17.1 Strong-Acids / Strong Bases Titration

17.1 Strong-Acids / Strong Bases Titration

17.2 Strong-Acids / StrongBases Titration Dr. Fred Omega Garces Chemistry 201 Miramar College Strong-Acids / Strong Bases Titration 2/27/20 Reaction Between Acid - Base What is the pH of a solution when an acid is mixed with a base? Stoichiometry Problem: HA H3O+ + AMOH OH- + M+ MOH + HA H2O + MA Stoichiometry StoichiometryProblem: Problem: + The +or OH Theamount amountof ofHH3O 3O or OH remaining remainingafter afteraaportion portionisis neutralize determines neutralize determinesthe thepH pH of the solution. of the solution. In an acid - base reaction, H+ & OH- always combine together to form water and an ionic compound (a salt): Neutralization Reaction. HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Analysis is a Stoichiometry problem only if a strong acid is combined with a strong base. Strong-Acids / Strong Bases Titration 2/27/20 Titration Acid/Base Titration A technique of chemical analysis to determine the amount of a substance in a sample. i.e., What is the acidic content of Lake Miramar? A sample can be tested by titration. In a titration experiment, a known volume of a standard concentrated solution (the titrant) is used to analyze a sample (the analyte). One is usually an acid, the other a base. An indicator is added to the

analyte to signal when the titration is complete. This is called the endpoint. When the moles of acid(H3O+) and moles of base (OH-) are equal in a titration experiment, the stoichiometric equivalent point is reached. This is called the equivalent point. Indicator changes color @ endpoint moles titrant = moles analyte. @ equivalent point. Strong-Acids / Strong Bases Titration 2/27/20 Acid-Base Indicator Some common acid-base indicators. The color changes occur over a range of pH values. Notice that a few indicators have two color changes over two different pH ranges. Mechanism for phenolphthalein indicator. At Low pH phenolphthalein is colorless and has a structure in which there is a five membered ring. In the presence of excess acid the five membered ring is broken and the resulting change in conformation gives rise to a compound which is pink. Strong-Acids / Strong Bases Titration 2/27/20 S Acid- S Base Analysis Indicator changes color @ endpoint. Indicator is chosen so that endpoint occurs at equivalent pt. The following is true at the equivalence point. moles H3O+ = moles OHMacid Vacid = Mbase Vbase For monoprotic A & B Macid Vacid = Mbase Vbase but For polyprotic A & B NacidVacid = NbaseVbase for monoprotic and polyprotic A & B eq H3O+ L H3O+ = eq OH- L OHL Soln L Soln For Acid base calculation at Equivalence pt: Sometimes: Macid Vacid = Mbase Vbase but always: NacidVacid = NbaseVbase Strong-Acids / Strong Bases Titration 2/27/20 S Acid- S Base Calculation Titration of HCl with NaOH (Reger 14.1) A titration is used to detm conc. of HCl solution. Exactly 20.00mL of the acid solution was placed in a flask, with phenolphthalein added. 18.34mL of 0.0982 M NaOH was needed to reach the endpoint. What is the conc. of the HCl ? Rxn: Rxn: NaOH NaOH OH OH 100% 100%

HCl H+ 100% HCl H+ 100% + Net H + OH H O Net H+ + OH- H2 O 2 Conc ConcHH3O+: 3O+: = 0.0982 mol OH- 18.34 mL . 1 mol H+ = 0.0982 mol OH- 18.34 mL . 1 mol H+ 1 1 20.00 mL HCl L L 1mol OH1mol OH- 1 1 20.00 ml 20.00 ml [HCl] [HCl]==0.09005 0.09005M M 18.34 mL NaOH @ end pt. Strong-Acids / Strong Bases Titration 2/27/20 Titration: Thought Experiment Consider the titration of a Strong acid with a Strong Base. What is the pH after incremental addition of some moles of base to the acid. Amt Amt Amt Remaining NaOHHCl H2O pH (mol) or

