IB Chemistry on Acid Base Dissociation Constant, pH scale and Kw of water

• 1. Tutorial on Acid Base, pH scale, Acid/BaseDissociation Constant, Ka, Kb and Kw Prepared by Lawrence Kok http://lawrencekok.blogspot.com

2. pH measurement of Acidity of solution pH is the measure of acidity of a solution in logarithmic scalepH = power of hydrogen or minus logarithm to base ten of hydrogen ion concentration Acidic pH < 7 Alkaline pH > 7 pH with Conc H+pH = -log [H+][H+] = 0.0000001MpH = -log [0.0000001]pH = -log1010-7pH = 7 (Neutral)7 pOH with Conc OH-pOH = -log [OH-] [OH-] = 0.0000001M pOH = -log [0.0000001] pOH = -log1010-7 pOH = 7 pH + pOH = 14 pH + 7 = 14 pH = 7 (Neutral) 3. pH measurement of Acidity of solution pH is the measure of acidity of a solution in logarithmic scalepH = power of hydrogen or minus logarithm to base ten of hydrogen ion concentration Acidic pH < 7 Alkaline pH > 7 pOH with Conc OH-pH with Conc H+ pOH = -log [OH-]pH = -log [H+][OH-] = 0.1M[H+] = 0.0000001M pOH = -log[0.1]pH = -log [0.0000001] pOH = 1pH = -log1010-7 pH + pOH = 14pH = 7 (Neutral)pH + 1 = 14pH = 13 (Alkaline)7 pOH with Conc OH-pOH = -log [OH-]pH with Conc H+ [OH-] = 0.0000001MpH = -log [H+] pOH = -log [0.0000001][H+] = 0.01M2pOH = -log1010-7pH = -log [0.01] pOH = 7pH = -log1010-2 pH + pOH = 14pH = 2 (Acidic) pH + 7 = 14 pH = 7 (Neutral) 4. pH measurement of Acidity of solution pH is the measure of acidity of a solution in logarithmic scalepH = power of hydrogen or minus logarithm to base ten of hydrogen ion concentration Acidic pH < 7Alkaline pH > 7 pOH with Conc OH-pH with Conc H+pOH = -log [OH-]pH = -log [H+] [OH-] = 0.1M[H+] = 0.0000001MpOH = -log[0.1]pH = -log [0.0000001]pOH = 1pH = -log1010-7pH + pOH = 14pH = 7 (Neutral) pH + 1 = 14 pH = 13 (Alkaline) 7pOH with Conc OH- pOH = -log [OH-]pH with Conc H+[OH-] = 0.0000001MpH = -log [H+]pOH = -log [0.0000001][H+] = 0.01M 2pOH = -log1010-7pH = -log [0.01]pOH = 7pH = -log1010-2pH + pOH = 14pH = 2 (Acidic)pH + 7 = 14pH = 7 (Neutral) Conc H+ increase by 10x pH decrease by 1 unit Conc OH- increase by 10x pH increase by 1 unit Easier using pH scale than Conc [H+] Conc H+ increase 10x from 0.0001(10-4) to 0.001(10-3)- pH change by 1 unit from pH 4 to 3 pH 3 is (10x) more acidic than pH 4 1 unit change in pH is 10 fold change in Conc [H+] 5. pH measurement of Acidity of solution pH is the measure of acidity of a solution in logarithmic scalepH = power of hydrogen or minus logarithm to base ten of hydrogen ion concentration Acidic pH < 7Alkaline pH > 7 pOH with Conc OH-pH with Conc H+pOH = -log [OH-]pH = -log [H+] [OH-] = 0.1M[H+] = 0.0000001MpOH = -log[0.1]pH = -log [0.0000001]pOH = 1pH = -log1010-7pH + pOH = 14pH = 7 (Neutral) pH + 1 = 14 pH = 13 (Alkaline) 7pOH with Conc OH- pOH = -log [OH-]pH with Conc H+[OH-] = 0.0000001MpH = -log [H+]pOH = -log [0.0000001][H+] = 0.01M 2pOH = -log1010-7pH = -log [0.01]pOH = 7pH = -log1010-2pH + pOH = 14pH = 2 (Acidic)pH + 7 = 14pH = 7 (Neutral) Conc H+ increase by 10x pH decrease by 1 unit Conc OH- increase by 10x pH increase by 1 unit Easier using pH scale than Conc [H+] Conc H+ increase 10x from 0.0001(10-4) to 0.001(10-3)- pH change by 1 unit from pH 4 to 3 Easier scale pH 3 is (10x) more acidic than pH 4 1 unit change in pH is 10 fold change in Conc [H+] 6. pH measurement of Acidity of solutionH2O dissociate forming H3O+ and