1. Tutorial on Acidic, Basic Buffer and BufferPreparation Prepared by Lawrence Kok http://lawrencekok.blogspot.com

2. Acidic Buffer SolutionAcidic 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)CH3COOH CH3COO- +H+ CH3COONa CH3COO- + Na+ Acid part Salt part Acidic buffer 3. 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)Base OH- added Acid H+ addedNeutralize each other Neutralize each otherCH3COOH CH3COO- +H+CH3COONa CH3COO- + Na+Acid partSalt partAcidic buffer 4. 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)Base OH- added Acid H+ addedNeutralize each other Neutralize each otherClick here on acidic buffer simulationCH3COOH CH3COO- +H+CH3COONa CH3COO- + Na+Acid partSalt partClick here on acidic buffer simulationAcidic bufferAcidic 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- 5. Basic Buffer SolutionBasic 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) NH3 + H2O NH4+ + OH_NH4CI NH4+ + CI_ Base partSalt partSalt part Base part Basic buffer 6. Basic Buffer Solution 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)Acid H+ added Base OH- addedNeutralize each other Neutralize each other NH3 + H2O NH4+ + OH_NH4CI NH4+ + CI_Base part Salt part Salt partBase partBasic buffer 7. Basic Buffer Solution 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)Acid H+ added Base OH- addedNeutralize each other Neutralize each otherClick here on basic buffer simulation NH3 + H2O NH4+ + OH_NH4CI NH4+ + CI_Base part Salt part Salt partBase partContain weak base and its saltBasic buffer 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+ 8. How to prepare acidic/ basic buffer Basic BufferAcidic BufferAcidic Buffer FormulaBasic Buffer Formula Mixture Weak acid + Salt/Conjugate base Mixture Weak base + Salt/Conjugate acid CH3COOH CH3COO- + H+ (dissociate partially) NH3 + H2O NH4+ + OH_ (dissociate partially) CH3COONa CH3COO- + Na+ (dissociate fully) NH4CI NH4+ + CI_ (dissociate fully) Acid Dissociation constantBase Dissociation constant CH3COOH + H2O CH3COO- + H3O+ multiply -lgNH3 + H2O NH4+ + OH- Ka = (CH3COO-) (H3O+)both sidesKb = (NH4+) (OH-) (CH3COOH) (NH3) -lgKa = -lgH+ -lg (CH3COO-)-lgKb = -lgOH- -lg (NH4+)(CH3COOH) (NH3) -lgH+ = -lg Ka + lg (CH3COO-)-lgOH- = -lgKb + lg (NH4+) (CH3COOH) salt (NH3) salt pH = pKa + lg (CH3COO-)pOH = pKb + lg (NH4+) (CH3COOH) acid(NH3) base pH = pKa + lg (salt)pH = pKa - lg (acid) pOH = pKb + lg (salt) pOH = pKb - lg (base) (acid) (salt) (base) (salt)Henderson Hasselbalch EquationHenderson Hasselbalch Equation 9. Acidic Buffer Calculation1 Find pH buffer - 0.10M CH3COOH(acid), 0.25M CH3COONa(salt) Ka = 1.8 x 10-5M2 Find pH buffer - 0.20 mol CH3COONa(salt) add to 0.5dm3, 0.10M CH3COOH(acid)Ka = 1.8 x 10-5M3Find 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 10. Acidic Buffer Calculation1 Find pH buffer - 0.10M CH3COOH(acid), 0.25M CH3COONa(salt) Ka = 1.8 x 10-5M Convert Ka to pKa1st method (formula)pH = pKa -lg(acid)Ka = (1.8 x 10-5) (salt)pKa = -lgKapH = 4.74 lg (0.10)pKa = -lg(1.8 x 10-5 ) (0.25)pKa = 4.74pH = 5.142 Find pH buffer - 0.20 mol CH3COONa(salt) add to 0.5dm3, 0.10M CH3COOH(acid)Ka = 1.8 x 10-5M Convert Ka to pKaFind conc salt1st method (formula)pH = pKa -lg(acid) Conc CH3COO- = Moles/volumeKa = (1.8 x 10-5)(salt)= 0.20/0.5pKa = -lgKapH = 4.74 lg (0.10) = 0.40MpKa = -lg(1.8 x 10-5 )(0.40)pKa = 4.74pH = 5.343Find conc of CH3CH2COONa(salt) added to 1.0dm3 of 1.0M CH3CH2COOH(acid) Ka = 1.8 x 10-5M, pKa = 4.74 , pH 4.51st method (formula) pH = pKa -lg (acid) (salt) 4.5 = 4.74 lg (1.0)(salt) lg (1.0) = 0.24 (salt) Conc (salt) = 0.0578M 11. Acidic Buffer Calculation 1 Find pH buffer - 0.10M CH3COOH(acid), 0.25M CH3COONa(salt)2nd method (Ka)Ka = 1.8 x 10-5MCH3COOH CH3COO- + H+Ka = (CH3COO-)(H+) 1st method (formula) Convert Ka to pKa (CH3COOH)1.8 x 10-5 = 0.25 x (H+) pH = pKa -lg(acid) 0.10 Ka = (1.8 x 10-5)(salt)H+ = 7.2 x 10-6 pKa = -lgKa pH = 4.74 lg (0.10)pH = -lg H+ pKa = -lg(1.8 x 10-5 )(0.25)pH = -lg(7.2 x 10-6) pKa = 4.74 pH = 5.14pH = 5.142Find pH buffer - 0.20 mol CH3COONa(salt) add to 0.5dm3, 0.10M CH3COOH(acid) 2nd method (Ka) Ka = 1.8 x 10-5MCH3COOH CH3COO- + H+ Ka = (CH3COO-)(H+)Convert Ka to pKa Find conc salt 1st method (formula)(CH3COOH) 1.8 x 10-5 = 0.40 x (H+) pH = pKa -lg(acid) Conc CH3COO- = Moles/volume Ka = (1.8 x 10-5)(0.10) (salt)= 0.20/0.5 pKa = -lgKa H+ = 4.5 x 10-6 pH = 4.74 lg (0.10) = 0.40M pKa = -lg(1.8 x 10-5 )pH = -lg H+ (0.40) pKa = 4.74pH = -lg(4.5 x 10-6) pH = 5.34 pH = 5.34 3Find 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 1st method (formula)2nd method (Ka)pH = pKa -lg (acid) CH3CH2COOH CH3CH2COO- + H+ Click here for videos from Khan Academy pH = -lg(H+)(salt)Ka = (CH3CH2COO-)(H+) 4.5 = -lg(H+)4.5 = 4.74 lg (1.0) (CH3CH2COOH) H+ = 3.16 x 10-5 (salt) 1.8 x 10-5 = CH3CH2COO- x (3.16 x 10-5)lg (1.0) = 0.24(1.0)(salt)CH3CH2COO- = 0.0578MConc (salt) = 0.0578MClick here for detail explanation from chem guide 12. Basic Buffer Calculation1 Find pH buffer - 0.50M NH3 (base), 0.32M NH4CI (salt) Kb = 1.8 x 10-5M2 Find pH buffer - 4.28g NH4CI (salt) add to 0.25dm3, 0.50NH3(base)Kb = 1.8 x 