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

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IB Chemistry on pH scale, Acid Base Dissociation constant, pKa, pKb and Kw water.

IB Chemistry on pH scale, Acid Base Dissociation constant, pKa, pKb and Kw water.

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  • 1. pH measurement of Acidity of solution • pH is the measure of acidity of a solution in logarithmic scale •pH = 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 = 14 pH = 13 (Alkaline) pOH with Conc OH- pOH = -log [OH-]pH with Conc H+ [OH-] = 0.0000001MpH = -log [H+] pOH = -log [0.0000001][H+] = 0.01M pOH = -log1010-7pH = -log [0.01] pOH = 7pH = -log1010-2 pH + pOH = 14pH = 2 (Acidic) pH + 7 = 14 pH = 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 increase 10x from 0.0001(10-4) to 0.001(10-3) H+ - 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+]
  • 2. pH measurement of Acidity of solution H2O dissociate forming H3O+ and OH- (equilibrium exist) • H2O + H2O ↔ H3O+ + OH− • Kc = [H3O+][OH−]/[H2O]2 • Kc [H2O]2 = [H3O+][OH−] Kw - dissociation constant water Dissociation H2O small, conc [H2O] is constant - ionic product water Kc [H2O]2 is constant called Kw = [H3O+][OH−] Kw = 1.0 x 10-14 mol2 dm-6 - dissociation constant water at -25C Kw = [H3O+][OH−] Alkaline Alkaline 1.0 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 = 2 pH + pOH = 14 pH + 12 = 14 pH = 2 Click here on pH animation Click here to acid/base simulation
  • 3. Formula for acid/base calculation pH = -log10[H+] Kw = [H+][OH-] pKa = - lg10Ka pOH = -log10[OH-] Ka x Kb = Kw pKb = - lg10Kb pH + pOH = pKw Ka x Kb = 1 x 10-14 pKa + pKb = pKw pH + pOH = 14 pKa + pKb = 14 Weak Acid Weak Base Weak acid Weak acid Weak 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 H CH3COOH (acid) - CH3COO- (conjugate base) Video on how Ka x Kb = Kw derived Ka x Kb = Kw Dissociation 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- ) = Kw Ka x Kb = Kw (CH3COOH) (CH3COO-)
  • 4. Formula for acid/base calculationKa /Kb measures the equilibrium position Ka /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/base Ka /Kb small ↓ – pKa /pKb high ↑– weak acid/base ↓ Kb →↑ pKb , pKb = - lg10Kb↑ Ka → ↓ pKa , pKa = - lg10KaStrong acid Weak base Large ↑ Ka Small ↓KbWeak acid Strong base Large ↑ Kb Small ↓ Ka↓ Ka → ↑ pKa , pKa = - lg10Ka ↑ Kb → ↓ pKb , pKb = - lg10Kb
  • 5. Weak acid Animation •Dissociate 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 10-7 Click here on weak acid dissociation Animation on weak acid Click here on weak acid dissociation animation Click here on CH3COOH dissociation animation
  • 6. Strong Acid/Base calculation Strong acid Strong base • 100% dissociation (complete) • 100% dissociation (complete) • HCI → H+ + CI- • NaOH → Na+ + OH- Animation on Acid Dissociation Find pH of 0.001M NaOH Find pH of 0.10M HCI NaOH → Na+ + OH- HCI → H+ + CI- 0.001 0.001 0.10 0.10 (H+)(OH-) = 1 x 10-14 pH = -lg(H+) = -lg(0.100) H+ = (1 x 10-14)/1 x 10-3 = 1 x 10-11 pH = 1.00 pH = -lg(H+) = -lg(1 x 10-11) pH = 11.0 HCI → H+ + CI- NaOH → Na+ + OH- 1st Method Click here strong acid ionization Find pH of 0.001M NaOH Find pH of 0.100M H2SO4 NaOH → Na+ + OH- H2SO4 → 2H+ + SO42- 0.001 0.001 0.100 0.200 pOH = -lg(OH-) pH = -lg(H+) = -lg(0.200) pOH = -lg(0.001) = 3 pH = 0.700 pH + pOH = 14 pH + 3 = 14 pH = 11.0 H2SO4 → 2H+ +SO42- NaOH → Na+ + OH- Click here on proton equilibria 2nd Methodhttp://www.clker.com/clipart-lab-beaker.html
