ is produced for each HCl dissociating so [H+] = 0.02M = 2 x 10-2 mol dm-3

pH = - log [H+] = 1.7

Calculate the pH of 0.1M NaOH

NaOH completely dissociates in aqueous solution NaOH Na+ + OH¯

One OH¯ is produced for each NaOH dissociating [OH¯] = 0.1M = 1 x 10-1 mol dm-3

The ionic product of water (at 25°C) Kw = [H+][OH¯] = 1 x 10-14 mol2 dm-6

therefore [H+] = Kw / [OH¯] = 1 x 10-13 mol dm-3

pH = - log [H+] = 13

WORKEDEXAMPLEWORKEDEXAMPLE

• Calculate the pH of

• 0.01 M HCl

• 0.1 M HNO3

• 0.02 M HCl

0.2mol dm-3 nitric acid[H+] = 1.2 x 10-2mol dm-3[H+] = 6.7 x 10-2mol dm-3[H+] = 9.1 x 10-2mol dm-3[H+] = 6.8 x 10-3mol dm-3[H+] = 1.0 x 10-7mol dm-3[H+] = 5.4 x 10-9mol dm-3[H+] = 9.9 x 10-6mol dm-3

Calculating pH - Calculating pH - weak acidsweak acids

A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)

A weak acid is one which only partially dissociates in aqueous solution

Calculating pH - Calculating pH - weak acidsweak acids

A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)

Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)

[HA(aq)]

A weak acid is one which only partially dissociates in aqueous solution

Calculating pH - Calculating pH - weak acidsweak acids

A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)

Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)

[HA(aq)]

The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]

therefore Ka = [H+(aq)]2 (3)

[HA(aq)]

A weak acid is one which only partially dissociates in aqueous solution

Calculating pH - Calculating pH - weak acidsweak acids

A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)

Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)

[HA(aq)]

The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]

therefore Ka = [H+(aq)]2 (3)

[HA(aq)]

Rearranging (3) gives [H+(aq)]2

= [HA(aq)] Ka

therefore [H+(aq)] = [HA(aq)] Ka

A weak acid is one which only partially dissociates in aqueous solution

Calculating pH - Calculating pH - weak acidsweak acids

A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)

Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)

[HA(aq)]

The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]

therefore Ka = [H+(aq)]2 (3)

[HA(aq)]

Rearranging (3) gives [H+(aq)]2

= [HA(aq)] Ka

therefore [H+(aq)] = [HA(aq)] Ka

pH = [H+(aq)]

A weak acid is one which only partially dissociates in aqueous solution

Calculating pH - Calculating pH - weak acidsweak acids

A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)

Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)

[HA(aq)]

The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]

therefore Ka = [H+(aq)]2 (3)

[HA(aq)]

Rearranging (3) gives [H+(aq)]2

= [HA(aq)] Ka

therefore [H+(aq)] = [HA(aq)] Ka

pH = -log [H+(aq)]

ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration.

A weak acid is one which only partially dissociates in aqueous solution

Calculating pH - Calculating pH - weak acidsweak acids

Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )

HX dissociates as follows HX(aq) H+(aq) + X¯(aq)

WORKEDEXAMPLEWORKEDEXAMPLE

Calculating pH - Calculating pH - weak acidsweak acids

Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )

HX dissociates as follows HX(aq) H+(aq) + X¯(aq)

Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3

[HX(aq)]

WORKEDEXAMPLEWORKEDEXAMPLE

Calculating pH - Calculating pH - weak acidsweak acids

Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )

HX dissociates as follows HX(aq) H+(aq) + X¯(aq)

Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3

[HX(aq)]

Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3

equal amounts and then rearrange equation

WORKEDEXAMPLEWORKEDEXAMPLE

Calculating pH - Calculating pH - weak acidsweak acids

Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )

HX dissociates as follows HX(aq) H+(aq) + X¯(aq)

Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3

[HX(aq)]

Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3

equal amounts and the rearrange equation

ASSUMPTIONHA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration

WORKEDEXAMPLEWORKEDEXAMPLE

Calculating pH - Calculating pH - weak acidsweak acids

Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )

HX dissociates as follows HX(aq) H+(aq) + X¯(aq)

Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3

[HX(aq)]

Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3

equal amounts and the rearrange equation

ASSUMPTIONHA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration

[H+(aq)] = 0.1 x 4 x 10-5 mol dm-3

= 4.00 x 10-6 mol dm-3

= 2.00 x 10-3 mol dm-3

ANSWER pH = - log [H+(aq)] = 2.699

WORKEDEXAMPLEWORKEDEXAMPLE

CALCULATING THE pH OF MIXTURESCALCULATING THE pH OF MIXTURES

The method used to calculate the pH of a mixture of an acid and an alkali depends on...

• whether the acids and alkalis are STRONG or WEAK

• which substance is present in excess

STRONG ACID and STRONG BASE - EITHER IN EXCESSSTRONG ACID and STRONG BASE - EITHER IN EXCESS

WEAK ACID and EXCESS STRONG BASEWEAK ACID and EXCESS STRONG BASE

STRONG BASE and EXCESS WEAK ACIDSTRONG BASE and EXCESS WEAK ACID

pH of mixturespH of mixtures

Strong acids and strong alkalis (either in excess)Strong acids and strong alkalis (either in excess)

1. Calculate the initial number of moles of H+ and OH¯ ions in the solutions

2. As H+ and OH¯ ions react in a 1:1 ratio; calculate unreacted moles species in excess

3. Calculate the volume of solution by adding the two original volumes

4. Convert volume to dm3 (divide cm3 by 1000)

5. Divide moles by volume to find concentration of excess the ion in mol dm-3

6. Convert concentration to pH

If the excess is H+ pH = - log[H+]

If the excess is OH¯ pOH = - log[OH¯] then

pH + pOH = 14

or use Kw = [H+] [OH¯] = 1 x 10-14 at 25°C therefore

[H+] = Kw / [OH¯] then

pH = - log[H+]

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl

pH of mixturespH of mixtures

Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)

WORKEDEXAMPLEWORKEDEXAMPLE

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl

1. Calculate the number of moles of H+ and OH¯ ions present

pH of mixturespH of mixtures

Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)

25cm3 of

0.1M NaOH

20cm3 of

0.1M HCl

2.5 x 10-3 moles

2.0 x 10-3 moles

moles of OH ¯

= 0.1 x 25/1000= 2.5 x 10-3

moles of H+

= 20 x 20/1000= 2.0 x 10-3

WORKEDEXAMPLEWORKEDEXAMPLE

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species

pH of mixturespH of mixtures

Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)

The reaction taking place is… HCl + NaOH NaCl + H2O

or in its ionic form H+ + OH¯ H2O (1:1 molar ratio)

25cm3 of

0.1M NaOH

20cm3 of

0.1M HCl

2.5 x 10-3 moles

2.0 x 10-3 moles

WORKEDEXAMPLEWORKEDEXAMPLE

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species

pH of mixturespH of mixtures

Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)

The reaction taking place is… HCl + NaOH NaCl + H2O

or in its ionic form H+ + OH¯ H2O (1:1 molar ratio)

2.0 x 10-3 moles of H+ will react with the same number of moles of OH¯

this leaves 2.5 x 10-3 - 2.0 x 10-3 = 5.0 x 10-4 moles of OH¯ in excess

5.0 x 10-4

moles of OH¯

UNREACTED

25cm3 of

0.1M NaOH

20cm3 of

0.1M HCl

2.5 x 10-3 moles

2.0 x 10-3 moles

WORKEDEXAMPLEWORKEDEXAMPLE

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species

3. Calculate the volume of the solution by adding the two individual volumes

pH of mixturespH of mixtures

Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)

the volume of the solution is 25 + 20 = 45cm3

WORKEDEXAMPLEWORKEDEXAMPLE

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species

3. Calculate the volume of the solution by adding the two individual volumes

4. Convert volume to dm3 (divide cm3 by 1000)

pH of mixturespH of mixtures

Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)

the volume of the solution is 25 + 20 = 45cm3

there are 1000 cm3 in 1 dm3

volume = 45/1000 = 0.045dm3

WORKEDEXAMPLEWORKEDEXAMPLE

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species

3. Calculate the volume of the solution by adding the two individual volumes

4. Convert volume to dm3 (divide cm3 by 1000)

5. Divide moles by volume to find concentration of excess ion in mol dm-3

pH of mixturespH of mixtures

Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)