10-5M3 Find mass of CH3COONa added to 500ml, 0.10M CH3COOH(acid)pH = 4.5, Ka = 1.8 x 10-5M, pKa = 4.74 13. Basic Buffer Calculation1 Find pH buffer - 0.50M NH3 (base), 0.32M NH4CI (salt) Kb = 1.8 x 10-5M1st method (formula)pOH = pKb -lg(base) (salt)pOH = 4.74 lg(0.50) (0.32)pOH = 4.55pH + pOH = 14pH = 9.452 Find pH buffer - 4.28g NH4CI (salt) add to 0.25dm3, 0.50NH3(base)Kb = 1.8 x 10-5MFind conc salt1st method (formula)pOH = pKb -lg(base)Moles, NH4CI = mass/RMM (salt) = 4.28 / 53.5pOH = 4.74 lg(0.50) = 0.08mol (0.32)pOH = 4.55Conc of NH4CI = moles/volpH + pOH = 14= 0.08/0.25pH = 9.45= 0.32M3 Find mass of CH3COONa added to 500ml, 0.10M CH3COOH(acid)pH = 4.5, Ka = 1.8 x 10-5M, pKa = 4.74 1st method (formula) 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 14. Basic Buffer Calculation1 Find pH buffer - 0.50M NH3 (base), 0.32M NH4CI (salt) Kb = 1.8 x 10-5M1st method (formula)2nd method (Kb)pOH = pKb -lg(base)NH3 + H2O NH4+ +OH- (salt)Kb = (NH4+) (OH-)pOH = 4.74 lg(0.50)(NH3) pOH = -lgOH- (0.32)1.8 x 10-5 = 0.32 x OH- pOH = -lg 2.81 x 10-5 Click here addition base to bufferpOH = 4.55 0.50 pOH = 4.55pH + pOH = 14OH- = 0.50 x 1.8 x 10-5 pH + pOH = 14pH = 9.450.32 pH = 9.45 OH- = 2.81 x 10-52 Find pH buffer - 4.28g NH4CI (salt) add to 0.25dm3, 0.50NH3(base)Kb = 1.8 x 10-5M Find conc salt2nd method (Kb)Click here addition acid to buffer1st method (formula) NH3 + H2O NH4+ +OH-pOH = pKb -lg(base) Moles, NH4CI = mass/RMMKb = (NH4+) (OH-) (salt)= 4.28 / 53.5 (NH3)pOH = 4.74 lg(0.50)= 0.08mol pOH = -lgOH- 1.8 x 10-5 = 0.32 x OH- (0.32)pOH = -lg 2.81 x 10-5 Conc of NH4CI = moles/vol 0.50pOH = 4.55 pOH = 4.55= 0.08/0.25OH- = 0.50 x 1.8 x 10-5pH + pOH = 14pH + pOH = 14= 0.32M0.32pH = 9.45pH = 9.45 OH- = 2.81 x 10-5 Click here acidic buffer simulation3 Find mass of CH3COONa added to 500ml, 0.10M CH3COOH(acid)pH = 4.5, Ka = 1.8 x 10-5M, pKa = 4.74 1st method (formula)2nd method (Ka) pH = pKa -lg (acid)CH3COOH CH3COO- + H+(salt)Ka = (CH3COO-) (H+)4.5 = 4.74 lg (0.10)(CH3COOH)(salt)1.8 x 10-5 = CH3COO- x (10-4.5) lg(0.10) = 4.74 -4.5(0.10)(salt)CH3COO- = 0.10 x 1.8 x 10-5 pH = -lg H+ lg (0.10) = 0.2410-4.5 4.5 = -lg(H+)Conc salt = 0.0578M x RMM (82) 4.74g in 1000ml(salt)H+ = 10-4.5 2.37g in 500ml Conc salt = 0.0578MConc salt = 0.0578M 15. Buffer Preparation Acidic Buffer PreparationBasic Buffer PreparationPrepare Acidic Buffer at pH = 5.2Buffer solution Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 Choose pKa acid closest to pH 5.2 Choose pKb base closest to pOH = 4.5 pKa = 4.74 (ethanoic acid) chosen pKb = 4.74 (NH3) chosen pH = pKa -lg [acid] pOH = pKb -lg [base][salt] [salt] 5.2 = 4.74 lg [acid] 4.5 = 4.74 lg [base][salt] [salt] [acid] = 0.35 [base] = 1.74[salt] [salt]Ratio of [acid] = 0.35 Ratio of [base] = 1.74[salt] [salt] 3 ways to prepare buffer 3 ways to prepare buffer 1Use same conc of acid/salt but different vol ratio 1 Use same conc of base/salt but different vol ratio 1M, 35ml (acid ) = 0.35 or 0.1M, 35ml (acid) = 0.35 1M, 174ml (base) = 1.74 or 0.1M, 174ml (base) = 1.741M, 100ml (salt) 0.1M, 100ml (salt)1M, 100ml (salt) 0.1M, 100ml (salt) 2Use same vol of acid/salt but different conc ratio 2 Use same vol of base/salt but different conc ratio 3.5M, 10ml (acid ) = 0.35 or 0.35M, 10ml (acid) = 0.35 1.74M, 10ml (base) = 1.74 or 0.174M, 10ml (base) = 1.7410.0M, 10ml (salt) 1.00M, 10ml (salt)1.00M, 10ml (salt) 0.10M, 10ml (salt) 3Use fix vol, 1dm3 and use different mole ratio (Acid/salt) 3 Use fix vol, 1dm3 and use different mole ratio (base/salt) 0.35 mole acid + 1 mole salt to 1 dm3 solvent= 0.35 1.74 mole base + 1 mole salt to 1 dm3 solvent= 1.74 0.035 mole acid + 0.1 mole salt to 1 dm3 solvent = 0.35 0.174 mole base + 0.1 mole salt to 1 dm3 solvent = 1.74Buffer capacity Buffer capacity Adding water will not change the pH of acidic buffer Adding water will not change the pH of basic buffer Ratio of acid/salt still the same Ratio of base/salt still the samehttp://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/ 16. Buffering CapacityAcidic Buffer Preparation Basic Buffer Preparation Prepare Acidic Buffer at pH = 5.2 Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 Choose pKa acid closest to pH 5.2 Buffer solution Choose pKb base closest to pOH = 4.5 pKa = 4.74 (ethanoic acid) chosen pKb = 4.74 (NH3) chosen pH = pKa -lg [acid] pOH = pKb -lg [base] [salt] [salt] 5.2 = 4.74 lg [acid] 4.5 = 4.74 lg [base] [salt] [salt] [acid] = 0.35 [base] = 1.74 [salt] [salt] Ratio of [acid] = 0.35Ratio of [base] = 1.74 [salt][salt] Which has greater buffering capacity ? Which has greater buffering capacity ? 