IB Chemistry on Acidic, Basic Buffer and pH Calculation
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Transcript of IB Chemistry on Acidic, Basic Buffer and pH Calculation
Acidic Buffer Solution
Acidic Buffer • Resist a change in pH when acid/base is added. • Mixing weak acid + salt/conjugate base • CH3COOH ↔ CH3COO- + H+ (dissociate partially)
• CH3COONa → CH3COO- + Na+ (dissociate fully)
Acidic buffer
CH3COOH ↔ CH3COO- +H+
Click here on acidic buffer simulation
Click here on acidic buffer simulation
Base OH- added Acid H+ added
Acid part
Neutralize each other
CH3COONa → CH3COO- + Na+
Neutralize each other
Acidic Buffer - contain weak acid and its salt • CH3COOH (weak acid) + CH3COONa (salt) • CH3COOH (weak acid) + CH3COO- (base) to neutralise added H+ or OH− • CH3COOH ↔ CH3COO- + H+ → H+ to neutralise added OH− • CH3COONa → CH3COO- + Na+ → CH3COO- to neutralise added H+ • Effective buffer have equal amt of weak acid CH3COOH and base CH3COO-
Salt part
Basic Buffer • Resist a change in pH when acid/base is added. • Mixing weak base + salt/conjugate acid • NH3 + H2O ↔ NH4
+ + OH_ (dissociate partially)
• NH4CI → NH4+ + CI_ (dissociate fully)
Basic Buffer Solution
Click here on basic buffer simulation
Base part Salt part
Base OH- added Acid H+ added
Contain weak base and its salt • NH3(base) + NH4CI (salt) • NH3(base) + NH4
+ (acid) will neutralise added H+ or OH− • NH3 + H2O ↔ NH4
+ + OH− → NH3 molecule to neutralise added H+ • NH4CI → NH4
+ + CI− → NH4+ to neutralise added OH−
• Effective buffer have equal amt of weak base NH3 and conjugate acid NH4+
Neutralize each other Neutralize each other
Base part Salt part
Basic buffer
NH3 + H2O ↔ NH4+ + OH_ NH4CI → NH4
+ + CI_
How to prepare acidic/ basic buffer
Acid Dissociation constant CH3COOH + H2O ↔ CH3COO- + H3O+
Ka = (CH3COO-) (H3O+) (CH3COOH) -lgKa = -lgH+ -lg (CH3COO-) (CH3COOH) -lgH+ = -lg Ka + lg (CH3COO-) (CH3COOH) pH = pKa + lg (CH3COO-) (CH3COOH)
Acidic Buffer Formula • Mixture Weak acid + Salt/Conjugate base • CH3COOH ↔ CH3COO- + H+ (dissociate partially)
• CH3COONa → CH3COO- + Na+ (dissociate fully)
Basic Buffer Formula • Mixture Weak base + Salt/Conjugate acid • NH3 + H2O ↔ NH4
+ + OH_ (dissociate partially)
• NH4CI → NH4+ + CI_ (dissociate fully)
pH = pKa - lg (acid) (salt)
pH = pKa + lg (salt) (acid)
Base Dissociation constant NH3 + H2O ↔ NH4
+ + OH-
Kb = (NH4+) (OH-)
(NH3) -lgKb = -lgOH- -lg (NH4
+) (NH3) -lgOH- = -lgKb + lg (NH4
+) (NH3) pOH = pKb + lg (NH4
+) (NH3)
pOH = pKb + lg (salt) (base)
pOH = pKb - lg (base) (salt)
Basic Buffer Acidic Buffer
salt salt
acid base
Henderson Hasselbalch Equation Henderson Hasselbalch Equation
multiply -lg both sides
Acidic Buffer Calculation
Find pH buffer - 0.20 mol CH3COONa(salt) add to 0.5dm3, 0.10M CH3COOH(acid) Ka = 1.8 x 10-5M
CH3COOH ↔ CH3COO- + H+
Ka = (CH3COO-)(H+) (CH3COOH) 1.8 x 10-5 = 0.40 x (H+) (0.10) H+ = 4.5 x 10-6
pH = -lg H+ pH = -lg(4.5 x 10-6) pH = 5.34
pH = pKa -lg(acid) (salt) pH = 4.74 – lg (0.10) (0.40) pH = 5.34