  • 7. 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+)2 Ka = (H+)(A-) or (H+)2 (HA) (HA) (HA) (HA) pH = -lg(H+) Ka = (H+)2 /HA Ka = (H+)2 /HA 4.5 = -lg(H+) (H+)2 = Ka x HA HA = (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 0.100 H+ = 1.34 x 10-3 HA = (10-4.5)2/(4.1 x 10-6) pH = -lg(H+) HA = 2.4 x 10-4 pH = -lg (1.34 x 10-3) pH = 2.872 HA ↔ 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+ = 10-3.9 Ka (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 10-4 HF + H2O ↔ F- + H3O+ lose H HF (acid) - F- (conjugate base) HF + H2O ↔ F- + H3O+ HA ↔ H+ + A- Ka (HF) x Kb (F-) = Kw
  • 8. 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 10-4 B + 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/ Kb pOH = -lg[OH-] (OH-)2 = 1.80 x 10-5 x 0.01 3.2 = -lg[OH-] OH- = 4.24 x 10-4 OH- = 10-3.2 Animation on Base Dissociation 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 = 10.6 B + H2O ↔ BH+ + OH- B + H2O ↔ BH+ + OH- Click here on equilibria animation Find pH of 0.05 B, pKb = 3.40Find Kb of 0.03 B , pH= 10.0 B + H2O ↔ BH+ + OH- pH + pOH = 14 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 10-2 Kb = (OH-)2 /B OH- = 4.46 x 10-3 Kb = (10-4 x 10-4)/0.03 pOH = -lg(OH-) Kb = 3.33 x 10-7 pOH = -lg(4.46 x 10-3) = 2.40 pH = 14 – 2.40 pH = 11.6 B + H2O ↔ BH+ + OH- B + H2O ↔ BH+ + OH-
  • 9. 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 /HA pH + pOH = 14 pKb = 14 – 10.64 pOH = 14 – 10.8 pKb = 3.36 pOH = 3.2 pKb= -lg(Kb) pOH = -lg[OH-] 3.36 = -lg(Kb) Ka = (10-5 x 10-5)/0.01 3.2 = -lg[OH-] Kb = 10-3.36 Ka = 1.00 x 10-8 Simulation on Base Dissociation OH- = 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+)2 pKa = -lg(Ka) (HA) (HA) 4.2 = -lg(Ka) Ka = (H+)2 /HA Click 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 = 2.60 HA ↔ H+ + A- Click here on weak base simulation
  • 10. Acid/base Calculation Formula acid/base calculation pH = -log10[H+] pOH = -log10[OH-] pH + pOH = 14 Kw = [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 water HCI → H+ + CI- (100% dissociate) -43.o x 10 3.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 1M 1MKw = [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
  • 11. Questions on Acids and Base1 Which list contains only strong acids ? A. CH3COOH, H2CO3, H3PO4 B. HCI, HNO3, H2CO3 C. CH3COOH, HNO3, H2SO4 О D. 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 < KOH О B. HNO3 < CH3COOH < CH3CH2NH2 < KOH C. CH3CH2NH2 < HNO3 < CH3COOH < KOH D. 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. 1 B. 3 О C. 11 D. 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 B B. [A] > [B] О C. Concentration of H+ ions in A is higher than B D. Concentration of H+ ions in B is twice the concentration of H+ in A5 pH of a solution changes from pH =2 to pH =5. What happens to the concentration of H+ ions during this pH change? ОA. Decrease by factor of 1000 B. Increase by factor of 1000 C. Decrease by factor of 100 D. Increase by a factor of 1006 List 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 ↑
  • 12. Video on Acid/ BaseClick here on pH calculation Click here on pKa /pKb calculation How pH = pOH = 14 derived How Ka x Kb = Kw derived Simulation on Acid/ Base Click here on pH animation Click here to acid/base simulation Click here on weak base simulation Click here strong acid ionization Click here on weak acid dissociation