1 Use same conc of acid/salt but different vol ratio1 Use same conc of base/salt but different vol ratio Buffer A Buffer BBuffer A Buffer B 1M, 35ml (acid ) = 0.35 or 0.1M, 35ml (acid) = 0.35 1M, 174ml (base) = 1.74 or 0.1M, 174ml (base) = 1.74 1M, 100ml (salt) 0.1M, 100ml (salt) 1M, 100ml (salt) 0.1M, 100ml (salt)A B A B1M, 35ml1M, 100ml0.1M, 35ml0.1M, 100ml 1M, 174ml 1M, 100ml 0.1M, 174ml 0.1M, 100ml (acid )(base)(salt) (salt )(acid ) (salt)(base) (salt) 17. Buffering Capacity Acidic Buffer Preparation Basic Buffer PreparationPrepare Acidic Buffer at pH = 5.2Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 Choose pKa acid closest to pH 5.2Buffer solution Choose pKb base closest to pOH = 4.5 pKa = 4.74 (ethanoic acid) chosen pKb = 4.74 (NH3) chosen pH = pKa -lg [acid] pOH = pKb -lg [base][salt][salt] 5.2 = 4.74 lg [acid] 4.5 = 4.74 lg [base][salt][salt] [acid] = 0.35 [base] = 1.74[salt][salt]Ratio of [acid] = 0.35 Ratio of [base] = 1.74[salt] [salt] Which has greater buffering capacity ? Which has greater buffering capacity ? 1Use same conc of acid/salt but different vol ratio 1Use same conc of base/salt but different vol ratioBuffer A Buffer BBuffer A Buffer B 1M, 35ml (acid ) = 0.35 or 0.1M, 35ml (acid) = 0.35 1M, 174ml (base) = 1.74 or 0.1M, 174ml (base) = 1.741M, 100ml (salt) 0.1M, 100ml (salt) 1M, 100ml (salt) 0.1M, 100ml (salt) ABA B1M, 35ml1M, 100ml 0.1M, 35ml0.1M, 100ml 1M, 174ml 1M, 100ml 0.1M, 174ml 0.1M, 100ml (acid ) (base)(salt) (salt ) (acid ) (salt)(base) (salt) Buffer A > Buffer B Buffer A > Buffer BStronger buffering capacityStronger buffering capacity Amt of acid/salt is higher to neutralise added H+ or OH- Amt of base/salt is higher to neutralise added H+ or OH- Ratio acid/salt same, pH buffer same but buffering capacity diff Ratio acid/salt same, pH buffer same but buffering capacity diff Higher buffer conc Higher buffering capacity Higher buffer conc Higher buffering capacityhttp://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/ 18. Buffering CapacityAcidic Buffer PreparationBasic Buffer Preparation Prepare Acidic Buffer at pH = 5.2Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 Choose pKa acid closest to pH 5.2Buffer solution Choose pKb base closest to pOH = 4.5 pKa = 4.74 (ethanoic acid) chosen pKb = 4.74 (NH3) chosen pH = pKa -lg [acid] pOH = pKb -lg [base] [salt][salt] 5.2 = 4.74 lg [acid] 4.5 = 4.74 lg [base] [salt][salt] [acid] = 0.35 [base] = 1.74 [salt][salt] Ratio of [acid] = 0.35 Ratio of [base] = 1.74 [salt] [salt] Which has greater buffering capacity ?Which has greater buffering capacity ? 2 Use same vol of acid/salt but different conc ratio 2 Use same vol of base/salt but different conc ratioBuffer ABuffer B Buffer ABuffer B 3.5M, 10ml (acid ) = 0.35 or 0.35M, 10ml (acid) = 0.35 1.74M, 10ml (base) = 1.74 or 0.174M, 10ml (base) = 1.74 10M, 10ml (salt) 1.00M, 10ml (salt)1.00M, 10ml (salt) 0.10M, 10ml (salt)A BA B3.5M, 10ml10M, 10ml0.35M, 10ml 1M, 10ml 1.74M, 10ml1M, 10ml 0.174M, 10ml 0.1M, 10ml (acid )(salt ) (base) (salt) (acid ) (salt)(base) (salt) 19. Buffering Capacity Acidic Buffer Preparation Basic Buffer PreparationPrepare Acidic Buffer at pH = 5.2Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 Choose pKa acid closest to pH 5.2Buffer solution Choose pKb base closest to pOH = 4.5 pKa = 4.74 (ethanoic acid) chosen pKb = 4.74 (NH3) chosen pH = pKa -lg [acid] pOH = pKb -lg [base][salt][salt] 5.2 = 4.74 lg [acid] 4.5 = 4.74 lg [base][salt][salt] [acid] = 0.35 [base] = 1.74[salt][salt]Ratio of [acid] = 0.35 Ratio of [base] = 1.74[salt] [salt] Which has greater buffering capacity ? Which has greater buffering capacity ? 2Use same vol of acid/salt but different conc ratio 2Use same vol of base/salt but different conc ratio Buffer ABuffer BBuffer ABuffer B 3.5M, 10ml (acid ) = 0.35 or 0.35M, 10ml (acid) = 0.35 1.74M, 10ml (base) = 1.74 or 0.174M, 10ml (base) = 1.7410M, 10ml (salt) 1.00M, 10ml (salt) 1.00M, 10ml (salt) 0.10M, 10ml (salt) ABA B3.5M, 10ml10M, 10ml0.35M, 10ml 1M, 10ml1.74M, 10ml 1M, 10ml 0.174M, 10ml 0.1M, 10ml (acid )(salt ) (base) (salt) (acid ) (salt) (base)(salt) Buffer A > Buffer B Buffer A > Buffer BStronger buffering capacityStronger buffering capacity Amt of acid/salt is higher to neutralise added H+ or OH- Amt of base/salt is higher to neutralise added H+ or OH- Ratio acid/salt same, pH buffer same but buffering capacity diff Ratio acid/salt same, pH buffer same but buffering capacity diff Higher buffer conc Higher buffering capacity Higher buffer conc Higher buffering capacityhttp://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/ 20. Buffering CapacityAcidic Buffer PreparationBasic Buffer Preparation Prepare Acidic Buffer at pH = 5.2 Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 Choose pKa acid closest to pH 5.2 Buffer solution Choose pKb base closest to pOH = 4.5 pKa = 4.74 (ethanoic acid) chosen pKb = 4.74 (NH3) chosen pH = pKa -lg [acid] pOH = pKb -lg [base] [salt][salt] 5.2 = 4.74 lg [acid] 4.5 = 4.74 lg [base] [salt] [salt] [acid] = 0.35 [base] = 1.74 [salt][salt] Ratio of [acid] = 0.35Ratio of [base] = 1.74 [salt][salt] Which has greater buffering capacity ?Which has greater buffering capacity ?3 Use fix vol, 1dm3 but diff mole ratio (acid/salt) 3 Use fix vol, 1dm3 but diff mole ratio (base/salt)Buffer ABuffer BBuffer A Buffer B 0.35mol (acid ) = 0.35 or 0.035mol (acid) = 0.35 1.74mol (base) = 1.74 or 0.174mol (base) = 1.741.00mol (salt) 0.100mol (salt)1.00mol (salt) 0.100mol (salt)AB AB0.35mol 1.00mol 0.174mol0.10mol0.035mol0.10mol 1.74mol1.00mol (acid )(salt )(base)(salt) 1dm3(acid )1dm3(salt)(base)1dm3 (salt)1dm3 21. Buffering Capacity Acidic Buffer Preparation Basic Buffer PreparationPrepare Acidic Buffer at pH = 5.2Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 Choose pKa acid closest to pH 5.2Buffer solution Choose pKb base closest to pOH = 4.5 pKa = 4.74 (ethanoic acid) chosen pKb = 4.74 (NH3) chosen pH = pKa -lg [acid] pOH = pKb -lg [base][salt] [salt] 5.2 = 4.74 lg [acid] 4.5 = 4.74 lg [base][salt][salt] [acid] = 0.35 [base] = 1.74[salt] [salt]Ratio of [acid] = 0.35 Ratio of [base] = 1.74[salt] [salt] Which has greater buffering capacity ?Which has greater buffering capacity ?3 Use fix vol, 1dm3 but diff mole ratio (acid/salt) 3 Use fix vol, 1dm3 but diff mole ratio (base/salt)Buffer ABuffer BBuffer A Buffer B 0.35mol (acid ) = 0.35 or 0.035mol (acid) = 0.35 1.74mol (base) = 1.74 or 0.174mol (base) = 1.741.00mol (salt) 0.100mol (salt)1.00mol (salt) 0.100mol (salt) ABA B0.35mol 1.00mol 0.174mol0.10mol0.035mol0.10mol 1.74mol1.00mol (acid )(salt )(base)(salt)1dm3 (acid )1dm3(salt)(base)1dm3 (salt)1dm3 Buffer A > Buffer B Buffer A > Buffer BStronger buffering capacityStronger buffering capacity Amt of acid/salt is higher to neutralise added H+ or OH- Amt of base/salt is higher to neutralise added H+ or OH- Ratio acid/salt same, pH buffer same but buffering capacity diff Ratio acid/salt same, pH buffer same but buffering capacity diff Higher buffer conc Higher buffering capacity Higher buffer conc Higher buffering capacityhttp://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/ 22. Buffering Capacity Acidic Buffer Preparation Basic Buffer Preparation Prepare Acidic Buffer at pH = 5.2 Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 Choose pKa acid closest to pH 5.2 Buffer solution Choose pKb base closest to pOH = 4.5 pKa = 4.74 (ethanoic acid) chosen pKb = 4.74 (NH3) chosen pH = pKa -lg [acid] pOH = pKb -lg [base] [salt][salt] 5.2 = 4.74 lg [acid] 4.5 = 4.74 lg [base] [salt][salt] [acid] = 0.35 [base] = 1.74 [salt][salt] Ratio of [acid] = 0.35Ratio of [base] = 1.74 [salt][salt]Will pH changeWill pH change4 Same mole ratio (acid/salt) but different total volume4Same mole ratio (base/salt) but different total volume Buffer ABuffer BBuffer ABuffer B 0.35mol (acid)= 0.35 in 1dm3 or 0.35mol (acid) = 0.35 in 2dm3 1.74mol (base)= 1.74 in 1dm3 or 1.74mol (base) = 1.74 in 2dm31.00mol (salt) 1.00mol (salt)1.00mol (salt)1.00mol (salt) ABA B0.35mol1.00mol 0.35mol1.00mol1.74mol1.00mol1.74mol1.00mol (acid ) (salt )(acid ) (salt)(base) (salt)1dm3(base)1dm3 (salt)2dm3 2dm3 23. Buffering Capacity Acidic Buffer PreparationBasic Buffer PreparationPrepare Acidic Buffer at pH = 5.2Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 Choose pKa acid closest to pH 5.2Buffer solution Choose pKb base closest to pOH = 4.5 pKa = 4.74 (ethanoic acid) chosen pKb = 4.74 (NH3) chosen pH = pKa -lg [acid] pOH = pKb -lg [base][salt] [salt] 5.2 = 4.74 lg [acid] 4.5 = 4.74 lg [base][salt] [salt] [acid] = 0.35 [base] = 1.74[salt] [salt]Ratio of [acid] = 0.35 Ratio of [base] = 1.74[salt] [salt]Will pH changeWill pH change4 Same mole ratio (acid/salt) but different total volume4Same mole ratio (base/salt) but different total volume Buffer ABuffer BBuffer ABuffer B 0.35mol (acid)= 0.35 in 1dm3 or 0.35mol (acid) = 0.35 in 2dm3 1.74mol (base)= 1.74 in 1dm3 or 1.74mol (base) = 1.74 in 2dm31.00mol (salt) 1.00mol (salt)1.00mol (salt)1.00mol (salt) ABA B0.35mol 1.00mol 0.35mol 1.00mol1.74mol 1.00mol1.74mol1.00mol (acid )(salt )(acid )(salt)(base)(salt)1dm3(base)1dm3 (salt)2dm32dm3pH Buffer A = pH Buffer B pH Buffer A = pH Buffer B Same pH Same pH Adding water will not change the pH Adding water will not change the pH Amt of acid/salt still the same Amt of acid/salt still the same Ratio conc acid/salt same, pH buffer same Ratio conc base/salt same, pH buffer samehttp://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/ 24. Buffering Capacity Acidic Buffer Preparation Acidic Buffer PreparationPrepare Acidic Buffer at pH = 4.74 Prepare Acidic Buffer at pH = 5.2 Choose pKa acid closest to pH 4.74 Buffer solution Choose pKa acid closest to pH 5.2 pKa = 4.74 (ethanoic acid) chosen pKa = 4.74 (ethanoic acid) chosen pH = pKa -lg [acid] pH = pKa -lg [acid][salt] [salt] 4.74 = 4.74 lg [acid] 5.2 = 4.74 lg [acid] [salt][salt] [acid] = 1.00 [acid] = 0.35[salt] [salt]Ratio of [acid] = 1.00 Ratio of [acid] = 0.35[salt] [salt] Which has greater buffering capacity ? Which has greater buffering capacity ? 5 Same conc ratio (acid/salt) in 1dm3 5 Different conc ratio (acid/salt) in 1dm3Buffer ABuffer B 1.00mol (acid ) = 1.00 0.35mol (acid ) = 0.35 1.00mol (salt) 1.00mol (salt) A B1.00mol 1.00mol 0.35mol 1.00mol(acid)(salt)1dm31dm3(acid) (salt) Concentration ratio Concentration ratio [acid]/[salt] = 1 [acid]/[salt] ratio < 1http://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/ 25. Buffering Capacity Acidic Buffer Preparation Acidic Buffer PreparationPrepare Acidic Buffer at pH = 4.74Prepare Acidic Buffer at pH = 5.2 Choose pKa acid closest to pH 4.74 Buffer solution Choose pKa acid closest to pH 5.2 pKa = 4.74 (ethanoic acid) chosen pKa = 4.74 (ethanoic acid) chosen pH = pKa -lg [acid] pH = pKa -lg [acid][salt][salt] 4.74 = 4.74 lg [acid] 5.2 = 4.74 lg [acid] [salt] [salt] [acid] = 1.00 [acid] = 0.35[salt][salt]Ratio of [acid] = 1.00Ratio of [acid] = 0.35[salt][salt] Which has greater buffering capacity ?Which has greater buffering capacity ? 