Conc CH3COO- = Moles/volume = 0.20/0.5 = 0.40M
Ka = (1.8 x 10-5) pKa = -lgKa
pKa = -lg(1.8 x 10-5 ) pKa = 4.74
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Find conc of CH3CH2COONa(salt) added to 1.0dm3 of 1.0M CH3CH2COOH(acid) Ka = 1.8 x 10-5M, pKa = 4.74 , pH 4.5
pH = pKa -lg (acid) (salt) 4.5 = 4.74 – lg (1.0) (salt) lg (1.0) = 0.24 (salt) Conc (salt) = 0.0578M
CH3CH2COOH ↔ CH3CH2COO- + H+
Ka = (CH3CH2COO-)(H+) (CH3CH2COOH) 1.8 x 10-5 = CH3CH2COO- x (3.16 x 10-5) (1.0) CH3CH2COO- = 0.0578M
pH = -lg(H+) 4.5 = -lg(H+) H+ = 3.16 x 10-5
CH3COOH ↔ CH3COO- + H+
Ka = (CH3COO-)(H+) (CH3COOH) 1.8 x 10-5 = 0.25 x (H+) 0.10 H+ = 7.2 x 10-6
pH = -lg H+ pH = -lg(7.2 x 10-6) pH = 5.14
pH = pKa -lg(acid) (salt) pH = 4.74 – lg (0.10) (0.25) pH = 5.14
Find pH buffer - 0.10M CH3COOH(acid), 0.25M CH3COONa(salt) Ka = 1.8 x 10-5M
1st method (formula)
1
Convert Ka to pKa
Ka = (1.8 x 10-5) pKa = -lgKa
pKa = -lg(1.8 x 10-5 ) pKa = 4.74
2nd method (Ka)
2
1st method (formula) Convert Ka to pKa
2nd method (Ka)
3
1st method (formula)
Find conc salt
2nd method (Ka)
Click here for detail explanation from chem guide
Find pH buffer - 0.50M NH3 (base), 0.32M NH4CI (salt) Kb = 1.8 x 10-5M
Basic Buffer Calculation
Find pH buffer - 4.28g NH4CI (salt) add to 0.25dm3, 0.50NH3(base) Kb = 1.8 x 10-5M
Moles, NH4CI = mass/RMM = 4.28 / 53.5 = 0.08mol
Conc of NH4CI = moles/vol = 0.08/0.25 = 0.32M
NH3 + H2O ↔ NH4+ + OH-
Kb = (NH4+) (OH-)
(NH3) 1.8 x 10-5 = 0.32 x OH- 0.50 OH- = 0.50 x 1.8 x 10-5 0.32 OH- = 2.81 x 10-5
pOH = pKb -lg(base) (salt) pOH = 4.74 – lg(0.50) (0.32) pOH = 4.55 pH + pOH = 14 pH = 9.45
1
1st method (formula) 2nd method (Kb)
pOH = -lgOH-
pOH = -lg 2.81 x 10-5 pOH = 4.55 pH + pOH = 14 pH = 9.45
1st method (formula)
pOH = pKb -lg(base) (salt) pOH = 4.74 – lg(0.50) (0.32) pOH = 4.55 pH + pOH = 14 pH = 9.45
NH3 + H2O ↔ NH4+ + OH-
Kb = (NH4+) (OH-)
(NH3) 1.8 x 10-5 = 0.32 x OH- 0.50 OH- = 0.50 x 1.8 x 10-5 0.32 OH- = 2.81 x 10-5
2
pOH = -lgOH-
pOH = -lg 2.81 x 10-5 pOH = 4.55 pH + pOH = 14 pH = 9.45
2nd method (Kb) Find conc salt
Find mass of CH3COONa added to 500ml, 0.10M CH3COOH(acid) pH = 4.5, Ka = 1.8 x 10-5M, pKa = 4.74
pH = pKa -lg (acid) (salt) 4.5 = 4.74 – lg (0.10) (salt) lg(0.10) = 4.74 -4.5 (salt) lg (0.10) = 0.24 (salt) Conc salt = 0.0578M
CH3COOH ↔ CH3COO- + H+
Ka = (CH3COO-) (H+) (CH3COOH) 1.8 x 10-5 = CH3COO- x (10-4.5) (0.10) CH3COO- = 0.10 x 1.8 x 10-5 10-4.5 Conc salt = 0.0578M
pH = -lg H+ 4.5 = -lg(H+) H+ = 10-4.5
Conc salt = 0.0578M → x RMM (82) → 4.74g in 1000ml 2.37g in 500ml
3
2nd method (Ka) 1st method (formula)
Click here addition base to buffer
Click here acidic buffer simulation
Click here addition acid to buffer
Basic Buffer Preparation Acidic Buffer Preparation
Prepare Acidic Buffer at pH = 5.2 • Choose pKa acid closest to pH 5.2 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 5.2 = 4.74 – lg [acid] [salt] • [acid] = 0.35 [salt] Ratio of [acid] = 0.35 [salt]