WORKEDEXAMPLEWORKEDEXAMPLE

[OH¯] = 5.0 x 10-4 / 0.045 = 1.11 x 10-2 mol dm-3

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species

3. Calculate the volume of the solution by adding the two individual volumes

4. Convert volume to dm3 (divide cm3 by 1000)

5. Divide moles by volume to find concentration of excess ion in mol dm-3

6. As the excess is OH¯ use pOH = - log[OH¯] then pH + pOH = 14

or Kw = [H+][OH¯] so [H+] = Kw / [OH¯]

pH of mixturespH of mixtures

Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)

Kw = 1 x 10-14

mol2 dm-6 (at 25°C)

Kw = 1 x 10-14

mol2 dm-6 (at 25°C)

WORKEDEXAMPLEWORKEDEXAMPLE

[OH¯] = 5.0 x 10-4 / 0.045 = 1.11 x 10-2 mol dm-3

[H+] = Kw / [OH¯] = 9.00 x 10-13 mol dm-3

pH = - log[H+] = 12.05

pH of mixturespH of mixtures

Weak acid and EXCESS strong alkaliWeak acid and EXCESS strong alkali

1. Calculate the initial number of moles of H+ and OH¯ ions in the solutions

2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of the excess OH¯

3. Calculate the volume of solution by adding the two original volumes

4. Convert volume to dm3 (divide cm3 by 1000)

5. Divide moles by volume to find concentration of excess OH¯ in mol dm-3

6. Convert concentration to pH

either using Kw = [H+] [OH¯] = 1 x 10-14 at 25°C therefore

[H+] = Kw / [OH¯] then

pH = - log[H+]

or pOH = - log[OH¯] and

pH + pOH = 14

pH of mixturespH of mixtures

Weak acid and EXCESS strong alkaliWeak acid and EXCESS strong alkali

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH

1. Calculate the number of moles of H+ and OH¯ ions present

25cm3 of

0.1M NaOH

2.5 x 10-3 moles

2.2 x 10-3 moles

22cm3 of 0.1M

CH3COOH

moles of OH ¯

= 0.1 x 25/1000= 2.5 x 10-3

moles of H+

= 22 x 20/1000= 2.2 x 10-3

WORKEDEXAMPLEWORKEDEXAMPLE

The reaction taking place is CH3COOH + NaOH CH3COONa + H2O

or in its ionic form H+ + OH¯ H2O (1:1 molar ratio)

