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Prepared by Lawrence Kok

Video Tutorial on Bronsted Lowry Conjugate acid base pair, buffer and Ka calculation.

IB Chemistry Conjugate acid base pair, Buffer calculation, Henderson Hasselbalch equation

Acid / Base and its Conjugate

• Brønsted-Lowry of acid-base theory

• Acid (proton donor) lose H+ become conjugate base

• Base accept H+ become conjugate acid

• Removal of H+ from acid produce its conjugate base

• Acception of H+ by base produce its conjugate acid

Conjugate Acid Base pairWeak Acid produce Strong Conjugate Base    Weak Base produce Strong Conjugate Acid

• CH3COOH (acid) + H2O (base) ↔ CH3COO- (conjugate base) + Cl− (conjugate acid)

• NH3 (base) + H2O (acid) ↔  NH4+ (conjugate acid) + OH− (conjugate base)

Strong / Weak Acid and its Conjugate Base 

Strong Acid forms Weak Conjugate Base• Strong acid HCI dissociate completely form Cl−

(weak conjugate base)

• Cl− weak base won't accept H+ to form back HCI

• HCI → H+ + Cl−  ( one way )

Weak Acid form Strong Conjugate Base• CH3COOH weak acid dissociate partially form CH3COO- (strong conjugate base)

• CH3COO- (strong base) will accept H+ to form back CH3COOH molecule.

• CH3COOH ↔ CH3COO- + H+ (reversible)

Strong / Weak Base and its Conjugate Acid

Strong Base form Weak Conjugate acid

• Strong Base OH− accept H+ form H2O (weak conjugate acid)• H2O (weak acid) won't lose H+ again to form OH−

• OH−  → H2O (1 way)

Weak Base form Strong Conjugate Acid

• NH3 weak base dissociate partially form NH4+ (strong conjugate acid)

• NH4+ (strong acid) lose H+ to form back NH3 molecule

• NH3 + H2O ↔ NH4+ +OH− (reversible)

Acidic vs Basic Buffer

Acidic Buffer • Contain weak acid and its salt

• CH3COOH (weak acid) + CH3COONa (salt)

• Buffer need weak acid CH3COOH and base CH3COO- to neutralise added H+ or OH−

• CH3COOH ↔ CH3COO- +  H+ ionise partially producing H+ to neutralise OH−

• CH3COONa → CH3COO-  + Na+ ionise fully producing CH3COO- to neutralise H+

• Effective buffer will have equal amt of weak acid CH3COOH and base CH3COO-

Basic Buffer

Contain weak base and its salt• NH3 (weak base) + NH4CI (salt)

• NH3 (base) and NH4+ (acid) will neutralise added H+ or OH−

• NH3 + H2O ↔ NH4+  +  OH−, ionise partially producing NH3 molecule to neutralise

added H+

• NH4CI → NH4+ + CI−, ionise completely producing NH4

+ to neutralise OH−

 

Click HERE for more info

Buffer Calculation using Henderson-Hasselbalch equation 

Acidic Buffer CalculationCH3COOH → CH3COO-  +  H+

Ka =[CH3COO-][H+] /[ CH3COOH]

1st method using Ka[H+] = Ka [CH3COOH] / [ CH3COO-]pH = - log [H+]

2nd method using formula (picture )pH = pKa + log [CH3COO-] / [CH3COOH] pH = pKa - log [CH3COOH] / [CH3COO-]Assume CH3COOH  from weak acid and CH3COO- from ethanoate salt

Acidic Buffer Calculation

• Find, pH of a buffer containing (0.20M Ethanoic acid + 0.05M Sodium ethanoate)CH3COOH ↔ CH3COO-  +  H+                (Ka = 1.74 x 10-5, pKa = 4.76 )Ka =[CH3COO-] [H+] / [ CH3COOH]

1st method using Ka[H+] = Ka [CH3COOH] / [CH3COO-][H+] = 1.74 x 10-5 x 0.20 / 0.05[H+] = 6.96 x 10-5

pH = -log [H+] = - log [6.96 x 10-5] pH = 4.16

2nd method using formulapH = pKa + log [CH3COO- ] / [CH3COOH] pH = 4.76 + log 0.05 / 0.20pH = 4.76 + log 0.25pH = 4.16

Assume all acid from CH3COOH and all CH3COO- from Ethanoate salt

Calculate Conc of sodium propanoate salt, used to make 1dm3 buffer of pH 4.50 containing 1.00M propanoic acid   (Ka = 1.35 x 10-5, pKa = 4.87)

C2H5COOH ↔  C2H5COO- + H+

Ka = [C2H5COO- ][H+] / [C2H5COOH ]