Use same conc of acid/salt but different vol ratio • 1M, 35ml (acid ) = 0.35 or 0.1M, 35ml (acid) = 0.35 1M, 100ml (salt) 0.1M, 100ml (salt)
Use same vol of acid/salt but different conc ratio • 3.5M, 10ml (acid ) = 0.35 or 0.35M, 10ml (acid) = 0.35 10.0M, 10ml (salt) 1.00M, 10ml (salt)
Buffer capacity • Adding water will not change the pH of acidic buffer • Ratio of acid/salt still the same
Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 • Choose pKb base closest to pOH = 4.5 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.5 = 4.74 – lg [base] [salt] • [base] = 1.74 [salt] Ratio of [base] = 1.74 [salt]
Use same conc of base/salt but different vol ratio • 1M, 174ml (base) = 1.74 or 0.1M, 174ml (base) = 1.74 1M, 100ml (salt) 0.1M, 100ml (salt)
Use same vol of base/salt but different conc ratio • 1.74M, 10ml (base) = 1.74 or 0.174M, 10ml (base) = 1.74 1.00M, 10ml (salt) 0.10M, 10ml (salt)
Buffer capacity • Adding water will not change the pH of basic buffer • Ratio of base/salt still the same
Buffer solution
Buffer Preparation
http://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/
1 1
2 2
3 Use fix vol, 1dm3 and use different mole ratio (Acid/salt) • 0.35 mole acid + 1 mole salt to 1 dm3 solvent = 0.35 • 0.035 mole acid + 0.1 mole salt to 1 dm3 solvent = 0.35
Use fix vol, 1dm3 and use different mole ratio (base/salt) • 1.74 mole base + 1 mole salt to 1 dm3 solvent = 1.74 • 0.174 mole base + 0.1 mole salt to 1 dm3 solvent = 1.74
3
3 ways to prepare buffer 3 ways to prepare buffer
Basic Buffer Preparation Acidic Buffer Preparation
Prepare Acidic Buffer at pH = 5.2 • Choose pKa acid closest to pH 5.2 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 5.2 = 4.74 – lg [acid] [salt] • [acid] = 0.35 [salt] Ratio of [acid] = 0.35 [salt]
Use same conc of acid/salt but different vol ratio Buffer A Buffer B • 1M, 35ml (acid ) = 0.35 or 0.1M, 35ml (acid) = 0.35 1M, 100ml (salt) 0.1M, 100ml (salt)
Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 • Choose pKb base closest to pOH = 4.5 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.5 = 4.74 – lg [base] [salt] • [base] = 1.74 [salt] Ratio of [base] = 1.74 [salt]
Use same conc of base/salt but different vol ratio Buffer A Buffer B • 1M, 174ml (base) = 1.74 or 0.1M, 174ml (base) = 1.74 1M, 100ml (salt) 0.1M, 100ml (salt)
Buffer solution
Buffering Capacity
http://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/
1 1
1M, 35ml
(acid ) 1M, 100ml
(salt )
0.1M, 35ml
(acid )
0.1M, 100ml
(salt)
B A
1M, 174ml
(base)
1M, 100ml
(salt)
0.1M, 174ml (base)
0.1M, 100ml (salt)
B A
Buffer A > Buffer B Stronger buffering capacity • Amt of acid/salt is higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff • Higher buffer conc – Higher buffering capacity
Buffer A > Buffer B Stronger buffering capacity • Amt of base/salt is higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff • Higher buffer conc – Higher buffering capacity
Which has greater buffering capacity ? Which has greater buffering capacity ?