2.2 x 10-3 moles of H+ will react with the same number of moles of OH¯

this leaves 2.5 x 10-3 - 2.2 x 10-3 = 3.0 x 10-4 moles of OH¯ in excess

pH of mixturespH of mixtures

Weak acid and EXCESS strong alkaliWeak acid and EXCESS strong alkali

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of excess OH¯

3.0 x 10-4

moles of OH¯

UNREACTED

25cm3 of

0.1M NaOH

2.5 x 10-3 moles

2.2 x 10-3 moles

22cm3 of 0.1M

CH3COOH

WORKEDEXAMPLEWORKEDEXAMPLE

pH of mixturespH of mixtures

Weak acid and EXCESS strong alkaliWeak acid and EXCESS strong alkali

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of excess OH¯

3. Calculate the volume of solution by adding the two individual volumes

the volume of the solution is 25 + 22 = 47cm3

WORKEDEXAMPLEWORKEDEXAMPLE

the volume of the solution is 25 + 22 = 47cm3

there are 1000 cm3 in 1 dm3

volume = 47/1000 = 0.047dm3

pH of mixturespH of mixtures

Weak acid and EXCESS strong alkaliWeak acid and EXCESS strong alkali

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of excess OH¯

3. Calculate the volume of solution by adding the two individual volumes

4. Convert volume to dm3 (divide cm3 by 1000)

WORKEDEXAMPLEWORKEDEXAMPLE

pH of mixturespH of mixtures

Weak acid and EXCESS strong alkaliWeak acid and EXCESS strong alkali

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of excess OH¯

3. Calculate the volume of solution by adding the two individual volumes

4. Convert volume to dm3 (divide cm3 by 1000)

5. Divide moles by volume to find concentration of excess ion in mol dm-3

the volume of the solution is 25 + 22 = 47cm3

there are 1000 cm3 in 1 dm3

volume = 47/1000 = 0.047dm3

[OH¯] = 3.0 x 10-4 / 0.047 = 6.38 x 10-3 mol dm-3

WORKEDEXAMPLEWORKEDEXAMPLE

Calculate the pH of a mixture of 25cm3 of 0.1M NaOH and 22cm3 of 0.1M CH3COOH

1. Calculate the number of moles of H+ and OH¯ ions present

2. As the ions react in a 1:1 ratio, calculate the unreacted moles of excess OH¯

3. Calculate the volume of solution by adding the two individual volumes

4. Convert volume to dm3 (divide cm3 by 1000)

5. Divide moles by volume to find concentration of excess ion in mol dm-3

6. As the excess is OH¯ use pOH = - log[OH¯] then pH + pOH = 14

or Kw = [H+][OH¯] so [H+] = Kw / [OH¯]

[OH¯] = 3x 10-4 / 0.045 = 6.38 x 10-3 mol dm-3

[H+] = Kw / [OH¯] = 1.57 x 10-12 mol dm-3

pH = - log[H+] = 11.8

pH of mixturespH of mixtures

Weak acid and EXCESS strong alkaliWeak acid and EXCESS strong alkali

WORKEDEXAMPLEWORKEDEXAMPLE

pH of mixturespH of mixtures

EXCESS Weak monoprotic acid and strong alkaliEXCESS Weak monoprotic acid and strong alkali

This method differs from the others because the excess substance is weak and as such is only PARTIALLY DISSOCIATED into ions. It is probably the hardest calculation to understand.

1. Calculate the initial number of moles of acid and OH¯ ions in the solutions

2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of the excess acid

3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed

4. Obtain the value of Ka for the weak acid and substitute the other values

5. Re-arrange the expression and calculate the value of [H+]

6. Convert concentration to pH using pH = - log[H+]

The following example shows you how to calculate the pH of the solutionproduced by adding 20cm3 of 0.1M NaOH to 25cm3 of 0.1M CH3COOH

pH of mixturespH of mixtures

EXCESS Weak monoprotic acid and strong alkaliEXCESS Weak monoprotic acid and strong alkali

1. Calculate the initial number of moles of acid and OH¯ ions in the solutions

20cm3 of

0.1M NaOH

25cm3 of 0.1M

CH3COOH

2.0 x 10-3 moles

2.5 x 10-3 moles

moles of OH ¯

= 0.1 x 20/1000= 2.0 x 10-3

moles of H+

= 25 x 20/1000= 2.5 x 10-3

WORKEDEXAMPLEWORKEDEXAMPLE

pH of mixturespH of mixtures

EXCESS Weak monoprotic acid and strong alkaliEXCESS Weak monoprotic acid and strong alkali

1. Calculate the initial number of moles of acid and OH¯ ions in the solutions

2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid

5.0 x 10-4

moles

20cm3 of

0.1M NaOH

2.0 x 10-3 moles

2.5 x 10-3 moles

unreacted

CH3COOH

2.0 x 10-3 moles of H+ will react with the same number of H+; this leaves

2.5 x 10-3 - 2.0 x 10-3 = 5.0 x 10-4 moles of CH3COOH in excess

The reaction taking place is CH3COOH + NaOH CH3COONa + H2O

WORKEDEXAMPLEWORKEDEXAMPLE

25cm3 of 0.1M

CH3COOH

pH of mixturespH of mixtures

EXCESS Weak monoprotic acid and strong alkaliEXCESS Weak monoprotic acid and strong alkali

