Calculations1112 II handout - Ústav lékařské chemie …

1 1pHpHcalculationscalculationsMUDr. Jan Pl ten k, PhDBrBr nstednsted--LowryLowryconceptconceptof of acidsacidsandandbasesbases Acid is a proton donor Base is a proton acceptorHCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)AcidBaseConjugateacidConjugatebas eH2O(l) + NH3(aq) NH4+(aq) + OH-(aq)BaseAcidConjugateacidConjugatebas e2 WhichWhichof of thethefollowingfollowingare are conjugateconjugateacidacid--base base pairspairs??A) HCl, NaOHB) H2O, OH-C) H2SO4, SO42-D) H2SO3, HSO3-E) HClO4, ClO3-F) H3C-NH2, H3C-NH3+AutoionizationAutoionizationof waterof waterWater is amphoteric as it can behave both as acid and base2 H2O(l)H3O+(aq) + OH-(aq) Kw= [H3O+][OH-] = [H+][OH-]Ion-product constant for water:In pure water at 25 C:[H+] = [OH-] = 10-7mol/LKw= ( 10-7mol/L) ( 10-7mol/L) = 10-14mol2/L2 Constant!

2 3pHpHpH= -log10(activity of H+)pOH= -log10(activity of OH-)Ion-product of water (constant!): pH + pOH = :pH=7 (neutral): [H+] = 10-7M = mol/lpH=1 (acidic): [H+] = 10-1M = mol/lpH=13 (alkaline): [H+] = 10-13M = mol/lActivity = f . c f is activity coefficient, f<1, c is molar concentrationStrong acidStrong HCl, HNO3, H2SO4In aqueous solution fully dissociates to H+and A pH of strong acid can be calculated as pH = log (f [H+])For HCl: [H+]= [HCl]For H2SO4: [H+]= 2 [H2SO4]4 Calculating the pH of strong acid solutionsCalculating the pH of strong acid solutionsExample: Calculate the pH of mol/L = = the pH of strong acid solutionsCalculating the pH of strong acid solutionsExample 2: Calculate the pH of mol/L = = = ~ the pH of strong acid solutionsCalculating the pH of strong acid solutionsExample 3: Calculate the pH of 10-10M = log(10-10) = alkaline?

3 Water contributes more protons than HCl in this case (10-7M),pH will be the same as in pure water, 7 Strong baseStrong NaOH, KOH, Ba(OH)2In aqueous solution fully dissociate to metal ion and OH pH of strong base can be calculated as pOH = log (f [OH-]) pH = 14 pOH = 14 ( log (f [OH-])For NaOH: [OH-]= [NaOH]For Ba(OH)2: [OH-]= 2 [Ba(OH)2]6 Calculating the pH of strong Calculating the pH of strong basebasesolutionssolutionsExample: a) Calculate the pH of NaOH = 14 ( ) = ~ ) If this solution is diluted 10-fold, what will bethe resulting pH? pH = ~ acidWeak H2CO3, CH3 COOHOnly some small fraction of molecules in solution dissociates to anion and proton:CH3 COOH CH3 COO + H+[CH3 COO ] [H+] Kd= [CH3 COOH] pH = pK log [AH]pK = log Kd7 If we know pK (Kd) and concentration of a weak acid solution, we can calculate (predict) pH of the solution: If we measure pH of a weak acid solution of a known concentration, we can determine its pK(Kd):pH = pK log [AH]pK = 2 pH + log [AH]Reading pK of weak acid from titration curveInflection point:If weak acid is just half-titrated, then pH = pK[CH3 COO ].

4 [H+] Kd= [CH3 COOH] 8 Calculating the pH of Calculating the pH of weakweakacid solutionsacid solutionsExample: Calculate the pH of mol/L acetic acid. Ka= = pK log [AH]pK = log( 10-5) = = == ( 1) = the pH of Calculating the pH of weakweakacid solutionsacid solutionsExample 2: Calculate the pH of mol/L hypochlorous acid. Ka= = pK log [AH]pK = log( 10-8) = = == ( ) = Weak NH3(aq), organic aminesA fraction of molecules in aqueous solution accepts proton from water:NH3(aq) + H2O NH4++ OH-[NH4+] [OH-] Kd= [NH3] pOH = pK log [B]pH = 14 pOH = 14 pK + log [B]Calculating the pH of Calculating the pH of weakweakbasebasesolutionssolutionsExampl e: Calculate the pH of 5 mol/L aqueous ammonia.

5 Kb= = 14 pKb+ log [B]pKb= log( 10-5) = = 14 + log5 == 14 + = of saltssaltsReaction of dissolved salts with water, :A) Anion from a strong acid, cation from a weak base, NH4Cl:NH4++ H2O NH3+ H3O+Cl-+ H2O no reactionB) Anion from a weak acid, cation from a strong base, NaHCO3:Na++ H2O no reactionHCO3-+ H2O H2CO3+ is is alkalineCalculate the pH of mol/L sodium hydrogen carbonate, NaHCO3. The Ka1of carbonic acid is = 14 pKb+ log [B]pKa= log( 10-7) = 14 = = 14 + == 14 + ( ) =