2] What is the difference between the end point and equivalence point for a monobasicmonoacid

Transcription

1 4 Titrations modified October 9, ] A solution of M AgNO 3 is used to titrate a ml solution of M KCl. The K sp of AgCl is 1.8e-11 a) What is pag if ml of the titrant is added to the KCl solution? b) What is pag if ml of the titrant is added? c) What is pag if ml of the titrant is added? 1 2] What is the difference between the end point and equivalence point for a monobasicmonoacid titration? 2 3] What is the final ph if solutions of mL of M NaOH(aq) and 75.0-mL of M CH 3 COOH(aq) are added together? 3 4] When titrating a weak acid with a strong base it is expected that the equivalence point will be 4 a) Slightly acidic, since the equilibrium HA = H + + A - predominates. b) Neutral since [OH - ] = [H + ] c) Slightly basic since the equilibrium A - + H 2 O = HA + OH - predominates. d) It is impossible to know since it could be acidic or basic depending on the K a of the acid. e) Strongly basic since excess OH - is present. 5] A sample solution of ml M oxalic acid (H 2 C 2 O 4 ) is titrated with ml of M of NaOH. Which of the following is true after the two solutions are mixed? 5 a) This is the first equivalence point. b) This a ph buffer region where [H 2 C 2 O 4 ] = [HC 2 O 4 - ] c) This is a metal buffer region d) This is the second equivalence point. e) This is the excess OH - region where the ph is strongly alkaline. 6] A sample solution of ml M oxalic acid (H 2 C 2 O 4 ) is titrated with ml of M of NaOH. Which of the following is true after the two solutions are mixed? 6 a) There is 2.50 mmol of H 2 C 2 O 4 present.

2 b) There is 1.50 mmol of OH - present. c) There is 2.50 mmol of HC 2 O 4 - present. d) There is 1.00 mmol of H 2 C 2 O 4 present. e) There is 1.00 mmol of C 2 O 4 2- present 7] What is the ph of a titration solution that consists of M CH 3 COOH (K a = 1.75e-5) and M NaOH? 7 8] Calculate the ph of a mixture of ml of M NaOH and ml of M H 3 AsO 4 (arsenic acid). 8 H 3 AsO 4 = H 2 AsO H + K a1 = 5.8e-3 H 2 AsO 4 - = HAsO H + K a2 = 1.10e-7 HAsO 4 2- = AsO H + K a3 = 3.2e-12 9] A sample solution of ml M oxalic acid (H 2 C 2 O 4 ) is titrated with ml of M of NaOH. Which of the following is true after the two solutions are mixed? 9 a) This is the first equivalence point. b) This a ph buffer region where [H 2 C 2 O 4 ] = [HC 2 O 4 - ] c) This is a metal buffer region d) This is the second equivalence point. This is the excess OH - region where the ph is strongly alkaline. 10] A solution of ml M oxalic acid (H 2 C 2 O 4 ) is mixed with ml of M of NaOH. What is the ph of the final solution? 10 a) 2.04 b) 3.55 c) d) e) ] 75.0-mL of M CH 3 COOH(aq) (pka = 4.757) is titrated with M NaOH(aq). What is the volume of NaOH solution required to reach the equivalence point?

3 What is the ph of the equivalence point? 11 12] A] Calculate ph of ml of M HA with pka = 6.15 before any base is added. B] ph when 3.00 ml of M NaOH is added. C] ph when ml of M NaOH is added. D] ph when ml of M NaOH is added ] A 50.0-mL, M solution of CH 3 COOH (aq) is titrated with 0.25 M NaOH(aq). Calculate the volume of titrant and the ph s needed to reach of 0.75, 1.00, and ] A g sample of a mixture of Na 2 CO 3, NaHCO 3, and inert impurities was titrated with M HCl, requiring 15.7 ml to reach the first end point and a total of 43.8 ml to reach the second endpoint. What is the percent Na 2 CO 3 and NaHCO 3 in the mixture? 14 Answers 1 Titration Rxn: Ag + + Cl - AgCl(s) a) Moles of excess Cl - = L * M KCl L * M AgNO 3 = 5.00e-3 mol Cl - [Cl - ] = 5.00e-3 mol Cl - / L = 3.33e-2 AgCl(s) Ag + + Cl - -- x x+3.33e-2 x(3.33e-2) 1.8e-11 x = 5.4e-10 M pag = 9.27 b) This is the equivalence point. AgCl(s) Ag + + Cl - -- x x x 2 = 1.8e-11 x = 4.24e-6 pag = 5.37 c) We are past the equivalence point. ( ) ml * M Ag + ( ml) = M Ag + or pag = 1.699