OH- (equilibrium exist) H2O + H2O H3O+ + OH Kc = [H3O+][OH]/[H2O]2 Kc [H2O]2 = [H3O+][OH]Kw - dissociation constant waterDissociation H2O small, conc [H2O] is constant - ionic product waterKc [H2O]2 is constant called Kw = [H3O+][OH]Kw = 1.0 x 10-14 mol2 dm-6 - dissociation constant water at -25CKw = [H3O+][OH] AlkalineAlkaline1.0 x 10-14 = [H3O+][OH]1.0 x 10-14 = [1.0 x 10-7][1.0 x 10-7] Conc [OH-] = 1 x 10-2Conc [H+] = 1 x 10-12[H3O+]= 1.0 x 10-7, [OH-] = 1.0 x 10-7 pOH = -log10[OH-]pH = -lg[H+]= -log1010-2pH = -lg [10-12] pOH = 2pH = 12 pH + pOH = 14 pH + 2 = 14 pH = 12 Using conc [H+] pH = -log10[H+] Using conc [OH-] pOH = -log10[OH-] Acidic Acidic Conc [OH-] = 1 x 10-12 Conc [H+] = 1 x 10-2 pOH = -log10[OH-] pH = -lg[H+] = -log1010-12 pH = -lg [10-2] pOH = 12 pH = 2pH + pOH = 14 pH + 12 = 14 pH = 2Click here on pH animationClick here to acid/base simulation 7. Formula for acid/base calculationpH = -log10[H+] Kw = [H+][OH-]pKa = - lg10KapOH = -log10[OH-] Ka x Kb = KwpKb = - lg10KbpH + pOH = pKwKa x Kb = 1 x 10-14 pKa + pKb = pKwpH + pOH = 14 pKa + pKb = 14 Weak Acid Weak Base Weak acid Weak acidWeak baseWeak baseHA H + A + - CH3COOH + H2O CH3COO- + H3O+ B + H2O BH+ + OH- NH3 + H2O NH4+ + OH-Ka = (H+) (A-) Ka = (CH3COO-) (H3O+) =(H+)2 Kb = (BH+) (OH-)Kb = (NH4+) (OH-) = (OH-)2(HA) (CH3COOH) (CH3COOH) (B)(NH3) (NH3) 8. Formula for acid/base calculationpH = -log10[H+] Kw = [H+][OH-]pKa = - lg10KapOH = -log10[OH-] Ka x Kb = KwpKb = - lg10KbpH + pOH = pKwKa x Kb = 1 x 10-14 pKa + pKb = pKwpH + pOH = 14 pKa + pKb = 14 Weak Acid Weak Base Weak acidWeak acidWeak base Weak baseHA H + A + - CH3COOH + H2O CH3COO- + H3O+ B + H2O BH+ + OH- NH3 + H2O NH4+ + OH-Ka = (H+) (A-) Ka = (CH3COO-) (H3O+) =(H+)2 Kb = (BH+) (OH-)Kb = (NH4+) (OH-) = (OH-)2(HA) (CH3COOH) (CH3COOH) (B)(NH3) (NH3)H2O (base) - H3O+ (conjugate acid)gain H CH3COOH + H2O CH3COO- + H3O+lose HCH3COOH (acid) - CH3COO- (conjugate base) 9. Formula for acid/base calculationpH = -log10[H+] Kw = [H+][OH-]pKa = - lg10KapOH = -log10[OH-] Ka x Kb = KwpKb = - lg10KbpH + pOH = pKwKa x Kb = 1 x 10-14 pKa + pKb = pKwpH + pOH = 14 pKa + pKb = 14 Weak Acid Weak Base Weak acidWeak acidWeak base Weak baseHA H + A + - CH3COOH + H2O CH3COO- + H3O+ B + H2O BH+ + OH- NH3 + H2O NH4+ + OH-Ka = (H+) (A-) Ka = (CH3COO-) (H3O+) =(H+)2 Kb = (BH+) (OH-)Kb = (NH4+) (OH-) = (OH-)2(HA) (CH3COOH) (CH3COOH) (B)(NH3) (NH3)H2O (base) - H3O+ (conjugate acid)gain H CH3COOH + H2O CH3COO- + H3O+lose HCH3COOH (acid) - CH3COO- (conjugate base)Ka x Kb = KwDissociation constant Weak acid CH3COOH + H2O CH3COO- + H3O+ Ka = (CH3COO-) (H3O+)(CH3COOH) 10. Formula for acid/base calculationpH = -log10[H+] Kw = [H+][OH-]pKa = - lg10KapOH = -log10[OH-] Ka x Kb = KwpKb = - lg10KbpH + pOH = pKwKa x Kb = 1 x 10-14 pKa + pKb = pKwpH + pOH = 14 pKa + pKb = 14 Weak Acid Weak Base Weak acidWeak acidWeak base Weak baseHA H + A + - CH3COOH + H2O CH3COO- + H3O+ B + H2O BH+ + OH- NH3 + H2O NH4+ + OH-Ka = (H+) (A-) Ka = (CH3COO-) (H3O+) =(H+)2 Kb = (BH+) (OH-)Kb = (NH4+) (OH-) = (OH-)2(HA) (CH3COOH) (CH3COOH) (B)(NH3) (NH3)H2O (base) - H3O+ (conjugate acid)gain H CH3COOH + H2O CH3COO- + H3O+lose HCH3COOH (acid) - CH3COO- (conjugate base)Ka x Kb = KwDissociation constant Weak acidDissociation constant Conjugate base CH3COOH + H2O CH3COO- + H3O+ CH3COO- + H2O CH3COOH + OH- Ka = (CH3COO-) (H3O+)Kb = (CH3COOH) (OH-)(CH3COOH) (CH3COO-) 11. Formula for acid/base calculationpH = -log10[H+]Kw = [H+][OH-]pKa = - lg10KapOH = -log10[OH-]Ka x Kb = KwpKb = - lg10KbpH + pOH = pKw Ka x Kb = 1 x 10-14 pKa + pKb = pKwpH + pOH = 14pKa + pKb = 14Weak Acid Weak Base Weak acid Weak acidWeak base Weak baseHA H + A + -CH3COOH + H2O CH3COO- + H3O+ B + H2O BH+ + OH- NH3 + H2O NH4+ + OH-Ka = (H+) (A-)Ka = (CH3COO-) (H3O+) =(H+)2 Kb = (BH+) (OH-)Kb = (NH4+) (OH-) = (OH-)2(HA)(CH3COOH) (CH3COOH) (B)(NH3) (NH3) H2O (base) - H3O+ (conjugate acid) gain HCH3COOH + H2O CH3COO- + H3O+ lose H CH3COOH (acid) - CH3COO- (conjugate base) Ka x Kb = KwDissociation constant Weak acid Dissociation constant Conjugate base CH3COOH + H2O CH3COO- + H3O+CH3COO- + H2O CH3COOH + OH- Ka = (CH3COO-) (H3O+) Kb = (CH3COOH) (OH-)(CH3COOH)(CH3COO-) (CH3COO-) (H3O+) x (CH3COOH) (OH-)(CH3COOH)(CH3COO-) 12. Formula for acid/base calculationpH = -log10[H+]Kw = [H+][OH-]pKa = - lg10KapOH = -log10[OH-]Ka x Kb = KwpKb = - lg10KbpH + pOH = pKw Ka x Kb = 1 x 10-14 pKa + pKb = pKwpH + pOH = 14pKa + pKb = 14Weak Acid Weak Base Weak acid Weak acidWeak base Weak baseHA H + A + -CH3COOH + H2O CH3COO- + H3O+ B + H2O BH+ + OH-NH3 + H2O NH4+ + OH-Ka = (H+) (A-)Ka = (CH3COO-) (H3O+) =(H+)2 Kb = (BH+) (OH-) Kb = (NH4+) (OH-) = (OH-)2(HA)(CH3COOH) (CH3COOH) (B) (NH3) (NH3) H2O (base) - H3O+ (conjugate acid) gain HCH3COOH + H2O CH3COO- + H3O+ lose H CH3COOH (acid) - CH3COO- (conjugate base) Video on how Ka x Kb = Kw derived Ka x Kb = KwDissociation constant Weak acid Dissociation constant Conjugate base CH3COOH + H2O CH3COO- + H3O+CH3COO- + H2O CH3COOH + OH- Ka = (CH3COO-) (H3O+) Kb = (CH3COOH) (OH-)(CH3COOH)(CH3COO-) (CH3COO-) (H3O+) x (CH3COOH) (OH-) = (H3O+)(OH- ) = KwKa x Kb = Kw(CH3COOH)(CH3COO-) 13. Formula for acid/base calculationKa /Kb measures the equilibrium positionKa /Kb measures the equilibrium positionKa /Kb large dissociation product favour shift to right Ka /Kb small dissociation reactant favour shift to leftKa /Kb large pKa /pKb small strong acid/baseKa /Kb small pKa /pKb high weak acid/base 14. Formula for acid/base calculationKa /Kb measures the equilibrium positionKa /Kb measures the equilibrium positionKa /Kb large dissociation product favour shift to right Ka /Kb small dissociation reactant favour shift to leftKa /Kb large pKa /pKb small strong acid/baseKa /Kb small pKa /pKb high weak acid/base Ka pKa , pKa = - lg10KaStrong acid Large KaStrong base Large Kb Kb pKb , pKb = - lg10Kb 15. Formula for acid/base calculationKa /Kb measures the equilibrium positionKa /Kb measures the equilibrium positionKa /Kb large dissociation product favour shift to right Ka /Kb small dissociation reactant favour shift to leftKa /Kb large pKa /pKb small strong acid/baseKa /Kb small pKa /pKb high weak acid/base Kb pKb , pKb = - lg10Kb Ka pKa , pKa = - lg10KaStrong acid Weak base Large KaSmall KbStrong baseWeak acid Small KaLarge Kb Ka pKa , pKa = - lg10Ka Kb pKb , pKb = - lg10Kb 16. Weak acid AnimationDissociate partially, used Ka /Kb value used Animation on weak acid dissociation -H2O dissociate forming H3O+ and OH (equilibrium exist) H2O + H2O H3O+ + OH Kw = [H3O+][OH]/[H2O]2Dissociation H2O is small, conc [H2O] is constant - Kw = [H3O+][OH]Kw = 1.0 x 10-14 mol2 dm-6 - dissociation constant water at -25C Kw = [H3O+][OH]1.0 x 10-14 = [H3O+][OH]1.0 x 10-14 = [1.0 x 