p H calculations p H calculations CONTENTS What

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p. H calculations

p. H calculations

p. H calculations CONTENTS • What is p. H? - a reminder • Calculating

p. H calculations CONTENTS • What is p. H? - a reminder • Calculating the p. H of strong acids and bases • Calculating the p. H of weak acids • Calculating the p. H of mixtures - strong acid and strong alkali

p. H calculations Before you start it would be helpful to… • know the

p. H calculations Before you start it would be helpful to… • know the differences between strong and weak acid and bases • be able to calculate p. H from hydrogen ion concentration • be able to calculate hydrogen ion concentration from p. H • know the formula for the ionic product of water and its value at 25°C

What is p. H? p. H = - log 10 [H+(aq)] where [H+] is

What is p. H? p. H = - log 10 [H+(aq)] where [H+] is the concentration of hydrogen ions in mol dm-3 to convert p. H into hydrogen ion concentration IONIC PRODUCT OF WATER [H+(aq)] = antilog (-p. H) Kw = [H+(aq)] [OH¯(aq)] mol 2 dm-6 = 1 x 10 -14 mol 2 dm-6 (at 25°C)

Calculating p. H - strong acids and alkalis WORKED EXAMPLE Strong acids and alkalis

Calculating p. H - strong acids and alkalis WORKED EXAMPLE Strong acids and alkalis completely dissociate in aqueous solution It is easy to calculate the p. H; you only need to know the concentration Calculate the p. H of 0. 02 M HCl completely dissociates in aqueous solution HCl H+ + Cl¯ One H+ is produced for each HCl dissociating so [H+] = 0. 02 M = 2 x 10 -2 mol dm-3 p. H = - log [H+] = 1. 7

Calculating p. H - strong acids and alkalis WORKED EXAMPLE Strong acids and alkalis

Calculating p. H - strong acids and alkalis WORKED EXAMPLE Strong acids and alkalis completely dissociate in aqueous solution It is easy to calculate the p. H; you only need to know the concentration Calculate the p. H of 0. 02 M HCl completely dissociates in aqueous solution HCl H+ + Cl¯ One H+ is produced for each HCl dissociating so [H+] = 0. 02 M = 2 x 10 -2 mol dm-3 p. H = - log [H+] = 1. 7 Calculate the p. H of 0. 1 M Na. OH completely dissociates in aqueous solution Na. OH One OH¯ is produced for each Na. OH dissociating [OH¯] = 0. 1 M = 1 x 10 -1 mol dm-3 Kw = [H+][OH¯] = 1 x 10 -14 mol 2 dm-6 The ionic product of water (at 25°C) therefore Na+ + OH¯ [H+] = Kw / [OH¯] = 1 x 10 -13 mol dm-3 p. H = - log [H+] = 13

Calculating p. H - weak acids A weak acid is one which only partially

Calculating p. H - weak acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq) (1)

Calculating p. H - weak acids A weak acid is one which only partially

Calculating p. H - weak acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as follows HA(aq) Applying the Equilibrium Law Ka = H+(aq) + A¯(aq) [H+(aq)] [A¯(aq)] [HA(aq)] mol dm-3 (1) (2)

Calculating p. H - weak acids A weak acid is one which only partially

Calculating p. H - weak acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as follows HA(aq) Applying the Equilibrium Law Ka = H+(aq) + A¯(aq) [H+(aq)] [A¯(aq)] mol dm-3 (1) (2) [HA(aq)] The ions are formed in equal amounts, so therefore [H+(aq)] = [A¯(aq)] Ka = [H+(aq)]2 [HA(aq)] (3)

Calculating p. H - weak acids A weak acid is one which only partially

Calculating p. H - weak acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as follows HA(aq) Applying the Equilibrium Law Ka = H+(aq) + A¯(aq) [H+(aq)] [A¯(aq)] mol dm-3 (1) (2) [HA(aq)] The ions are formed in equal amounts, so therefore [H+(aq)] = [A¯(aq)] Ka = [H+(aq)]2 [HA(aq)] Rearranging (3) gives therefore [H+(aq)]2 = [H+(aq)] = [HA(aq)] Ka (3)

Calculating p. H - weak acids A weak acid is one which only partially

Calculating p. H - weak acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as follows HA(aq) Applying the Equilibrium Law Ka = H+(aq) + A¯(aq) [H+(aq)] [A¯(aq)] mol dm-3 (1) (2) [HA(aq)] The ions are formed in equal amounts, so therefore [H+(aq)] = [A¯(aq)] Ka = [H+(aq)]2 [HA(aq)] Rearranging (3) gives therefore [H+(aq)]2 = [H+(aq)] = p. H [HA(aq)] Ka = [H+(aq)] (3)

Calculating p. H - weak acids A weak acid is one which only partially

Calculating p. H - weak acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as follows HA(aq) Applying the Equilibrium Law Ka = H+(aq) + A¯(aq) [H+(aq)] [A¯(aq)] mol dm-3 (1) (2) [HA(aq)] The ions are formed in equal amounts, so therefore [H+(aq)] = [A¯(aq)] Ka = [H+(aq)]2 (3) [HA(aq)] Rearranging (3) gives therefore [H+(aq)]2 = [H+(aq)] = p. H ASSUMPTION [HA(aq)] Ka = [H+(aq)] 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.