Basic Buffer Preparation Acidic Buffer Preparation
Prepare Acidic Buffer at pH = 5.2 • Choose pKa acid closest to pH 5.2 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 5.2 = 4.74 – lg [acid] [salt] • [acid] = 0.35 [salt] Ratio of [acid] = 0.35 [salt]
Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 • Choose pKb base closest to pOH = 4.5 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.5 = 4.74 – lg [base] [salt] • [base] = 1.74 [salt] Ratio of [base] = 1.74 [salt]
Buffer solution
Buffering Capacity
http://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/
2 2
Buffer A > Buffer B Stronger buffering capacity • Amt of acid/salt is higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff • Higher buffer conc – Higher buffering capacity
3.5M, 10ml
(acid )
10M, 10ml
(salt ) 0.35M, 10ml
(acid )
1M, 10ml
(salt)
B A
1.74M, 10ml
(base)
1M, 10ml
(salt)
0.174M, 10ml (base)
0.1M, 10ml (salt)
B A
Use same vol of acid/salt but different conc ratio Buffer A Buffer B • 3.5M, 10ml (acid ) = 0.35 or 0.35M, 10ml (acid) = 0.35 10M, 10ml (salt) 1.00M, 10ml (salt)
Buffer A > Buffer B Stronger buffering capacity • Amt of base/salt is higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff • Higher buffer conc – Higher buffering capacity
Use same vol of base/salt but different conc ratio Buffer A Buffer B • 1.74M, 10ml (base) = 1.74 or 0.174M, 10ml (base) = 1.74 1.00M, 10ml (salt) 0.10M, 10ml (salt)
Which has greater buffering capacity ? Which has greater buffering capacity ?
Basic Buffer Preparation Acidic Buffer Preparation
Prepare Acidic Buffer at pH = 5.2 • Choose pKa acid closest to pH 5.2 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 5.2 = 4.74 – lg [acid] [salt] • [acid] = 0.35 [salt] Ratio of [acid] = 0.35 [salt]
Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 • Choose pKb base closest to pOH = 4.5 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.5 = 4.74 – lg [base] [salt] • [base] = 1.74 [salt] Ratio of [base] = 1.74 [salt]
Buffer solution
Buffering Capacity
http://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/
3 3
Buffer A > Buffer B Stronger buffering capacity • Amt of acid/salt is higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff • Higher buffer conc – Higher buffering capacity
0.35mol
(acid )
1.00mol
(salt ) 0.035mol
(acid )
0.10mol
(salt)
B A
1.74mol
(base)
1.00mol
(salt)
0.174mol (base)
0.10mol (salt)
B A
Buffer A > Buffer B Stronger buffering capacity • Amt of base/salt is higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff • Higher buffer conc – Higher buffering capacity
Use fix vol, 1dm3 but diff mole ratio (acid/salt) Buffer A Buffer B • 0.35mol (acid ) = 0.35 or 0.035mol (acid) = 0.35 1.00mol (salt) 0.100mol (salt)
1dm3 1dm3 1dm3 1dm3
Use fix vol, 1dm3 but diff mole ratio (base/salt) Buffer A Buffer B • 1.74mol (base) = 1.74 or 0.174mol (base) = 1.74 1.00mol (salt) 0.100mol (salt)
Which has greater buffering capacity ? Which has greater buffering capacity ?
Basic Buffer Preparation Acidic Buffer Preparation
Prepare Acidic Buffer at pH = 5.2 • Choose pKa acid closest to pH 5.2 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 5.2 = 4.74 – lg [acid] [salt] • [acid] = 0.35 [salt] Ratio of [acid] = 0.35 [salt]
Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 • Choose pKb base closest to pOH = 4.5 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.5 = 4.74 – lg [base] [salt] • [base] = 1.74 [salt] Ratio of [base] = 1.74 [salt]
Buffer solution
Buffering Capacity
http://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/
4 4
Will pH change
pH Buffer A = pH Buffer B • Same pH • Adding water will not change the pH • Amt of acid/salt still the same • Ratio conc acid/salt same, pH buffer same
0.35mol
(acid )
1.00mol
(salt )
0.35mol
(acid )
1.00mol
(salt)
B A
1.74mol
(base)
1.00mol
(salt)
1.74mol (base)
1.00mol (salt)
B A
Same mole ratio (acid/salt) but different total volume Buffer A Buffer B • 0.35mol (acid)= 0.35 in 1dm3 or 0.35mol (acid) = 0.35 in 2dm3
1.00mol (salt) 1.00mol (salt)
1dm3
2dm3 1dm3
Will pH change
Same mole ratio (base/salt) but different total volume Buffer A Buffer B • 1.74mol (base)= 1.74 in 1dm3 or 1.74mol (base) = 1.74 in 2dm3
1.00mol (salt) 1.00mol (salt)
2dm3
pH Buffer A = pH Buffer B • Same pH • Adding water will not change the pH • Amt of acid/salt still the same • Ratio conc base/salt same, pH buffer same
Acidic Buffer Preparation Acidic Buffer Preparation
Prepare Acidic Buffer at pH = 4.74 • Choose pKa acid closest to pH 4.74 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 4.74 = 4.74 – lg [acid] [salt] • [acid] = 1.00 [salt] Ratio of [acid] = 1.00 [salt]
Buffer solution
Buffering Capacity
http://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/
5 5
Which has greater buffering capacity ?