5 Same conc ratio (acid/salt) in 1dm3 5 Different conc ratio (acid/salt) in 1dm3Buffer ABuffer B 1.00mol (acid ) = 1.00 0.35mol (acid ) = 0.35 1.00mol (salt) 1.00mol (salt) AB1.00mol 1.00mol0.35mol 1.00mol(acid)(salt) 1dm31dm3 (acid) (salt) Concentration ratioConcentration ratio [acid]/[salt] = 1[acid]/[salt] ratio < 1 Buffer A > Buffer BBuffer A > Buffer B Conc ratio [acid]/[salt] equal to 1 Further the conc ratio [acid]/[salt] from 1, lower buffer capacityBuffer has highest buffering capacity when pH = pKa Conc acid = Conc salt highest buffering capacityhttp://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/ 26. Acidic Buffer preparation CH3COOH (acid)/ CH3COO-(salt)pH acidic buffer depend on Ka and ratio [acid]/[salt] CH3COOH (acid) CH3COO- + H+CH3COONa (salt) CH3COO- + Na+ pH = pKa -lg (acid)(salt)2 ways to prepare buffer 1st calculate ratio of [acid]/[salt] 2nd titration weak acid with strong base 27. Acidic Buffer preparation CH3COOH (acid)/ CH3COO-(salt) pH acidic buffer depend on Ka and ratio [acid]/[salt]CH3COOH (acid) CH3COO- + H+CH3COONa (salt) CH3COO- + Na+pH = pKa -lg (acid) (salt) 2 ways to prepare buffer 1st calculate ratio of [acid]/[salt]1 Prepare Acidic Buffer at pH = 5.2 2nd titration weak acid with strong base2 Titration bet strong base with weak acid Choose pKa acid closest to pH 5.2 CH3COOH + NaOH CH3COONa + H2O pKa = 4.74 (ethanoic acid) chosen pH = pKa -lg [acid][salt]Strong base NaOH 5.2 = 4.74 lg [acid] 0.5mol[salt] [acid] = 0.35[salt]Ratio of [acid] = 0.35 Weak acid CH3COOH[salt]1mol Click here buffer simulation 28. Acidic Buffer preparation CH3COOH (acid)/ CH3COO-(salt) pH acidic buffer depend on Ka and ratio [acid]/[salt] CH3COOH (acid) CH3COO- + H+ CH3COONa (salt) CH3COO- + Na+ pH = pKa -lg (acid)(salt) 2 ways to prepare buffer 1st calculate ratio of [acid]/[salt] 1 Prepare Acidic Buffer at pH = 5.2 2nd titration weak acid with strong base 2 Titration bet strong base with weak acid Choose pKa acid closest to pH 5.2 CH3COOH + NaOH CH3COONa + H2O pKa = 4.74 (ethanoic acid) chosen pH = pKa -lg [acid] [salt]Strong base NaOH 5.2 = 4.74 lg [acid] 0.5mol [salt] [acid] = 0.35 [salt] Ratio of [acid] = 0.35 Weak acid CH3COOH [salt]1mol Use same conc of acid/salt but diff vol ratioClick here buffer simulation 1M, 35ml (acid ) = 0.35Prepare Acidic Buffer at pH = 4.74 Choose pKa acid closest to pH 4.74 1M, 100ml (salt) pKa = 4.74 (ethanoic acid) chosen pH = pKa -lg [acid] Use same vol of acid/salt but diff conc ratio[salt] 0.35M, 10ml (acid ) = 0.35 4.74 = 4.74 lg [acid] 1.00M, 10ml (salt) Buffer region at half[salt] equivalent point [acid] = 1.00 (amt acid = amt salt) Amt acid = Amt salt[salt]CH3COOH + NaOHCH3COONa+ H2OInitial 1 mol 0.5 mol added 0.1M, 35ml 1M, 100ml 0.35M, 10ml 1M, 10ml Change (1 0.5)mol0 mol0.5mol form (acid )(salt )(acid )(salt )Final0.5mol left0 mol0.5mol formAt half equivalent point : Amt acid = Amt salt : ( 0.5 = 0.5) Same conc ratio Same conc ratio pH = pKa -lg [acid] pH = pKa 4.74 = 4.74 [salt] Same conc Same VolumeBuffer at pH = 4.74 form when half amt of acid neutralise by base Diff Volume Diff Conc or at half equivalent point when amt acid = amt salt 29. 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) 1Titration bet weak acid + strong base 2Titration bet strong base + weak acid CH3COOH + NaOH CH3COONa + H2OCH3COOH + NaOH CH3COONa + H2OWeak acid CH3COOHStrong base NaOH2mol0.5molStrong base NaOHWeak acid CH3COOH 1mol 1molClick here buffer simulationClick here buffer simulation 30. 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)1Titration bet weak acid + strong base 2 Titration bet strong base + weak acidCH3COOH + NaOH CH3COONa + H2O CH3COOH + NaOH CH3COONa + H2OWeak acid CH3COOH Strong base NaOH2mol 0.5molStrong base NaOHWeak acid CH3COOH 1mol 1molPrepare Acidic Buffer at pH = 4.74Click here buffer simulation Prepare Acidic Buffer at pH = 4.74Click here buffer simulation Choose pKa acid closest to pH 4.74 Choose pKa acid closest to pH 4.74 pKa = 4.74 (ethanoic acid) chosen pKa = 4.74 (ethanoic acid) chosen pH = pKa -lg [acid] pH = pKa -lg [acid][salt] [salt] 4.74 = 4.74 lg [acid] 4.74 = 4.74 lg [acid]Buffer regionBuffer region at half [salt] [salt]Amt acid = Amt salt equivalent point [acid] = 1.00 (amt acid = amt salt) [acid] = 1.00 (amt acid = amt salt) Amt acid = Amt salt[salt] [salt] 31. 