10-7][1.0 x 10-7][H3O+]= 1.0 x 10-7, [OH-] = 1.0 x 10-7Click here on weak acid dissociation Animation on weak acid Click here on weak acid dissociation animation Click here on CH3COOH dissociation animation 17. Strong Acid/Base calculation Strong acid 100% dissociation (complete) HCI H+ + CI-Find pH of 0.10M HCIHCI H+ + CI-Find pH of 0.100M H2SO4 H2SO4 2H+ +SO42-http://www.clker.com/clipart-lab-beaker.html 18. Strong Acid/Base calculation Strong acid 100% dissociation (complete) HCI H+ + CI-Find pH of 0.10M HCIHCI H+ + CI-0.10 0.10pH = -lg(H+) = -lg(0.100)pH = 1.00HCI H+ + CI-Find pH of 0.100M H2SO4H2SO4 2H+ + SO42- 0.1000.200pH = -lg(H+) = -lg(0.200)pH = 0.700 H2SO4 2H+ +SO42-http://www.clker.com/clipart-lab-beaker.html 19. Strong Acid/Base calculation Strong acidStrong base 100% dissociation (complete) 100% dissociation (complete) HCI H+ + CI- NaOH Na+ + OH-Find pH of 0.001M NaOHFind pH of 0.10M HCIHCI H+ + CI-0.10 0.10pH = -lg(H+) = -lg(0.100)pH = 1.00HCI H+ + CI-NaOH Na+ + OH- Find pH of 0.001M NaOHFind pH of 0.100M H2SO4H2SO4 2H+ + SO42- 0.1000.200pH = -lg(H+) = -lg(0.200)pH = 0.700 H2SO4 2H+ +SO42- NaOH Na+ + OH-http://www.clker.com/clipart-lab-beaker.html 20. Strong Acid/Base calculation Strong acid Strong base 100% dissociation (complete) 100% dissociation (complete) HCI H+ + CI- NaOH Na+ + OH-Find pH of 0.001M NaOHFind pH of 0.10M HCI NaOH Na+ + OH-HCI H+ + CI- 0.001 0.0010.10 0.10 (H+)(OH-) = 1 x 10-14pH = -lg(H+) = -lg(0.100) H+ = (1 x 10-14)/1 x 10-3 = 1 x 10-11pH = 1.00 pH = -lg(H+) = -lg(1 x 10-11)pH = 11.0HCI H+ + CI- NaOH Na+ + OH-1st Method Find pH of 0.001M NaOHFind pH of 0.100M H2SO4H2SO4 2H+ + SO42- 0.1000.200pH = -lg(H+) = -lg(0.200)pH = 0.700 H2SO4 2H+ +SO42-NaOH Na+ + OH-http://www.clker.com/clipart-lab-beaker.html 21. Strong Acid/Base calculation Strong acid Strong base 100% dissociation (complete) 100% dissociation (complete) HCI H+ + CI- NaOH Na+ + OH-Find pH of 0.001M NaOHFind pH of 0.10M HCI NaOH Na+ + OH-HCI H+ + CI- 0.001 0.0010.10 0.10 (H+)(OH-) = 1 x 10-14pH = -lg(H+) = -lg(0.100) H+ = (1 x 10-14)/1 x 10-3 = 1 x 10-11pH = 1.00 pH = -lg(H+) = -lg(1 x 10-11)pH = 11.0HCI H+ + CI- NaOH Na+ + OH-1st Method Find pH of 0.001M NaOHFind pH of 0.100M H2SO4 NaOH Na+ + OH-H2SO4 2H+ + SO42- 0.001 0.001 0.1000.200pOH = -lg(OH-)pH = -lg(H+) = -lg(0.200)pOH = -lg(0.001) = 3pH = 0.700 pH + pOH = 14 pH + 3 = 14 pH = 11.0 H2SO4 2H+ +SO42-NaOH Na+ + OH-2nd Methodhttp://www.clker.com/clipart-lab-beaker.html 22. Strong Acid/Base calculation Strong acidStrong base 100% dissociation (complete) 100% dissociation (complete) HCI H+ + CI- NaOH Na+ + OH- Animation on Acid Dissociation Find pH of 0.001M NaOHFind pH of 0.10M HCINaOH Na+ + OH-HCI H+ + CI-0.001 0.0010.10 0.10(H+)(OH-) = 1 x 10-14pH = -lg(H+) = -lg(0.100)H+ = (1 x 10-14)/1 x 10-3 = 1 x 10-11pH = 1.00pH = -lg(H+) = -lg(1 x 10-11) pH = 11.0HCI H+ + CI-NaOH Na+ + OH- 1st Method Click here strong acid ionizationFind pH of 0.001M NaOHFind pH of 0.100M H2SO4NaOH Na+ + OH-H2SO4 2H+ + SO42-0.001 0.001 0.1000.200 pOH = -lg(OH-)pH = -lg(H+) = -lg(0.200) pOH = -lg(0.001) = 3pH = 0.700pH + pOH = 14pH + 3 = 14pH = 11.0 H2SO4 2H+ +SO42- NaOH Na+ + OH-Click here on proton equilibria 2nd Methodhttp://www.clker.com/clipart-lab-beaker.html 23. Weak Acid calculationFind pH of 0.100M HA, Ka = 1.80 x 10-5 Find Conc of HA, pH = 4.5, Ka = 4.1 x 10-6 HA