PH calculations A guide for A level students KNOCKHARDY PUBLISHING

pH calculationspH calculationsA guide for A level studentsA guide for A level students

KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING

pH calculationspH calculations

INTRODUCTION

This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards.

Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available.

Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm

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CONTENTS

• What is pH? - a reminder

• Calculating the pH of strong acids and bases

• Calculating the pH of weak acids

• Calculating the pH of mixtures - strong acid and strong alkali

• Calculating the pH of mixtures - weak acid and excess strong alkali

• Calculating the pH of mixtures - strong alkali and excess weak acid

• Check list


Before you start it would be helpful to…

• know the differences between strong and weak acid and bases

• be able to calculate pH from hydrogen ion concentration

• be able to calculate hydrogen ion concentration from pH

• know the formula for the ionic product of water and its value at 25°C


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


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



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

[HA(aq)]





[HA(aq)]

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

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

[HA(aq)]





[HA(aq)]



[HA(aq)]

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

= [HA(aq)] Ka

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





[HA(aq)]



[HA(aq)]


= [HA(aq)] Ka


pH = [H+(aq)]





[HA(aq)]



[HA(aq)]


= [HA(aq)] Ka


pH = [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.



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)]






[HX(aq)]

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

equal amounts and then rearrange equation






[HX(aq)]


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






[HX(aq)]


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


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


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



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



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




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



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







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


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





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



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










there are 1000 cm3 in 1 dm3

volume = 45/1000 = 0.045dm3







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




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



Kw = 1 x 10-14

mol2 dm-6 (at 25°C)

Kw = 1 x 10-14

mol2 dm-6 (at 25°C)


[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


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


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



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


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


The reaction taking place is CH3COOH + NaOH CH3COONa + H2O


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





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







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





volume = 47/1000 = 0.047dm3



















volume = 47/1000 = 0.047dm3

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





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




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





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



25cm3 of 0.1M

CH3COOH






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


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


25cm3 of 0.1M

CH3COOH







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


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

(5 x 10-4)







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









[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









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


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

CAN YOU DO ALL OF THESE? CAN YOU DO ALL OF THESE? YES YES NONO

You need to go over the You need to go over the relevant topic(s) againrelevant topic(s) again

Click on the button toClick on the button toreturn to the menureturn to the menu

WELL DONE!WELL DONE!Try some past paper questionsTry some past paper questions


THE ENDTHE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING

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PH calculations A guide for A level students KNOCKHARDY PUBLISHING