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)1Titration bet weak acid + strong base 2 Titration bet strong base + weak acidCH3COOH + NaOH CH3COONa + H2O CH3COOH + NaOH CH3COONa + H2OWeak acid CH3COOH Strong base NaOH2mol 0.5molStrong base NaOHWeak acid CH3COOH 1mol 1molPrepare Acidic Buffer at pH = 4.74Click here buffer simulation Prepare Acidic Buffer at pH = 4.74Click here buffer simulation Choose pKa acid closest to pH 4.74 Choose pKa acid closest to pH 4.74 pKa = 4.74 (ethanoic acid) chosen pKa = 4.74 (ethanoic acid) chosen pH = pKa -lg [acid] pH = pKa -lg [acid][salt] [salt] 4.74 = 4.74 lg [acid] 4.74 = 4.74 lg [acid]Buffer regionBuffer region at half [salt] [salt]Amt acid = Amt salt equivalent point [acid] = 1.00 (amt acid = amt salt) [acid] = 1.00 (amt acid = amt salt) Amt acid = Amt salt[salt] [salt]CH3COOH +NaOH CH3COONa + H2O CH3COOH + NaOH CH3COONa+H2OInitial 2mol added 1.0 mol 0. Initial 1 mol0.5 mol added0.Change (2 1)mol0 mol1mol form Change (1 0.5)mol 0 mol0.5mol form Final1mol left (excess) 0 mol1mol formFinal0.5mol left 0 mol0.5mol formWhen excess 1 mol acid added: At half equivalent point : Amt acid = Amt salt : ( 1 = 1) Amt acid = Amt salt : ( 0.5 = 0.5) pH = pKa -lg [acid] pH = pKa 4.74 = 4.74 pH = pKa -lg [acid] pH = pKa 4.74 = 4.74[salt] [salt]Buffer at pH = 4.74 form when amt of acid added = amt salt form atBuffer at pH = 4.74 form when half amt of acid neutralise by baseequivalent point or at half equivalent point when amt acid = amt salt 32. Basic Buffer preparation NH3(base)/ NH4CI(salt) pH basic buffer depend on Kb and ratio [base]/[salt] NH3 (base) + H2O NH4+ + OH_ NH4CI (salt) NH4+ + CI_ pOH = pKb -lg (base)(salt) 2 ways to prepare buffer 1st calculate ratio of [base]/[salt] 2nd titration weak base with strong acid 33. Basic Buffer preparation NH3(base)/ NH4CI(salt)pH basic buffer depend on Kb and ratio [base]/[salt] NH3 (base) + H2O NH4+ + OH_ NH4CI (salt) NH4+ + CI_pOH = pKb -lg (base) (salt)2 ways to prepare buffer 1st calculate ratio of [base]/[salt]1 Prepare Buffer pH = 9.5 /pOH = 4.52Titration bet strong acid with weak base Choose pKb base closest to pOH = 4.5 2nd titration weak base with strong acidNH3 + HCI NH4CI + H2O pKb = 4.74 (NH3) chosen pOH = pKb -lg [base] [salt] Strong acid HCI 4.5 = 4.74 lg [base] 0.5mol [salt] [base] = 1.74[salt]Ratio of [base] = 1.74 Weak base NH3[salt] 1mol Click here buffer simulation 34. Basic Buffer preparation NH3(base)/ NH4CI(salt)pH basic buffer depend on Kb and ratio [base]/[salt] NH3 (base) + H2O NH4+ + OH_ NH4CI (salt) NH4+ + CI_pOH = pKb -lg (base) (salt)2 ways to prepare buffer 1st calculate ratio of [base]/[salt] 1 Prepare Buffer pH = 9.5 /pOH = 4.5 2 Titration bet strong acid with weak base Choose pKb base closest to pOH = 4.5 2nd titration weak base with strong acid NH3 + HCI NH4CI + H2O pKb = 4.74 (NH3) chosen pOH = pKb -lg [base][salt]Strong acid HCI 4.5 = 4.74 lg [base]0.5mol[salt] [base] = 1.74 [salt] Ratio of [base] = 1.74Weak base NH3 [salt]1mol Use same conc of base/salt but diff vol ratioClick here buffer simulation 1M, 174ml (base) = 1.74Prepare Buffer pH = 9.26 /pOH = 4.74 1M, 100ml (salt) Choose pKb base closest to pOH = 4.74 pKb = 4.74 (NH3) chosen Use same vol of base/salt but diff conc ratio pOH = pKb -lg [base] 1.74M, 10ml (base) = 1.74[salt] 1.00M, 10ml (salt) 4.74 = 4.74 lg [base] Buffer region at half[salt] equivalent point [base] = 1.00 (amt base = amt salt) Amt base = Amt salt[salt]NH3 + HCI NH4CI+ H2OInitial 1 mol 0.5 mol added0.1M, 174ml 1M, 100ml 1.74M, 10ml 1M, 10mlChange (1 0.5)mol0 mol0.5mol form (base ) (salt ) (base ) (salt ) Final0.5mol left0 mol0.5mol formAt half equivalent point : Amt base = Amt salt : (0.5 = 0.5)Same conc Same Volume pOH = pKb - lg [base] pOH = pKb 4.74 = 4.74Diff Volume Diff Conc[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 35. Basic Buffer preparation NH3(base)/ NH4CI(salt)Preparation basic buffer by titration 1st weak acid (burette) with strong base (flask)2nd strong acid (burette) with weak base (flask)1 Titration bet weak base with strong acid2 Titration bet strong acid with weak base NH3 + HCI NH4CI + H2O NH3 + HCI NH4CI + H2OWeak base NH3 Strong acid HCI 2mol0.5mol Weak base NH3Strong acid HCI 1mol 1mol Click here buffer simulationClick here buffer simulation 36. Basic Buffer preparation NH3(base)/ NH4CI(salt) Preparation basic buffer by titration 1st weak acid (burette) with strong base (flask)2nd strong acid (burette) with weak base (flask)1 Titration bet weak base with strong acid2Titration bet strong acid with weak base NH3 + HCI NH4CI + H2ONH3 + HCI NH4CI + H2OWeak base NH3 Strong acid HCI 2mol0.5mol Weak base NH3Strong acid HCI 1mol 1mol Click here buffer simulationClick here buffer simulationPrepare Buffer pH = 9.26 /pOH = 4.74 Prepare Buffer pH = 9.26 /pOH = 4.74 Choose pKb base closest to pOH = 4.74 Choose pKb base closest to pOH = 4.74 pKb = 4.74 (NH3) chosen pKb = 4.74 (NH3) chosen pOH = pKb -lg [base] pOH = pKb -lg [base] [salt] [salt] 4.74 = 4.74 lg [base]Buffer region 4.74 = 4.74 lg [base] Buffer region at half[salt] Amt base = Amt salt [salt] equivalent point [base] = 1.00 (amt base = amt salt) [base] = 1.00 (amt base = amt salt) Amt base = Amt salt[salt][salt] 37. Basic Buffer preparation NH3(base)/ NH4CI(salt) Preparation basic buffer by titration 1st weak acid (burette) with strong base (flask)2nd strong acid (burette) with weak base (flask)1 Titration bet weak base with strong acid 2 Titration bet strong acid with weak base NH3 + HCI NH4CI + H2ONH3 + HCI NH4CI + H2OWeak base NH3Strong acid HCI 2mol 0.5molWeak base NH3Strong acid HCI1mol 1mol Click here buffer simulation Click here buffer simulationPrepare Buffer pH = 9.26 /pOH = 4.74Prepare Buffer pH = 9.26 /pOH = 4.74 Choose pKb base closest to pOH = 4.74 Choose pKb base closest to pOH = 4.74 pKb = 4.74 (NH3) chosen pKb = 4.74 (NH3) chosen pOH = pKb -lg [base] pOH = pKb -lg [base] [salt][salt] 4.74 = 4.74 lg [base] Buffer region 4.74 = 4.74 lg [base] Buffer region at half[salt]Amt base = Amt salt [salt] equivalent point [base] = 1.00 (amt base = amt salt) [base] = 1.00 (amt base = amt salt) Amt base = Amt salt[salt] [salt]NH3+HCI NH4CI+ H2ONH3 + HCI NH4CI + H2OInitial2 mol added1.0mol0.Initial1 mol0.5 mol added 0.Change (2 1)mol0 mol1mol form Change(1 0.5)mol 0 mol0.5mol form Final1mol left (excess) 0 mol1mol formFinal 0.5mol left0 mol 0.5mol formWhen excess 1 mol base added: At half equivalent point : Amt base = Amt salt : (1 = 1 ) Amt base = Amt salt : (0.5 = 0.5) pOH = pKb - lg [base] pOH = pKb 4.74 = 4.74 pOH = pKb - lg [base] pOH = pKb 4.74 = 4.74[salt] [salt]Buffer at pOH = 4.74 form when amt of base added = amt salt formBuffer at pOH = 4.74 form when half amt of base neutralise by acidat equivalent pointor at half equivalent point when amt base = amt salt 38. Sample Buffer Calculation1 Find pH buffer prepared by titration by adding 18ml, 0.10M HCI to 32ml, 0.10M NH3 Titration bet strong acid with weak basepKb = 4.75NH3 + HCI NH4CI + H2O Number = (M x V) = 18 x 0.1Strong acidStrong acid HCImoles1000100018ml, 0.1M HCI added 1.8 x 10-3 molWeak base NH3Weak base3.2 x 10-3 mol 32ml, 0.1M NH3 Number = (M x V) = 32 x 0.1moles10001000Click here buffer simulationBuffer region Weak base + salt 39. Sample Buffer Calculation1 Find pH buffer prepared by titration by adding 18ml, 0.10M HCI to 32ml, 0.10M NH3Titration bet strong acid with weak basepKb = 4.75 NH3 + HCI NH4CI + H2O Number = (M x V) = 18 x 0.1 Strong acidStrong acid HCImoles10001000 18ml, 0.1M HCI added 1.8 x 10-3 mol Weak base NH3 Weak base 3.2 x 10-3 mol32ml, 0.1M NH3 Number = (M x V) = 32 x 0.1moles10001000 Click here buffer simulation pH calculation usingBuffer regionWeak base + salt 1st method (formula) NH3+ HCI NH4CI+ H2OInitial 3.2 x10-3 mol 1.8 x10-3 mol added0.0Change (3.2 1.8) x 10-3mol 0 mol1.8 x 10-3 mol formFinal 1.4 x 10-3 mol 0 mol1.8 x 10-3 mol formChange moles to Conc Molestotal volumeConc (1.4 x 10-3)/ 0.05(1.8 x 10-3)/ 0.05Conc2.8 x 10-2 M 3.6 x 10-2 M(base)(salt) pOH = pKb - lg [base]Total vol = 50ml or 0.05dm3[salt] pOH = 4.75 lg [2.8 x 10-2]/[3.6 x 10-2] pOH = 4.86pH + pOH = 14pH = 9.14 40. Sample Buffer Calculation1 Find pH buffer prepared by titration by adding 18ml, 0.10M HCI to 32ml, 0.10M NH3Titration bet strong acid with weak basepKb = 4.75 NH3 + HCI NH4CI + H2O Number = (M x V) = 18 x 0.1 Strong acidStrong acid HCImoles10001000 18ml, 0.1M HCI added 1.8 x 10-3 mol Weak base NH3 Weak base 3.2 x 10-3 mol32ml, 0.1M NH3 Number = (M x V) = 32 x 0.1moles10001000Click here buffer simulation pH calculation using Buffer region Weak base + salt 1st method (formula)2nd method (Kb) NH3+ HCI NH4CI+ H2ONH3 + H2O NH4+ + OH-Initial 3.2 x 10-3 mol 1.8 x 10-3 mol added 0.0Kb = (NH4+) (OH-)Change (3.2 1.8) x 10-3mol 0 mol1.8 x 10-3 mol form(NH3)pOH = -lgOH-Final 1.4 x 10-3 mol 0 mol1.8 x 10-3 mol form1.77 x 10-5 = 3.6 x 10-2 x OH-pOH = -lg 1.37 x 10-5Change moles to Conc Molestotal volume 2.8 x 10-2pOH = 4.86Conc (1.4 x 10-3)/ 0.05(1.8 x 10-3)/ 0.05OH- = 2.8 x 10-2 x 1.77 x 10-5pH + pOH = 14Conc2.8 x 10-2 M 3.6 x 10-2 M 3.6 x 10-2pH = 9.14(base)(salt) OH- = 1.37 x 10-5 pOH = pKb - lg [base]Total vol = 50ml or 0.05dm3[salt] pOH = 4.75 lg [2.8 x 10-2]/[3.6 x 10-2] pOH = 4.86pH + pOH = 14pH = 9.14 41. Buffer Calculation2 Find pH when 50ml 0.1M NaOH added to 100ml, 0.1M CH3COOHTitration bet strong base with weak acidKa CH3COOH = 1.8 x 10-5M, pKa = 4.74 NaOH + CH3COOH CH3COONa + H2ONumber = (M x V) = 50 x 0.1 Strong base NaOHStrong basemoles1000 1000 5 x 10-3 mol50ml, 0.1M NaOH added Weak acid Weak acid CH3COOH100ml, 0.1M CH3COOH 10 x 10-3 molNumber = (M x V) = 100 x 0.1 moles1000 1000 Click here buffer simulation Buffer region at halfequivalent point Amt base = Amt salt 42. Buffer Calculation2 Find pH when 50ml 0.1M NaOH added to 100ml, 0.1M CH3COOH Titration bet strong base with weak acidKa CH3COOH = 1.8 x 10-5M, pKa = 4.74NaOH + CH3COOH CH3COONa + H2ONumber = (M x V) = 50 x 0.1Strong base NaOHStrong basemoles1000 10005 x 10-3 mol50ml, 0.1M NaOH added Weak acidWeak acid CH3COOH100ml, 0.1M CH3COOH10 x 10-3 molNumber = (M x V) = 100 x 0.1 moles1000 1000Click here buffer simulation pH calculation using Buffer region at half equivalent pointAmt base = Amt salt 1st method (formula) NaOH + CH3COOH CH3COONa H2OInitial 5x 10-3 mol added 10 xmol10-30.0Change 0 mol(10-5) x 10-3 mol5 x 10-3 mol formFinal0 mol 5 x 10-3 mol5 x 10-3 mol formChange moles to Conc Moles total volumeConc(5 x 10-3)/0.15 (5 x 10-3)/0.15Conc 3.3 x 10-2 M3.3 x 10-2 M(acid) (salt) pH = pKa - lg [acid]Total vol = 150ml or 0.15dm3[salt] pH = 4.74 lg [3.3 x 10-2]/[3.3 x 