H+ + A- HA H+ + A- Find Ka of 0.02M HA, pH= 3.9Determine Kb for F-, Ka HF = 6.8 x 10-4HF + H2O F- + H3O+ HA H+ + A- 24. Weak Acid calculationFind pH of 0.100M HA, Ka = 1.80 x 10-5 Find Conc of HA, pH = 4.5, Ka = 4.1 x 10-6 HA H+ + A-HA H+ + A- Ka = (H+)(A-) or (H+)2Ka = (H+)(A-) or (H+)2(HA)(HA) (HA) (HA)pH = -lg(H+) Ka = (H+)2 /HAKa = (H+)2 /HA 4.5 = -lg(H+)(H+)2 = Ka x HAHA = (H+)2/KaH+ = 10-4.5HA H+ + A- HA H+ + A- Find Ka of 0.02M HA, pH= 3.9 HA H+ + A- Determine Kb for F-, Ka HF = 6.8 x 10-4 Ka = (H+)(A-) or (H+)2 pH = -lg(H+)3.9 = -lg(H+) HF + H2O F- + H3O+(HA)(HA)Ka HF = 6.8 x 10-4 Ka = (H+)2 /HA H+ = 10-3.9 HF + H2O F- + H3O+ HA H+ + A- 25. Weak Acid calculationFind pH of 0.100M HA, Ka = 1.80 x 10-5 Find Conc of HA, pH = 4.5, Ka = 4.1 x 10-6 HA H+ + A-HA H+ + A- Ka = (H+)(A-) or (H+)2Ka = (H+)(A-) or (H+)2(HA)(HA) (HA) (HA) pH = -lg(H+) Ka = (H+)2 /HAKa = (H+)2 /HA4.5 = -lg(H+)(H+)2 = Ka x HAHA = (H+)2/Ka H+ = 10-4.5(H+)2 = Ka x HA HA = (H+)2/Ka(H+)2 = 1.80 x 10-5 x 0.100H+ = 1.34 x 10-3 HA = (10-4.5)2/(4.1 x 10-6)pH = -lg(H+) HA = 2.4 x 10-4pH = -lg (1.34 x 10-3)pH = 2.872HA H+ + A- HA H+ + A- Find Ka of 0.02M HA, pH= 3.9 HA H+ + A- Determine Kb for F-, Ka HF = 6.8 x 10-4 Ka = (H+)(A-) or (H+)2 pH = -lg(H+)3.9 = -lg(H+) HF + H2O F- + H3O+(HA)(HA)Ka HF = 6.8 x 10-4 Ka = (H+)2 /HA H+ = 10-3.9Ka (HF) x Kb (F-) = Kw Ka = (H+)2 /HA Kb F- = Kw = 1.0 x 10-14 Ka = (10-3.9 x 10-3.9)/0.02Ka HF 6.8 x 10-4 Ka = 7.92 x 10-7 Kb F- = 3.98 x 10-4 HF + H2O F- + H3O+ HA H+ + A- 26. Weak Acid calculationFind pH of 0.100M HA, Ka = 1.80 x 10-5 Find Conc of HA, pH = 4.5, Ka = 4.1 x 10-6 HA H+ + A-HA H+ + A- Ka = (H+)(A-) or (H+)2Ka = (H+)(A-) or (H+)2(HA)(HA) (HA) (HA) pH = -lg(H+) Ka = (H+)2 /HAKa = (H+)2 /HA4.5 = -lg(H+)(H+)2 = Ka x HAHA = (H+)2/Ka H+ = 10-4.5(H+)2 = Ka x HA HA = (H+)2/Ka(H+)2 = 1.80 x 10-5 x 0.100H+ = 1.34 x 10-3 HA = (10-4.5)2/(4.1 x 10-6)pH = -lg(H+) HA = 2.4 x 10-4pH = -lg (1.34 x 10-3)pH = 2.872HA H+ + A- HA H+ + A- Find Ka of 0.02M HA, pH= 3.9 HA H+ + A- Determine Kb for F-, Ka HF = 6.8 x 10-4 Ka = (H+)(A-) or (H+)2 pH = -lg(H+)3.9 = -lg(H+) HF + H2O F- + H3O+(HA)(HA)Ka HF = 6.8 x 10-4 Ka = (H+)2 /HA H+ = 10-3.9Ka (HF) x Kb (F-) = Kw Ka = (H+)2 /HA H2O (base) - H3O+ (conjugate acid)Kb F- = Kw = 1.0 x 10-14 Ka = (10-3.9 x 10-3.9)/0.02 gain H Ka HF 6.8 x 10-4 Ka = 7.92 x 10-7 Kb F- = 3.98 x 10-4HF + H2O F- + H3O+lose H HF (acid) - F- (conjugate base) HF + H2O F- + H3O+ HA H+ + A- Ka (HF) x Kb (F-) = Kw 27. Weak Acid calculationFind pH of 0.100M HA, Ka = 1.80 x 10-5 Find Conc of HA, pH = 4.5, Ka = 4.1 x 10-6 HA H+ + A-HA H+ + A- Ka = (H+)(A-) or (H+)2Ka = (H+)(A-) or (H+)2(HA)(HA) (HA) (HA) pH = -lg(H+) Ka = (H+)2 /HAKa = (H+)2 /HA4.5 = -lg(H+)(H+)2 = Ka x HAHA = (H+)2/Ka H+ = 10-4.5 Animation on Acid Dissociation(H+)2 = Ka x HA HA = (H+)2/Ka(H+)2 = 1.80 x 10-5 x 0.100H+ = 1.34 x 10-3 HA = (10-4.5)2/(4.1 x 10-6)pH = -lg(H+) HA = 2.4 x 10-4pH = -lg (1.34 x 10-3)pH = 2.872HA H+ + A- HA H+ + A- Click here weak acid ionization Find Ka of 0.02M HA, pH= 3.9 HA H+ + A- Determine Kb for F-, Ka HF = 6.8 x 10-4 Ka = (H+)(A-) or (H+)2 pH = -lg(H+)3.9 = -lg(H+) HF + H2O F- + H3O+(HA)(HA)Ka HF = 6.8 x 10-4 Ka = (H+)2 /HA H+ = 10-3.9Ka (HF) x Kb (F-) = Kw Ka = (H+)2 /HA H2O (base) - H3O+ (conjugate acid)Kb F- = Kw = 1.0 x 10-14 Ka = (10-3.9 x 10-3.9)/0.02 gain H Ka HF 6.8 x 10-4 Ka = 7.92 x 10-7 Kb