4 2 The equivalence point is where mol acid = mol base and the end point is where indicator changes color. 3 mol CH 3 COOH = 75.0-mL*0.100 M = 7.50 mmol mol OH - = mL* M= 10.0 mmmol excess OH- = mmol = 2.50 mmol [OH - ] = 2.50 mmol/275-ml = 9.09e-3 poh = ph = C 5 D 6 e: 1.00 mmol past 1st eq. pt. Initial HA- = 2.5 mmol HA - + OH - =A 2- + H mmol A 2-7 ph = pka = nd eq. pt where all H 3 AsO 4 is titrated to HAsO 2-4 which is intermediate of H 2 AsO - 4 and AsO 3-4. These two equilibria become important: H 2 AsO - 4 = HAsO H + K a2 = 1.10e-7 HAsO 2-4 = AsO H + K a3 = 3.2e-12

5 Therefore ph = ½(pK a2 + pk a3 ) = ½ ( ) = This is the second equivalence point. 10 2*0.05*50 = 5 mmol H 2 C 2 O *60 = 6 mmol NaOH excess OH - = 6 5 mmol = 1 mmol [OH-] = 1mmol/110 ml = 9.09e-3 poh = ph = eq. pt. mol CH 3 COOH = mol OH - mol CH 3 COOH = 75.0-mL*0.100 M = 7.50 mmol vol NaOH = 7.50 mmol/ M = 150-mL total volume = 150-mL+75.0-mL = 225-mL [CH 3 COO - ] = 7.50 mmol/225-ml = 3.33e-2 M CH 3 COO - + H 2 O = CH 3 COOH + OH e x -- + x + x K b = K w /K a = 5.71e-10 = x 2 /(3.33e-2-x) x 2 /(3.33e-2) x = 4.36e-6 poh = ph = A] HA= H + + A -

6 x +x +x x 2 / x = 7.08e-7 x = 1.19e-4 [H + ] = 3.92 B] Initial HA = * = 1.00 mmol OH- added = mmol Rxn: HA+ OH - = A - + H 2 O 1.00 mmol Buffer system K a = [H + ][A - ] / [HA] [H + ] = K a [HA] / [A - ] = 7.08e-7*0.700 / ph = 5.78 C] OH- added = 1.00 mmol we are at the eq. pt. Rxn: HA+ OH - = A - + H 2 O mmol mmol mmol [A - ] = 1.00 mmol / ml = M Rxn: A - + H 2 O= HA+ OH x +x +x K b = K w / K a = 1.41e-8 = x 2 / ( x)

7 x = 1.533e-5 ph = 9.19 D] Excess OH - = 0.10 ml * M = mmol ph = = C b /C a = mol b /mol a [OH - ] = mmol / ml = 1.66e-4 M mol a = L*0. 150M = 7.50e-3 mol CH 3 COOH for = 0.75 mol b = 5.63e-3 mol vol NaOH = 5.63e-3 mol * 1/0.25 M = 2.25e-2 L = 23-mL for ph Rxn: CH 3 COOH + OH - = CH 3 COO - + H 2 O 7.50e e-3 0mol -5.63e e e-3 mol 1.87e e-3 mol [CH 3 COOH] = 1.87e-3 mol/0.073 L = 2.6e-2M [CH 3 COO - ] = 5.63e-3 mol/0.073 L = 7.7e-2 M Ka = 1.75e-5 = [H + ][7.7e-2]/[2.6e-2] [H + ] = 5.9e-6 ph = 5.23 for = 1.00 mol b = 7.5e-3 mol all HA is consumed since mol b = mol a therefore vol NaOH = 7.5e-3 mol*(1/0.25 M) = 3.0e-2 L or 30-mL [A - ] = 7.5e-3 mol/ L = M

8 A - + H 2 O = HA + OH x x x x 2 /0.094-x = K b = K w /K a = 1e-14/1.75e-5 = 5.71e-10 x = 7.3e-6 poh = 5.14 ph = 8.86 for = 1.25 mol b = 9.4e-3 mol vol NaOH = 9.4e-3 mol*(1/0.25m) = 38-mL region of excess OH - mol OH- = 9.4e-3 7.5e-3 = 1.9e-3 [OH - ] = 1.9e-3 mol/0.088 L = 2.2e-2pOH = 1.67 ph = The first end point is due to: H + + CO mol of CO 3 2- = mol/l * 15.7 ml = 1.71 mmols The 2 nd endpoint is due to: H + + HCO 3-2CO 3 Remember that the second endpoint is due to HCO 3 - from the first endpoint (Na 2 CO 3 ) and from HCO 3 - from NaHCO 3. Therefore HCO 3 - from NaHCO 3 is calculated as: mol/l * 43.8 ml - 2 * 1.71 mmols = 1.35 mmol % Na 2 CO 3, = 1.71e-3 mol * g/mol * 100/0.527g = 34.4% % NaHCO 3 = 1.35e-3 mol * 84.0 g/mol * 100/0.527g = 21.5%