Buffer A > Buffer B • Conc ratio [acid]/[salt] equal to 1 •Buffer has highest buffering capacity when pH = pKa • Conc acid = Conc salt → highest buffering capacity
Concentration ratio [acid]/[salt] = 1
1.00mol
(acid)
1.00mol
(salt)
A
1.00mol
(salt)
B
Buffer A > Buffer B • Further the conc ratio [acid]/[salt] from 1, lower buffer capacity
Same conc ratio (acid/salt) in 1dm3 Buffer A • 1.00mol (acid ) = 1.00 1.00mol (salt)
1dm3 1dm3
Prepare Acidic Buffer at pH = 5.2 • Choose pKa acid closest to pH 5.2 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 5.2 = 4.74 – lg [acid] [salt] • [acid] = 0.35 [salt] Ratio of [acid] = 0.35 [salt]
Different conc ratio (acid/salt) in 1dm3 Buffer B • 0.35mol (acid ) = 0.35 1.00mol (salt)
Which has greater buffering capacity ?
0.35mol
(acid)
Concentration ratio [acid]/[salt] ratio < 1
Acidic Buffer preparation CH3COOH (acid)/ CH3COO-(salt)
pH = pKa -lg (acid) (salt)
CH3COOH (acid) ↔ CH3COO- + H+
CH3COONa (salt) → CH3COO- + Na+
pH acidic buffer depend on Ka and ratio [acid]/[salt]
2 ways to prepare buffer • 1st – calculate ratio of [acid]/[salt] • 2nd – titration weak acid with strong base 1 Prepare Acidic Buffer at pH = 5.2
• Choose pKa acid closest to pH 5.2 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 5.2 = 4.74 – lg [acid] [salt] • [acid] = 0.35 [salt] Ratio of [acid] = 0.35 [salt]
Use same conc of acid/salt but diff vol ratio • 1M, 35ml (acid ) = 0.35 1M, 100ml (salt)
Use same vol of acid/salt but diff conc ratio • 0.35M, 10ml (acid ) = 0.35 1.00M, 10ml (salt)
1M, 35ml
(acid )
1M, 100ml
(salt )
0.35M, 10ml
(acid )
1M, 10ml
(salt )
Titration bet strong base with weak acid CH3COOH + NaOH → CH3COONa + H2O
2
Click here buffer simulation
Strong base NaOH
Weak acid CH3COOH
CH3COOH + NaOH → CH3COONa + H2O
Initial 1 mol 0.5 mol added 0.
Change (1 – 0.5)mol 0 mol 0.5mol form
Final 0.5mol left 0 mol 0.5mol form
At half equivalent point : • Amt acid = Amt salt : ( 0.5 = 0.5) • pH = pKa -lg [acid] → pH = pKa → 4.74 = 4.74
[salt] Buffer at pH = 4.74 form when half amt of acid neutralise by base or at half equivalent point when amt acid = amt salt
Prepare Acidic Buffer at pH = 4.74 • Choose pKa acid closest to pH 4.74 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 4.74 = 4.74 – lg [acid] [salt] • [acid] = 1.00 (amt acid = amt salt) [salt]
Same conc Diff Volume
Same Volume Diff Conc
Same conc ratio Same conc ratio
1mol
0.5mol
Buffer region at half equivalent point
Amt acid = Amt salt
Acidic Buffer preparation CH3COOH (acid)/ CH3COO-(salt)
Preparation acidic buffer by titration • 1st – weak acid (burette) with strong base (flask) 2nd – strong base (burette) with weak acid (flask)
1 Titration bet strong base + weak acid CH3COOH + NaOH → CH3COONa + H2O
2
Click here buffer simulation
Strong base NaOH
Weak acid CH3COOH
CH3COOH + NaOH → CH3COONa + H2O
Initial 1 mol 0.5 mol added 0.