F- = 3.98 x 10-4HF + H2O F- + H3O+lose H HF (acid) - F- (conjugate base) HF + H2O F- + H3O+ HA H+ + A- Ka (HF) x Kb (F-) = Kw 28. Weak Base calculationFind pH of 0.01 B, Kb = 1.80 x 10-5 Find Conc of B, pH = 10.8, Kb = 4.36 x 10-4B + H2O BH+ + OH-B + H2O BH+ + OH- Find pH of 0.05 B, pKb = 3.40Find Kb of 0.03 B , pH= 10.0B + H2O BH+ + OH- B + H2O BH+ + OH- 29. Weak Base calculationFind pH of 0.01 B, Kb = 1.80 x 10-5 Find Conc of B, pH = 10.8, Kb = 4.36 x 10-4B + H2O BH+ + OH- B + H2O BH+ + OH-Kb = (BH+)(OH-) or (OH-)2 Kb = (OH-)2 pH + pOH = 14 (B)(B) (B) pOH = 14 10.8 = 3.2(OH-)2 = Kb x B B = (OH-)2/ KbpOH = -lg[OH-] (OH-)2 = 1.80 x 10-5 x 0.013.2 = -lg[OH-]OH- = 4.24 x 10-4 OH- = 10-3.2B + H2O BH+ + OH-B + H2O BH+ + OH- Find pH of 0.05 B, pKb = 3.40Find Kb of 0.03 B , pH= 10.0 pH + pOH = 14 B + H2O BH+ + OH- Kb = (OH-)2 pKb = -lg(Kb)B + H2O BH+ + OH- pOH = 4(B)3.40 = -lg(Kb)Kb = (OH-)2 pOH = -lg(OH-)Kb = (OH-)2 / B Kb = 10-3.40(B)4 = -lg(OH-) (OH-)2 = Kb x B Kb = 3.98 x 10-4OH- = 10-4B + H2O BH+ + OH- B + H2O BH+ + OH- 30. Weak Base calculationFind pH of 0.01 B, Kb = 1.80 x 10-5 Find Conc of B, pH = 10.8, Kb = 4.36 x 10-4B + H2O BH+ + OH- B + H2O BH+ + OH-Kb = (BH+)(OH-) or (OH-)2 Kb = (OH-)2 pH + pOH = 14 (B)(B) (B) pOH = 14 10.8 = 3.2(OH-)2 = Kb x B B = (OH-)2/ KbpOH = -lg[OH-] (OH-)2 = 1.80 x 10-5 x 0.01 3.2 = -lg[OH-]OH- = 4.24 x 10-4OH- = 10-3.2 pOH = -lg(OH-) B = (OH-)2/ Kb pOH = -lg(4.24 x 10-4 ) pOH = 3.37 B = (10-3.2)2/ 4.36 x 10-4 pH + pOH = 14 B = 9.13 x 10-4 M pH = 14 3.37 pH = 10.6B + H2O BH+ + OH-B + H2O BH+ + OH- Find pH of 0.05 B, pKb = 3.40Find Kb of 0.03 B , pH= 10.0 pH + pOH = 14 B + H2O BH+ + OH- Kb = (OH-)2 pKb = -lg(Kb)B + H2O BH+ + OH- pOH = 4(B)3.40 = -lg(Kb)Kb = (OH-)2 pOH = -lg(OH-)Kb = (OH-)2 / B Kb = 10-3.40(B) 4 = -lg(OH-)(OH-)2 = Kb x B Kb = 3.98 x 10-4 OH- = 10-4(OH-)2 = 3.98 x 10-4 x 5.00 x 10-2Kb = (OH-)2 /BOH- = 4.46 x 10-3Kb = (10-4 x 10-4)/0.03 pOH = -lg(OH-)Kb = 3.33 x 10-7pOH = -lg(4.46 x 10-3) = 2.40pH = 14 2.40pH = 11.6B + H2O BH+ + OH- B + H2O BH+ + OH- 31. Weak Base calculationFind pH of 0.01 B, Kb = 1.80 x 10-5Find Conc of B, pH = 10.8, Kb = 4.36 x 10-4B + H2O BH+ + OH-B + H2O BH+ + OH-Kb = (BH+)(OH-) or (OH-)2Kb = (OH-)2 pH + pOH = 14 (B)(B)(B) pOH = 14 10.8 = 3.2(OH-)2 = Kb x BB = (OH-)2/ KbpOH = -lg[OH-] (OH-)2 = 1.80 x 10-5 x 0.013.2 = -lg[OH-]OH- = 4.24 x 10-4 OH- = 10-3.2Animation on Base Dissociation pOH = -lg(OH-)B = (OH-)2/ Kb pOH = -lg(4.24 x 10-4 ) pOH = 3.37B = (10-3.2)2/ 4.36 x 10-4 pH + pOH = 14B = 9.13 x 10-4 M pH = 14 3.37 pH = 10.6 B + H2O BH+ + OH-B + H2O BH+ + OH- Click here on equilibria animationFind pH of 0.05 B, pKb = 3.40Find Kb of 0.03 B , pH= 10.0 pH + pOH = 14B + H2O BH+ + OH-Kb = (OH-)2 pKb = -lg(Kb)B + H2O BH+ + OH- pOH = 4 (B)3.40 = -lg(Kb)Kb = (OH-)2 pOH = -lg(OH-) Kb = (OH-)2 / B Kb = 10-3.40(B) 4 = -lg(OH-) (OH-)2 = Kb x B Kb = 3.98 x 10-4 OH- = 10-4 (OH-)2 = 3.98 x 10-4 x 5.00 x 10-2Kb = (OH-)2 /B OH- = 4.46 x 10-3Kb = (10-4 x 10-4)/0.03pOH = -lg(OH-)Kb = 3.33 x 10-7 pOH = -lg(4.46 x 10-3) = 2.40 pH = 14 2.40 pH = 11.6B + H2O BH+ + OH-B + H2O BH+ + OH- 32. Weak Acid /Base CalculationFind Ka of 0.01 HA, pH= 5.00 Find Conc of B, pH = 10.8, pKa = 10.64 HA H+ + A- Find pH of 0.1 HA, pKa = 4.20 B + H2O BH+ + OH-HA H+ + A- 33. Weak Acid /Base CalculationFind Ka of 0.01 HA, pH= 5.00 Find Conc of B, pH = 10.8, pKa = 10.64HA