WORKED EXAMPLE Calculating p. H - weak acids Calculate the p. H of a

WORKED EXAMPLE Calculating p. H - weak acids Calculate the p. H of a weak acid HX of concentration 0. 1 M ( Ka = 4 x 10 -5 mol dm-3 ) HX dissociates as follows HX(aq) H+(aq) + X¯(aq)

WORKED EXAMPLE Calculating p. H - weak acids Calculate the p. H of a

WORKED EXAMPLE Calculating p. H - weak acids Calculate the p. H of a weak acid HX of concentration 0. 1 M ( Ka = 4 x 10 -5 mol dm-3 ) HX dissociates as follows Dissociation constant for a weak acid HX(aq) Ka H+(aq) = + X¯(aq) [H+(aq)] [X¯(aq)] [HX(aq)] mol dm-3

WORKED EXAMPLE Calculating p. H - weak acids Calculate the p. H of a

WORKED EXAMPLE Calculating p. H - weak acids Calculate the p. H of a weak acid HX of concentration 0. 1 M ( Ka = 4 x 10 -5 mol dm-3 ) HX dissociates as follows Dissociation constant for a weak acid HX(aq) Ka H+(aq) = + X¯(aq) [H+(aq)] [X¯(aq)] mol dm-3 [HX(aq)] Substitute for X¯ as ions are formed in equal amounts and then rearrange equation [H+(aq)] = [HX(aq)] Ka mol dm-3

WORKED EXAMPLE Calculating p. H - weak acids Calculate the p. H of a

WORKED EXAMPLE Calculating p. H - weak acids Calculate the p. H of a weak acid HX of concentration 0. 1 M ( Ka = 4 x 10 -5 mol dm-3 ) HX dissociates as follows Dissociation constant for a weak acid HX(aq) Ka H+(aq) = + X¯(aq) [H+(aq)] [X¯(aq)] mol dm-3 [HX(aq)] Substitute for X¯ as ions are formed in equal amounts and the rearrange equation [H+(aq)] = [HX(aq)] Ka mol dm-3 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

WORKED EXAMPLE Calculating p. H - weak acids Calculate the p. H of a

WORKED EXAMPLE Calculating p. H - weak acids Calculate the p. H of a weak acid HX of concentration 0. 1 M ( Ka = 4 x 10 -5 mol dm-3 ) HX dissociates as follows HX(aq) Dissociation constant for a weak acid Ka H+(aq) = + X¯(aq) [H+(aq)] [X¯(aq)] mol dm-3 [HX(aq)] Substitute for X¯ as ions are formed in equal amounts and the rearrange equation [H+(aq)] = [HX(aq)] Ka mol dm-3 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 [H+(aq)] ANSWER = 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 p. H = - log [H+(aq)] = 2. 699

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TITLE CONTENTS 1 CONTENTS 2 CONTENTS 3 CONTENTSTITLE CONTENTS 1 CONTENTS 2 CONTENTS 3 CONTENTS
CONTENTS 1 CONTENTS 2 CONTENTS 3 CONTENTS 4CONTENTS 1 CONTENTS 2 CONTENTS 3 CONTENTS 4
contents 1 contents 2 contents 3 contents 4contents 1 contents 2 contents 3 contents 4
Variables Constants and Calculations Calculations Although the calculationsVariables Constants and Calculations Calculations Although the calculations
CMS Contents Managements System 0 CONTENTS CMS ContentsCMS Contents Managements System 0 CONTENTS CMS Contents
CELL CONTENTS Cell Contents Cell contents with whichCELL CONTENTS Cell Contents Cell contents with which
Percentage Calculations Percentage Calculations Dont let percentages foolPercentage Calculations Percentage Calculations Dont let percentages fool
Mole Calculations 1 Chemical Calculations Atoms and moleculesMole Calculations 1 Chemical Calculations Atoms and molecules
Acids and Bases Jeopardy Acid Calculations Base CalculationsAcids and Bases Jeopardy Acid Calculations Base Calculations
Inverse Normal Calculations Inverse Normal Calculations Consider aInverse Normal Calculations Inverse Normal Calculations Consider a
Dry Cargo Calculations Dry Cargo Calculations It wouldDry Cargo Calculations Dry Cargo Calculations It would
Medication Calculations Randall Stennett RN MSN Medication CalculationsMedication Calculations Randall Stennett RN MSN Medication Calculations
LAY OUT CALCULATIONS TYPICAL YARD Why Layout CalculationsLAY OUT CALCULATIONS TYPICAL YARD Why Layout Calculations
Calculations with gases Calculations With Gases Objective toCalculations with gases Calculations With Gases Objective to
Charge calculations in pyrometallurgical processes Charge calculations areCharge calculations in pyrometallurgical processes Charge calculations are