Change (1 – 0.5)mol 0 mol 0.5mol form
Final 0.5mol left 0 mol 0.5mol form
At half equivalent point : • Amt acid = Amt salt : ( 0.5 = 0.5) • pH = pKa -lg [acid] → pH = pKa → 4.74 = 4.74
[salt] Buffer at pH = 4.74 form when half amt of acid neutralise by base or at half equivalent point when amt acid = amt salt
Prepare Acidic Buffer at pH = 4.74 • Choose pKa acid closest to pH 4.74 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 4.74 = 4.74 – lg [acid] [salt] • [acid] = 1.00 (amt acid = amt salt) [salt]
Buffer region at half equivalent point
Amt acid = Amt salt
Strong base NaOH
Weak acid CH3COOH
Click here buffer simulation
Titration bet weak acid + strong base CH3COOH + NaOH → CH3COONa + H2O
Prepare Acidic Buffer at pH = 4.74 • Choose pKa acid closest to pH 4.74 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 4.74 = 4.74 – lg [acid] [salt] • [acid] = 1.00 (amt acid = amt salt) [salt]
Buffer region Amt acid = Amt salt
CH3COOH + NaOH → CH3COONa + H2O
Initial 2mol added 1.0 mol 0.
Change (2 – 1)mol 0 mol 1mol form
Final 1mol left (excess) 0 mol 1mol form
When excess 1 mol acid added: • Amt acid = Amt salt : ( 1 = 1) • pH = pKa -lg [acid] → pH = pKa → 4.74 = 4.74
[salt] Buffer at pH = 4.74 form when amt of acid added = amt salt form at equivalent point
1mol
0.5mol
1mol
2mol
Basic Buffer preparation NH3(base)/ NH4CI(salt)
pOH = pKb -lg (base) (salt)
pH basic buffer depend on Kb and ratio [base]/[salt] • NH3 (base) + H2O ↔ NH4
+ + OH_
• NH4CI (salt) → NH4+ + CI_
2 ways to prepare buffer • 1st – calculate ratio of [base]/[salt] • 2nd – titration weak base with strong acid
Prepare Buffer pH = 9.5 /pOH = 4.5 • Choose pKb base closest to pOH = 4.5 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.5 = 4.74 – lg [base] [salt] • [base] = 1.74 [salt] Ratio of [base] = 1.74 [salt]
Use same conc of base/salt but diff vol ratio • 1M, 174ml (base) = 1.74 1M, 100ml (salt)
Use same vol of base/salt but diff conc ratio • 1.74M, 10ml (base) = 1.74 1.00M, 10ml (salt)
1
1M, 100ml
(salt )
1M, 174ml
(base )
1M, 10ml
(salt )
1.74M, 10ml
(base )
2 Titration bet strong acid with weak base NH3 + HCI → NH4CI + H2O
Strong acid HCI
NH3 + HCI → NH4CI + H2O
Initial 1 mol 0.5 mol added 0.
Change (1 – 0.5)mol 0 mol 0.5mol form
Final 0.5mol left 0 mol 0.5mol form
At half equivalent point : • Amt base = Amt salt : (0.5 = 0.5) • pOH = pKb - lg [base] → pOH = pKb → 4.74 = 4.74 [salt] Buffer at pOH = 4.74 form when half amt of base neutralise by acid or at half equivalent point when amt base = amt salt
Weak base NH3
Click here buffer simulation
Prepare Buffer pH = 9.26 /pOH = 4.74 • Choose pKb base closest to pOH = 4.74 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.74 = 4.74 – lg [base] [salt] • [base] = 1.00 (amt base = amt salt) [salt]
Same conc Diff Volume
Same Volume Diff Conc
Buffer region at half equivalent point
Amt base = Amt salt
1mol
0.5mol
Same conc ratio Same conc ratio
Basic Buffer preparation NH3(base)/ NH4CI(salt)
1 2 Titration bet strong acid with weak base NH3 + HCI → NH4CI + H2O
Strong acid HCI
NH3 + HCI → NH4CI + H2O
Initial 1 mol 0.5 mol added 0.
Change (1 – 0.5)mol 0 mol 0.5mol form
Final 0.5mol left 0 mol 0.5mol form
At half equivalent point : • Amt base = Amt salt : (0.5 = 0.5) • pOH = pKb - lg [base] → pOH = pKb → 4.74 = 4.74 [salt] Buffer at pOH = 4.74 form when half amt of base neutralise by acid or at half equivalent point when amt base = amt salt
Weak base NH3
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Prepare Buffer pH = 9.26 /pOH = 4.74 • Choose pKb base closest to pOH = 4.74 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.74 = 4.74 – lg [base] [salt] • [base] = 1.00 (amt base = amt salt) [salt]
Buffer region at half equivalent point
Amt base = Amt salt
Preparation basic buffer by titration • 1st – weak acid (burette) with strong base (flask) 2nd – strong acid (burette) with weak base (flask)
Titration bet weak base with strong acid NH3 + HCI → NH4CI + H2O
Strong acid HCI
Weak base NH3
1mol
0.5mol
Prepare Buffer pH = 9.26 /pOH = 4.74 • Choose pKb base closest to pOH = 4.74 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.74 = 4.74 – lg [base] [salt] • [base] = 1.00 (amt base = amt salt) [salt]
NH3 + HCI → NH4CI + H2O
Initial 2 mol added 1.0mol 0.