H+ + A- B + H2O BH+ + OH- pH = -lg(H+)Ka = (H+)(A-) or (H+)2 Kb = (OH-)2 5.0 = -lg(H+)pKa + pKb = 14 (HA)(HA)(B) H+ = 1 x 10-5pKb = 14 pKaKa = (H+)2 /HApH + pOH = 14 pKb = 14 10.64pOH = 14 10.8 pKb = 3.36pOH = 3.2 pKb= -lg(Kb)pOH = -lg[OH-]3.36 = -lg(Kb)3.2 = -lg[OH-]Kb = 10-3.36OH- = 10-3.2Kb = 4.36 x 10-4 HA H+ + A- Find pH of 0.1 HA, pKa = 4.20 HA H+ + A- Ka = (H+)(A-) or (H+)2pKa = -lg(Ka)(HA)(HA)4.2 = -lg(Ka) Ka = (H+)2 /HAB + H2O BH+ + OH-Ka = 6.31 x 10-5 (H+)2 = Ka x HAHA H+ + A- 34. Weak Acid /Base CalculationFind Ka of 0.01 HA, pH= 5.00 Find Conc of B, pH = 10.8, pKa = 10.64HA H+ + A- B + H2O BH+ + OH- pH = -lg(H+)Ka = (H+)(A-) or (H+)2 Kb = (OH-)2 5.0 = -lg(H+)pKa + pKb = 14 (HA)(HA)(B) H+ = 1 x 10-5 pKb = 14 pKaKa = (H+)2 /HApH + pOH = 14pKb = 14 10.64pOH = 14 10.8pKb = 3.36pOH = 3.2pKb= -lg(Kb)pOH = -lg[OH-] 3.36 = -lg(Kb)Ka = (10-5 x 10-5)/0.01 3.2 = -lg[OH-] Kb = 10-3.36Ka = 1.00 x 10-8OH- = 10-3.2 Kb = 4.36 x 10-4 B = (OH-)2 / Kb B = (OH- )2/4.36 x 10-4 HA H+ + A-B = (10-3.2)2/ 4.36 x 10-4 B = 9.13 x 10-4 Find pH of 0.1 HA, pKa = 4.20 HA H+ + A- Ka = (H+)(A-) or (H+)2pKa = -lg(Ka)(HA)(HA)4.2 = -lg(Ka) Ka = (H+)2 /HAB + H2O BH+ + OH-Ka = 6.31 x 10-5 (H+)2 = Ka x HA (H+)2 = Ka x HA (H+)2 = 6.31 x 10-5 x 0.1 H+ = 2.51 x 10-3 pH = -lg(H+) pH = -lg (2.51 x 10-3) pH = 2.60HA H+ + A- 35. Weak Acid /Base CalculationFind Ka of 0.01 HA, pH= 5.00Find Conc of B, pH = 10.8, pKa = 10.64HA H+ + A-B + H2O BH+ + OH- pH = -lg(H+)Ka = (H+)(A-) or (H+)2Kb = (OH-)2 5.0 = -lg(H+) pKa + pKb = 14 (HA)(HA) (B) H+ = 1 x 10-5pKb = 14 pKaKa = (H+)2 /HA pH + pOH = 14pKb = 14 10.64 pOH = 14 10.8pKb = 3.36 pOH = 3.2pKb= -lg(Kb) pOH = -lg[OH-] 3.36 = -lg(Kb)Ka = (10-5 x 10-5)/0.013.2 = -lg[OH-] Kb = 10-3.36Ka = 1.00 x 10-8Simulation on Base DissociationOH- = 10-3.2 Kb = 4.36 x 10-4B = (OH-)2 / KbB = (OH- )2/4.36 x 10-4 HA H+ + A- B = (10-3.2)2/ 4.36 x 10-4B = 9.13 x 10-4 Find pH of 0.1 HA, pKa = 4.20 HA H+ + A- Ka = (H+)(A-) or (H+)2pKa = -lg(Ka)(HA)(HA)4.2 = -lg(Ka) Ka = (H+)2 /HAClick here on weak base simulation B + H2O BH+ + OH-Ka = 6.31 x 10-5 (H+)2 = Ka x HA (H+)2 = Ka x HA (H+)2 = 6.31 x 10-5 x 0.1 H+ = 2.51 x 10-3 pH = -lg(H+) pH = -lg (2.51 x 10-3) pH = 2.60HA H+ + A- Click here on weak base simulation 36. Acid/base CalculationFormula acid/base calculation pH = -log10[H+]pOH = -log10[OH-]pH + pOH = 14 Kw = [H+][OH-] What is the pH for [H+] = 10-12 MWhat is the pH for [OH-] = 0.1 MWhat is the conc of H+ in solution with pH 3?Calculate conc of OH- and pH for 0.001 HCI. HCI H+ + CI- (100% dissociate)Calculate conc of OH- when 3.o x 10-4 [H+] was added to pure waterHCI H+ + CI- (100% dissociate) -43.o x 103.o x 10-4What is the pH of 1.0M NaOH ?NaOH Na+ + OH- (100% dissociate)1M1M 1M 37. Acid/base CalculationFormula acid/base calculation pH = -log10[H+]pOH = -log10[OH-] pH + pOH = 14Kw = [H+][OH-] What is the pH for [H+] = 10-12 M What is the pH for [OH-] = 0.1 M What is the conc of H+ in solution with pH 3? pH = -lg[H+]pOH = -lg[OH-] pH = -lg[H+] pH = -lg [10-12]pOH = -lg [0.1] 3 = -lg[H+] pH = 12 pOH = 1[H+] = 10 pH pH + pOH = 14[H+] = 10 -3 pH = 14 1 = 13Calculate conc of OH- and pH for 0.001 HCI. HCI H+ + CI- (100% dissociate)0.001 0.001Kw = [H+][OH]= 10-14 (assume all H+ from HCI and H+ from water is negligible) [0.001][OH-]= 10-14 [OH-] = 10-14/0.001 = 10 -11pH = -log1o[H]+ =-log10o.oo1pH = 3Calculate