ib chemistry on acids bases, conjugate acid base pair, buffer and ka calculation
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IB Chemistry on Acids Bases, Conjugate Acid Base Pair, Buffer and Ka calculationTRANSCRIPT
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http://lawrencekok.blogspot.com
Prepared by Lawrence Kok
Video Tutorial on Bronsted Lowry Conjugate acid base pair, buffer and Ka calculation.
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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
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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)
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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)
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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)
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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-
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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
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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
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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
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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
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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 ]
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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
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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
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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
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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!!
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Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com