ph calculations. what is ph? ph = - log 10 [h + (aq) ] where [h + ] is the concentration of hydrogen...
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![Page 1: PH calculations. What is pH? pH = - log 10 [H + (aq) ] where [H + ] is the concentration of hydrogen ions in mol dm -3 to convert pH into hydrogen ion](https://reader036.vdocuments.site/reader036/viewer/2022062407/56649e865503460f94b8936b/html5/thumbnails/1.jpg)
pH calculationspH calculations
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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)
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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
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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
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• 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
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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
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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
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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
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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
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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
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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
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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)
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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)]
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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
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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
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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
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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
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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+]
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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)
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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
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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
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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
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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
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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
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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
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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)
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[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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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