1. Calculate the initial number of moles of acid and OH¯ ions in the solutions

2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid

3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed

2.0 x 10-3 moles of H+ will produce the same number of CH3COONa

this produces 2.0 x 10-3 moles of the anion CH3COO

The reaction taking place is CH3COOH + NaOH CH3COONa + H2O

2.0 x 10-3

moles

20cm3 of

0.1M NaOH

2.0 x 10-3 moles

2.5 x 10-3 moles

CH3COONa

produced

WORKEDEXAMPLEWORKEDEXAMPLE

25cm3 of 0.1M

CH3COOH

pH of mixturespH of mixtures

EXCESS Weak monoprotic acid and strong alkaliEXCESS Weak monoprotic acid and strong alkali

1. Calculate the initial number of moles of acid and OH¯ ions in the solutions

2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid

3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed

4. Obtain the value of Ka for the weak acid and substitute the other values

Ka = [H+(aq)] [CH3COO¯(aq)] mol dm-3

[CH3COOH(aq)]

Substitute the number of moles of unreacted acid here

Substitute the number of moles of anion produced here... it will be the same as the number of moles of H+ used up

Substitute

the Ka value

WORKEDEXAMPLEWORKEDEXAMPLE

1.7 x 10-5 = [H+(aq)] x (2 x 10-3) mol dm-3

(5 x 10-4)

pH of mixturespH of mixtures

EXCESS Weak monoprotic acid and strong alkaliEXCESS Weak monoprotic acid and strong alkali

1. Calculate the initial number of moles of acid and OH¯ ions in the solutions

2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid

3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed

4. Obtain the value of Ka for the weak acid and substitute the other values

Substitute the number of moles of unreacted acid here

Substitute the number of moles of anion produced here... it will be the same as the number of moles of H+ used up

Substitute

the Ka value

WORKEDEXAMPLEWORKEDEXAMPLE

pH of mixturespH of mixtures

EXCESS Weak monoprotic acid and strong alkaliEXCESS Weak monoprotic acid and strong alkali

1. Calculate the initial number of moles of acid and OH¯ ions in the solutions

2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid

3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed

4. Obtain the value of Ka for the weak acid and substitute the other values

5. Re-arrange the expression and calculate the value of [H+]

[H+(aq)] = 1.7 x 10-5 x 5 x 10-4 mol dm-3

2 x 10-3

= 4.25 x 10-6 mol dm-3

WORKEDEXAMPLEWORKEDEXAMPLE

pH of mixturespH of mixtures

EXCESS Weak monoprotic acid and strong alkaliEXCESS Weak monoprotic acid and strong alkali

1. Calculate the initial number of moles of acid and OH¯ ions in the solutions

2. As H+ and OH¯ ions react in a 1:1 ratio, calculate unreacted moles of excess acid

3. Calculate moles of salt anion formed; 1 mol of anion is formed for every H+ removed

4. Obtain the value of Ka for the weak acid and substitute the other values

5. Re-arrange the expression and calculate the value of [H+]

6. Convert concentration to pH using pH = - log[H+]

[H+(aq)] = 1.7 x 10-5 x 5 x 10-4 mol dm-3

2 x 10-3

= 4.25 x 10-6 mol dm-3

pH = - log10[H+(aq)] = 5.37

WORKEDEXAMPLEWORKEDEXAMPLE

REVISION CHECKREVISION CHECK

What should you be able to do?

Calculate pH from hydrogen ion concentration

Calculate hydrogen ion concentration from pH

Write equations to show the ionisation in strong and weak acids

Calculate the pH of strong acids and bases knowing their molar concentration

Calculate the pH of weak acids knowing their Ka and molar concentration

Calculate the pH of mixtures of acids and bases