1st method using Ka[C2H5COO- ] = Ka [C2H5COOH ] / [ H+]       pH = -log [H+], 4.5 = -log[H+],  [H+]= 3.16 x 10-5

[C2H5COO-] = 1.35 x 10-5 x 1.00 /  3.16 x 10-5

[C2H5COO-] = 0.427M

2nd method using formulapH = pKa + log [C2H5COO-] / [C2H5COOH] 4.50 = 4.87 + log [C2H5COO-] / 1.00[C2H5COO-] = 0.427M

Assume all acid from C2H5COOH and all C2H5COO- from propanoate salt

Basic buffer calculation

NH3 + H2O ↔ NH4+  + OH−

Kb = [ NH4+ ][OH−] / [NH3 ]

1st method using Kb[OH−]  = Kb [ NH3 ]  / [ NH4

+ ]pOH = -log [OH−]

2nd method using formula (picture)pOH = pKb + log [ NH4

+ ] / [ NH3 ]

Calculate Kb for NH3 buffer of pH 9.3 containing 0.10M (NH3 + NH4CI )

NH3 + H2O ↔ NH4+  +  OH−

1st method using KbKb = [NH4

+] [OH−] / [NH3 ]    pOH = 14 - 9.3 = 4.7,  pOH = -log[OH] , [OH] = 10-4.7

Kb = 0.10 x 10-4.7 / 0.10Kb = 2.00 x 10-5

2nd method using formulapOH = pKb + log [ NH4

+ ] / [ NH3 ]pKb = pOH - log [ NH4

+ ] / [ NH3 ]pKb = 4.7 - log [ 0.10] / [ 0.10 ]pKb = 4.7, Kb = 10-4.7

Kb = 2.00 x 10-5

HCI  +  NH3  ↔   NH4CI      (Kb = 1.78 x 10-5 , pKa = 4.75)

Moles acid HCI  = MV = 18.0 x 0.10 /1000 = 1.80 x 10-3mol

Moles base NH3 = MV = 32.0 x 0.10 /1000 = 3.20 x 10-3 mol

1.80 x 10-3 mol acid reacts with 1.80 x 10-3 mol base forming 1.80 x 10-3 mol salt (NH4CI)

Amt base left = 3.2 x 10-3 - 1.80 x 10-3 = 1.60 x 10-3 mol       Amt of salt = 1.80 x 10-3 mol 

Conc of sol = Moles / Total Vol        Total vol = 0.05 dm3

Conc base = 1.60 x 10-3 / 0.05 dm3 = 3.20 x 10-2

Conc salt = 1.80 x 10-3/ 0.05 dm3 = 3.60 x 10-2

Calculate pH buffer when 18.0ml, 0.10M HCI added to 32.0ml, 0.10M NH3

2nd Method using formula

pOH = pKb + log [ NH4+ ] / [ NH3 ]

pOH = 4.75 + log [3.60 x 10-2] / [3.20 x 10-2] pOH = 4.80         pH + pOH = 14,         pH = 14 - 4.80 = 9.20    pH = 9.20

Video Tutorial on Buffer calculation. Click HERE to view

Key notes from tutorial on calculation

•Calculate pH buffer containing 0.10M (ethanoic acid + ethanoate salt)  and what is new pH after addition of 10ml 1.0M HCI to 1dm3 of buffer solution.   (pKa = 4.75)

CH3COOH ↔ CH3COO-  +  H+

Ka =[CH3COO-] [H+] /[ CH3COOH]

•Most effective buffer occurs when ratio of acid to salt is same, 1 : 1

•pH = pKa + log [CH3COO- ] / [CH3COOH] 

•pH = 4.75 + log [0.10] / [0.10]

•pH = 4.75

Initial pH of 0.10M (ethanoic acid + ethanoate salt) is 4.75 New pH after addition of 10ml 1.0M HCI.

Moles HCI added = 10 x 1.0 / 1000 = 0.01 mol HCI  +  CH3COO-    ↔  CH3COOH 0.01      0.01                    0.01

Amt of CH3COOH left = 0.10 + 0.01 = 0.11Amt of CH3COO- left = 0.10 - 0.01 = 0.09

Conc of CH3COOH = Mol / Total vol = 0.11 / 1010 = 1.09 x 10-4

Conc of CH3COO- = Mol / Total vol = 0.09 / 1010 = 8.91 x 10-5

2nd method using formulapH = pKa + log [CH3COO- ] / [CH3COOH] pH = 4.75 + log [8.91 x 10-5] / [1.09 x 10-4] pH = 4.75 + ( -0.087), pH  = 4.66Buffer resist a change in pH, pH drop only from 4.75 to 4.66 by only 0.09 unit!!

Acknowledgements

Thanks to source of pictures and video used in this presentation

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Prepared by Lawrence Kok

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