[OH- ] [H2O+] pOH 0 100 H+ - 100H+ pH = - log [100 / vT] 20 100-20 20 H2O 80 H+ - log [80 / vT] 50 100-50 50 H2O 50 H+ - log [50 / vT] 99 100-99 99 H2O 1 H+ - log [1 / vT] 100 100-100 100 H2O 0 H+ - log [0 / vT] 101 100-101 100 H2O 1 OH- pOH = - log [1 / vT] 200 100-200 100 H2O 100 OH-

Strong-Acids / Strong Bases Titration pH = ?* pOH = - log [100 / vT] 2/27/20 Titration S Acid - S Base: Example A 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH. What is the pH after addition of 0.00, 20.00, 49.00, 50.00, 51.00 and 60.00 mL of base. Titration curve between strong acid and strong base. Analyte is HCl and titrant is NaOH. Rxn: Rxn: NaOH NaOH OH OH 100% 100% HHCl H+ 100% Cl H+ 100% + Net H ++ OH - Net H + OH HH2O 2O ? Click for simulation Strong-Acids / Strong Bases Titration 2/27/20 Titration Strong Acid - Strong Base: Example ( 0.0 & 25.0 ml) i) 0.00 mL Base 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 0.00mL of base. pH based on the [H3O+] of HCl HCl is a strong acid, therefore [HCl] = [H3O+] [HCl] = [H3O+] = 0.100 M = 110-1 M pH = 1.00 ii) 25.00 mL Base: Vtotal = 75 ml solution 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 25.00mL of base mol HCl = 0.100M (50mL) =5.00 10-3 mol HCl or 5.0 mmol = H3O+ mol NaOH = 0.100M (25mL) = 2.5 10-3 mol NaOH or 2.5 mmol = OH -

pH based on the excess HCl remaining. Since HCl is a strong acid. HCl + NaOH H2 O s 5 mmol 2.5 mmol - R 2.5 2.5 2.5 2.5 2.5 f 2.5 0 2.5 2.5 2.5 [c] + - Na+ + Cl- - 2.5 mmol / 75.00 ml = 3.310-2 M [HCl] = [H3O+] = 3.310-2 M

pH = 1.48 Strong-Acids / Strong Bases Titration 2/27/20 Titration S Acid - S Base: Example ( 49.0 & 50.0 mL) iii) 49..00 mL Base: Vtotal = 99 ml solution 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 49.00mL of base. mol HCl = 0.100M (50mL) = 5.0 mmol HCl = H3O+ mol NaOH = 0.100M HCl + (49mL) = 4.9 mmol NaOH = OH - NaOH H2O s 5.0 mmol 4.9 mmol - R -4.9 - - f 0.1 mmol [c] -4.9 0 - + -

- Na + Cl- - - 0.1 mmol / 99.0 ml = 1.01 10-3 M [HCl] = [H3O+] = 1.01 10-3 M pH = 3.0 iv) 50.00 mL Base: Vtotal = 99 ml solution 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 50.00mL of base. mol HCl = 0.100M (50mL) = 5.0 mmol HCl = H3O+ mol NaOH = 0.100M (50mL) = 5.0 mmol NaOH = OH - HCl + NaOH s 5.0 mmol R -5.0 f 0.1 mmol [c] 0.1 mmol / 100.0 ml = 0 M No, -5.0 5.0 mmol H2O 0 -

- + - Na + Cl- - - - - Is the pH = zero ? Autoionization Water has [H3O+] = 1.00 10-7 M Strong-Acids / Strong Bases Titration pH = 7.0 2/27/20 Titration S Acid - S Base: Example ( 51.0 & 60.0 mL) v) 51.00 mL Base: Vtotal = 101 ml solution 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 51.00mL of base. mol HCl = 0.100M (50mL) = 5.0 mmol HCl = H3O+ mol NaOH = 0.100M HCl + (51mL) = 5.1 mmol NaOH = OH - NaOH H2O s 5.0 mmol 5.1 mmol -

R -5.0 - - f 0 mmol -5.0 - 0.1 [c] [NaOH] - + Na - - + Cl- - - 0.1 mmol / 101.0 ml = 9.90 10-4 M excess = [OH-] = 9.90 10-4 M pOH = 3.00 pH = 11.0 vi) 60.00 mL Base: Vtotal = 110 ml solution 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 60.00 mL of base. mol HCl = 0.100M (50 mL) = 5.0 mmol HCl = H3O+ mol NaOH = 0.100M