Change (2 – 1)mol 0 mol 1mol form
Final 1mol left (excess) 0 mol 1mol form
When excess 1 mol base added: • Amt base = Amt salt : (1 = 1 ) • pOH = pKb - lg [base] → pOH = pKb → 4.74 = 4.74 [salt] Buffer at pOH = 4.74 form when amt of base added = amt salt form at equivalent point
Click here buffer simulation
Buffer region Amt base = Amt salt
1mol
2mol
Sample Buffer Calculation
Find pH buffer prepared by titration by adding 18ml, 0.10M HCI to 32ml, 0.10M NH3 pKb = 4.75
1 Titration bet strong acid with weak base NH3 + HCI → NH4CI + H2O
Strong acid HCI 1.8 x 10-3 mol
Weak base NH3 3.2 x 10-3 mol
Buffer region Weak base + salt
NH3 + HCI → NH4CI + H2O
Initial 3.2 x 10-3 mol 1.8 x 10-3 mol added 0. 0
Change (3.2 – 1.8) x 10-3mol 0 mol 1.8 x 10-3 mol form
Final 1.4 x 10-3 mol 0 mol 1.8 x 10-3 mol form
Change moles to Conc → Moles ÷ total volume
Conc (1.4 x 10-3)/ 0.05 (1.8 x 10-3)/ 0.05
Conc 2.8 x 10-2 M 3.6 x 10-2 M
(base) (salt) • pOH = pKb - lg [base] [salt] • pOH = 4.75 – lg [2.8 x 10-2]/[3.6 x 10-2] • pOH = 4.86 pH + pOH = 14 pH = 9.14
Click here buffer simulation
Strong acid 18ml, 0.1M HCI added
Weak base
32ml, 0.1M NH3
Number = (M x V) = 18 x 0.1 moles 1000 1000
Number = (M x V) = 32 x 0.1 moles 1000 1000
Total vol = 50ml or 0.05dm3
NH3 + H2O ↔ NH4+ + OH-
Kb = (NH4+) (OH-)
(NH3) 1.77 x 10-5 = 3.6 x 10-2 x OH- 2.8 x 10-2 OH- = 2.8 x 10-2 x 1.77 x 10-5 3.6 x 10-2 OH- = 1.37 x 10-5
pOH = -lgOH-
pOH = -lg 1.37 x 10-5 pOH = 4.86 pH + pOH = 14 pH = 9.14
1st method (formula) 2nd method (Kb)
pH calculation using
2 Find pH when 50ml 0.1M NaOH added to 100ml, 0.1M CH3COOH Ka CH3COOH = 1.8 x 10-5M, pKa = 4.74
Strong base 50ml, 0.1M NaOH added
Weak acid 100ml, 0.1M CH3COOH
Number = (M x V) = 50 x 0.1 moles 1000 1000
Number = (M x V) = 100 x 0.1 moles 1000 1000
Strong base NaOH
5 x 10-3 mol
Weak acid CH3COOH
10 x 10-3 mol
Titration bet strong base with weak acid NaOH + CH3COOH → CH3COONa + H2O
Click here buffer simulation
Buffer region at half equivalent point
Amt base = Amt salt
Sample Buffer Calculation
pH calculation using
1st method (formula) 2nd method (Ka)
NaOH + CH3COOH → CH3COONa H2O
Initial 5 x 10-3 mol added 10 x 10-3 mol 0.