10-2] pH = 4.74 43. Buffer Calculation2 Find pH when 50ml 0.1M NaOH added to 100ml, 0.1M CH3COOH Titration bet strong base with weak acidKa CH3COOH = 1.8 x 10-5M, pKa = 4.74NaOH + CH3COOH CH3COONa + H2ONumber = (M x V) = 50 x 0.1Strong base NaOH Strong base moles1000 10005 x 10-3 mol 50ml, 0.1M NaOH added Weak acidWeak acid CH3COOH100ml, 0.1M CH3COOH10 x 10-3 molNumber = (M x V) = 100 x 0.1 moles1000 1000 Click here buffer simulation pH calculation usingBuffer region at halfequivalent point Amt base = Amt salt 1st method (formula) 2nd method (Ka) NaOH + CH3COOH CH3COONa H2OCH3COOH CH3COO- + H+Initial 5 x 10-3 mol added 10 x 10-3 mol0.0 Ka = (CH3COO-)(H+)Change0 mol (10-5) x 10-3 mol5 x 10-3 mol form(CH3COOH)Final 0 mol5 x 10-3 mol5 x 10-3 mol form1.8 x 10-5 = 3.3 x 10-2 x (H+)Change moles to Conc Moles total volume 3.3 x 10-2Conc(5 x 10-3)/0.15 (5 x 10-3)/0.15 H+ = 1.8 x 10-5Conc 3.3 x 10-2 M3.3 x 10-2 M pH = -lg H+(acid) (salt) pH = -lg(1.8 x 10-5 )pH = 4.74 pH = pKa - lg [acid]Total vol = 150ml or 0.15dm3[salt] pH = 4.74 lg [3.3 x 10-2]/[3.3 x 10-2] pH = 4.74 44. pH Calculation Titration bet strong base with strong acid 3Find pH when 50ml 0.1M NaOH added to 100ml, 0.1M HCINaOH + HCI NaCI + H2ONumber = (M x V) = 50 x 0.1Strong base NaOHStrong base moles1000 10005 x 10-3 mol50ml, 0.1M NaOH added Strong acid HCIStrong acid 10 x 10-3 mol 100ml, 0.1M HCINumber = (M x V) = 100 x 0.1 moles1000 1000 pH calculation Click here buffer simulationpH region4 Find pH of 0.10M NH4CI in water. Find pH of 0.50M NH3 in water. 4Kb NH3 = 1.8 x 10-5 MKb NH3 = 1.8 x 10-5 MAcid dissociation constantBase Dissociation constant0.50M 0.10M NH3 NH4CI 45. pH Calculation Titration bet strong base with strong acid 3Find pH when 50ml 0.1M NaOH added to 100ml, 0.1M HCINaOH + HCI NaCI + H2ONumber = (M x V) = 50 x 0.1Strong base NaOH Strong base moles1000 10005 x 10-3 mol 50ml, 0.1M NaOH added Strong acid HCIStrong acid 10 x 10-3 mol 100ml, 0.1M HCINumber = (M x V) = 100 x 0.1 moles1000 1000 pH calculation NaOH+ HCI NaCI+H2O Click here buffer simulation Initial 5 x 10-3 mol added 10 x 10-3 mol 00 Change0 mol(1 0 - 5) x 10-3 mol Final 0 mol5 x 10-3 molpH region Change moles to Conc Moles total volumeTotal vol = 150ml or 0.15dm3 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 = 1.484 Find pH of 0.10M NH4CI in water. Find pH of 0.50M NH3 in water. 4Kb NH3 = 1.8 x 10-5 MKb NH3 = 1.8 x 10-5 M Using KaUsing KbAcid dissociation constantBase Dissociation constantNH4+ + H2O NH3 + H3O+NH3 + H2O NH4+ + OH-Ka = (NH3)(H3O+) Kb = (NH4+) (OH-) (NH3) (NH4+)0.50M1.8 x 10-5 = (OH-)2 0.10M(H3O ) = Ka x NH4+ + 2 0.50 NH3 NH4CIH+ = 5.56 x 10-10 x 0.10OH- = 0.50 x 1.8 x 10-5H+ = 7.45 x 10-6 OH- = 3.0 x 10-3Ka (NH4) x Kb(NH3) = KwpH = -lg 7.45 x 10-6 pOH = -lg 3.0 x 10-3Ka = Kw /KbpH = 5.13pOH = 2.52Ka = 10-14/ 1.8 x 10-5 pH = 14 2.52Ka = 5.56 x 10-10pH = 11.48 46. Buffer Calculation5 Find pH buffer - by mixing 200ml, 0.60M NH3 (base) with 300ml, 0.30M NH4CI (salt)Kb = 1.8 x 10-5M, pKb = 4.74Conc after Conc before NH3 aft mixing = moles/vol = (200 x 0.60)/0.5 = 0.24M 300ml, 0.30M200ml, 0.60MNH4CINH4 aft mixingNH3 = moles/vol = (300 x 0.30)/0.5 = 0.18M Total Volume500ml or 0.5dm36 Find pH buffer by adding 3.20g CH3COONa to 1.00dm3, 0.01M CH3COOH(acid)Ka = 1.75 x 10-5M, pKa = 4.75Conc salt = 3.20g/dm3 RMM (82) 3.20g0.039M CH3COONa 1 dm3 , 0.01M CH3COOH 47. Buffer Calculation5 Find pH buffer - by mixing 200ml, 0.60M NH3 (base) with 300ml, 0.30M NH4CI (salt)Kb = 1.8 x 10-5M, pKb = 4.74Conc after pH buffer Conc before1st method (formula) NH3 aft mixing pOH = pKb -lg (base) = moles/vol (salt) = (200 x 0.60)/0.5 pOH = 4.74 lg (0.24) = 0.24M(0.18) 300ml, 0.30M200ml, 0.60M pOH = 4.62NH4CINH4 aft mixingNH3pH + pOH = 14 = moles/vol pH = 9.38 = (300 x 0.30)/0.5 = 0.18M 2nd method (Kb) Total VolumeNH3 + H2O NH4+ +OH-500ml or 0.5dm3Kb = (NH4+) (OH-) (NH3)pOH = -lgOH- 1.8 x 10-5 = 0.18 x OH-pOH = -lg 2.4 x 10-5 0.24pOH = 4.62 OH- = 0.24 x 1.8 x 10-5pH + pOH = 14 0.18pH = 9.386 Find pH buffer by adding 3.20g CH3COONa to 1.00dm3, 0.01M CH3COOH(acid)OH- = 2.4 x 10-5Ka = 1.75 x 10-5M, pKa = 4.75Conc salt = 3.20g/dm3 pH buffer RMM (82) 1st method (formula) 2nd method (Ka) 3.20g0.039MpH = pKa -lg (acid)CH3COOH CH3COO- + H+ CH3COONa(salt) Ka = (CH3COO-)(H+) 1 dm3 , 0.01M CH3COOHpH = 4.75 lg (0.01)(CH3COOH)(salt) 1.75 x 10-5 = 0.039 x (H+)pH = 4.75 lg (0.01)(0.01) (0.039) H+ = 0.01 x 1.75 x 10-5pH = 4.75 + 0.5910.039pH = 5.34 H+= 4.487 x 10 -6 pH = -lg(H+) = -lg(4.487 x 10 -6) pH = 5.34 48. Simulation and Animation on Buffer and TitrationsClick here acidic buffer simulationClick here acidic buffer simulationClick here basic buffer simulation Click here buffer simulation Click here acidic buffer animationClick here for videos from Khan Academy Click here on titration simulationClick here on titration animationClick here acidic buffer animation 49. AcknowledgementsThanks to source of pictures and video used in this presentationThanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/Prepared by Lawrence KokCheck out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com