HCl + (60 mL) = 6.0 mmol NaOH = OH - NaOH H2 O s 5.0 mmol 6.0 mmol - R -5.0 - - f 0 mmol [c] [NaOH] -5.0 1.0 - - + - Na + Cl- - - 1.0 mmol / 110.0 ml = 9.90 10-3 M excess = [OH-] = 9.90 10-3 M

Strong-Acids / Strong Bases Titration pOH = 2.04 pH = 11.96 2/27/20 Titration: Result Summary Summary of the titration of 0.100 M HCl with 0.100 M NaOH. What is the pH after incremental addition of some moles of base to the acid. Vol Amt Amt Net NaOH HCl-NaOH NaOH (mL) (mmol) (mmol) Conc pH H3O+ H+ or OH- or (mmol) (M) pOH 0 0 5mmol - 0 5 5 / 50=0.1 pH = 1.0 25 2.5 5 - 2.5 2.5 2.5/75=3.310-2 pH = 1.48

49 4.9 5 - 4.9 0.1 0.1/99=1.01 10-3 50 5.0 5-5 51 5.1 5.1 [OH] - 5 60 6.0 6 [OH] - 5 0 pH = 3.00 0/100 = 0 pH = 7 0 H+ 0.1/101=9.9e-4 pOH=3, pH=11 1 OH-1/110= 9.9e-3 Strong-Acids / Strong Bases Titration pOH=2, pH=12 2/27/20 Titration: Result Summary Summary of the titration of 0.100 M HCl with 0.100 M NaOH. Strong-Acids / Strong Bases Titration 2/27/20 Result: Titration Curve A 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH. What is the pH after addition of 0.00, 20.00, 49.00, 50.00, 51.00 and 60.00 mL of base. Titration curve between strong acid and strong base. Analyte is HCl and titrant is NaOH. Rxn: Rxn: NaOH

NaOH OH OH 100% 100% HHCl H+ 100% Cl H+ 100% + Net H + 2O Net H+ + OH OH- HH 2O Strong-Acids / Strong Bases Titration 2/27/20 Titration Curve In or Out of Class Exercise Design a problem in which NaOH (analyte) of some concentration and 50-mL volume is titrated with HCl of some concentration. Design the problem in such a way that the the following conditions are met. HCl Added pH soln 1 0.00 mL 13.500 2 10.00 mL 13.199 3 15.0 mL 12.988 4 20.0 ml 12.655 5 24.6 ml 11.530 6

25.4 ml 2.475 7 30.0 ml 1.403 8 35.0 ml 1.290 9 40.0 ml 0.977 10 50.0 ml 0.801 Strong-Acids / Strong Bases Titration 2/27/20 ... remember That a titration problem is nothing more than a Stoichiometry problem V ol (L) L iq u id p h ase D e n s ity (g / c c ) M ass (g ) C o n c . (m o l / L ) V ol (L ) P r e s s u r e ( a tm ) T e m p e ra tu re (K ) V o lu m e ( L ) # o f m o le c u le s / a to m s # o f m o le c u le s / a to m s 23 N N A v ( 6 .0 2 1 0 ) p a rtic le (a to m ic ) p h a s e M o la r M a s s ( g / m o l ) S o lid p h a s e A v

( 6 .0 2 1 0 2 3 ) p a r t i c l e (a to m ic ) p h a s e M o la r M a s s ( g / m o l ) S o lid p h a s e B a la n c e e q u a tio n M o l e s A M o l e s B S to ic . c o e ffic ie n t . A queous p h ase R (. 0 8 2 1 a tm L m ol K ) Strong-Acids / Strong Bases Titration R (.0821 M ass (g ) C o n c . (m o l / L ) A queous phase (g )G a s p h a s e G as p h ase V ol (L ) L iq u id (l) p h ase D e n s ity ( g / c c ) a tm L m ol K ) V ol (L ) P r e s s u r e ( a tm ) T e m p e ra tu re ( K ) V o lu m e ( L ) 2/27/20

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