conc of OH- when 3.o x 10-4 [H+] was added to pure waterHCI H+ + CI- (100% dissociate) -43.o x 103.o x 10-4Kw = [H+][OH] = 10-14 (assume all H+ from HCI and H+ from water is negligible)[OH] = 10-14 = 10 -14= 3.3 x 10 -11 M+ -4 [H ]3.o x 10What is the pH of 1.0M NaOH ?NaOH Na+ + OH- (100% dissociate)1M 1M1MKw = [H3O+][OH] = 10-14 (assume all OH- from NaOH and OH- from water is negligible)[H+] = 10-14 = 10 -14 = 1.0 x 10 -14[OH] 1.0pH = -log [H+]pH = -log [1.0 x 10 -14]pH = 14 38. Questions on Acids and Base1 Which list contains only strong acids ?A. CH3COOH, H2CO3, H3PO4B. HCI, HNO3, H2CO3C. CH3COOH, HNO3, H2SO4D. HCI, HNO3, H2SO42 When equal volume of four 1M solutions are arranged in order of increasing pH (lowest pH first), what is the correct order?A. CH3COOH < HNO3 < CH3CH2NH2 < KOHB. HNO3 < CH3COOH < CH3CH2NH2 < KOHC. CH3CH2NH2 < HNO3 < CH3COOH < KOHD. KOH < CH3CH2NH2 < CH3COOH < HNO33 100ml of NaOH solution of pH 12 is mixed with 900ml of water. What is the pH of resulting solution?A. 1B. 3C. 11D. 134 Solution of acid A has a pH of 1 and a solution of acid B has a pH of 2. Which statement is correct ?A. Acid A is stronger than acid BB. [A] > [B]C. Concentration of H+ ions in A is higher than BD. Concentration of H+ ions in B is twice the concentration of H+ in A5pH of a solution changes from pH =2 to pH =5. What happens to the concentration of H+ ions during this pH change? A. Decrease by factor of 1000 B. Increase by factor of 1000 C. Decrease by factor of 100 D. Increase by a factor of 1006List two ways to distinguish between strong and weak acid/base 39. Questions on Acids and Base1 Which list contains only strong acids ?A. CH3COOH, H2CO3, H3PO4B. HCI, HNO3, H2CO3C. CH3COOH, HNO3, H2SO4D. HCI, HNO3, H2SO42 When equal volume of four 1M solutions are arranged in order of increasing pH (lowest pH first), what is the correct order?A. CH3COOH < HNO3 < CH3CH2NH2 < KOHB. HNO3 < CH3COOH < CH3CH2NH2 < KOHC. CH3CH2NH2 < HNO3 < CH3COOH < KOHD. KOH < CH3CH2NH2 < CH3COOH < HNO33 100ml of NaOH solution of pH 12 is mixed with 900ml of water. What is the pH of resulting solution?A. 1B. 3C. 11D. 134 Solution of acid A has a pH of 1 and a solution of acid B has a pH of 2. Which statement is correct ?A. Acid A is stronger than acid BB. [A] > [B]C. Concentration of H+ ions in A is higher than BD. Concentration of H+ ions in B is twice the concentration of H+ in A5pH of a solution changes from pH =2 to pH =5. What happens to the concentration of H+ ions during this pH change?A. Decrease by factor of 1000 B. Increase by factor of 1000 C. Decrease by factor of 100 D. Increase by a factor of 1006List two ways to distinguish between strong and weak acid/base By Conductivity measurement By pH measurement1M Strong Acid Ionise completely More H+ ion Conductivity higher 1M Strong Acid Ionise completely More H+ ion pH lower 1M Weak Acid Ionise partially Less H+ ion Conductivity lower 1M Weak Acid Ionise partially Less H+ ion pH higher 40. Video on Acid/ BaseClick here on pH calculation Click here on pKa /pKb calculationHow pH = pOH = 14 derivedHow Ka x Kb = Kw derived Simulation on Acid/ Base Click here on pH animationClick here to acid/base simulation Click here on weak base simulationClick here strong acid ionization Click here on weak acid dissociation 41. 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