Change 0 mol (10-5) x 10-3 mol 5 x 10-3 mol form
Final 0 mol 5 x 10-3 mol 5 x 10-3 mol form
Change moles to Conc → Moles ÷ total volume
Conc (5 x 10-3)/0.15 (5 x 10-3)/0.15
Conc 3.3 x 10-2 M 3.3 x 10-2 M
(acid) (salt) • pH = pKa - lg [acid] [salt] • pH = 4.74 – lg [3.3 x 10-2]/[3.3 x 10-2] • pH = 4.74
Total vol = 150ml or 0.15dm3
CH3COOH ↔ CH3COO- + H+
Ka = (CH3COO-)(H+) (CH3COOH) 1.8 x 10-5 = 3.3 x 10-2 x (H+) 3.3 x 10-2 H+ = 1.8 x 10-5
pH = -lg H+ pH = -lg(1.8 x 10-5 ) pH = 4.74
Sample pH Calculation
3 Find pH when 50ml 0.1M NaOH added to 100ml, 0.1M HCI
Strong base 50ml, 0.1M NaOH added
Strong acid
100ml, 0.1M HCI
Number = (M x V) = 50 x 0.1 moles 1000 1000
Number = (M x V) = 100 x 0.1 moles 1000 1000
Strong base NaOH
5 x 10-3 mol
Strong acid HCI 10 x 10-3 mol
Click here buffer simulation
Titration bet strong base with strong acid NaOH + HCI → NaCI + H2O
pH region
pH calculation
NaOH + HCI → NaCI + H2O
Initial 5 x 10-3 mol added 10 x 10-3 mol 0
Change 0 mol (1 0 - 5) x 10-3 mol
Final 0 mol 5 x 10-3 mol
Change moles to Conc → Moles ÷ total volume
Conc (5 x 10-3)/ 0.15
Conc 3.3 x 10-2 M • pH = -lg[H+] • pH = -lg 3.3 x 10-2
pH = 1.48
Total vol = 150ml or 0.15dm3
NH4+ + H2O ↔ NH3 + H3O
+
Ka = (NH3)(H3O
+) (NH4
+) (H3O
+)2 = Ka x NH4+
H+ = √5.56 x 10-10 x 0.10 H+ = 7.45 x 10-6
pH = -lg 7.45 x 10-6 pH = 5.13
Find pH of 0.10M NH4CI in water. Kb NH3 = 1.8 x 10-5 M
4
Acid dissociation constant
0.10M NH4CI
Ka (NH4) x Kb(NH3) = Kw
Ka = Kw /Kb
Ka = 10-14/ 1.8 x 10-5 Ka = 5.56 x 10-10
Using Ka
Find pH of 0.50M NH3 in water. Kb NH3 = 1.8 x 10-5 M
4
NH3 + H2O ↔ NH4+ + OH-
Kb = (NH4+) (OH-)
(NH3) 1.8 x 10-5 = (OH-)2
0.50 OH- = √0.50 x 1.8 x 10-5 OH- = 3.0 x 10-3
pOH = -lg 3.0 x 10-3 pOH = 2.52 pH = 14 – 2.52 pH = 11.48
0.50M NH3
Base Dissociation constant
Using Kb
Sample Buffer Calculation
Find pH buffer - by mixing 200ml, 0.60M NH3 (base) with 300ml, 0.30M NH4CI (salt) Kb = 1.8 x 10-5M, pKb = 4.74
NH3 aft mixing = moles/vol = (200 x 0.60)/0.5 = 0.24M
NH4 aft mixing = moles/vol = (300 x 0.30)/0.5 = 0.18M
pOH = pKb -lg (base) (salt) pOH = 4.74 – lg (0.24) (0.18) pOH = 4.62 pH + pOH = 14 pH = 9.38
5
Find pH buffer by adding 3.20g CH3COONa to 1.00dm3, 0.01M CH3COOH(acid) Ka = 1.75 x 10-5M, pKa = 4.75
Conc salt = 3.20g/dm3 ↓ ÷ RMM (82) ↓ 0.039M
pH = pKa -lg (acid) (salt) pH = 4.75 – lg (0.01) (salt) pH = 4.75 – lg (0.01) (0.039) pH = 4.75 + 0.591 pH = 5.34
CH3COOH ↔ CH3COO- + H+
Ka = (CH3COO-)(H+) (CH3COOH) 1.75 x 10-5 = 0.039 x (H+) (0.01) H+ = 0.01 x 1.75 x 10-5 0.039 H+= 4.487 x 10 -6
pH = -lg(H+) = -lg(4.487 x 10 -6) pH = 5.34
200ml, 0.60M NH3
300ml, 0.30M NH4CI
Total Volume 500ml or 0.5dm3
Conc before Conc after pH buffer
1st method (formula)
NH3 + H2O ↔ NH4+ + OH-
Kb = (NH4+) (OH-)
(NH3) 1.8 x 10-5 = 0.18 x OH- 0.24 OH- = 0.24 x 1.8 x 10-5 0.18 OH- = 2.4 x 10-5
pOH = -lgOH-
pOH = -lg 2.4 x 10-5 pOH = 4.62 pH + pOH = 14 pH = 9.38
2nd method (Kb)
6
1 dm3 , 0.01M
CH3COOH
3.20g CH3COONa
pH buffer
1st method (formula) 2nd method (Ka)
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