November 9 2012 16 2 continued calculating pH changes in a buffer solution Addition of a small amount of strong acid to a buffer converts a stoichiometric amount of the base to the conjugate acid Addition of a small amount of strong base to a buffer converts a stoichiometric amount of the acid to the conj base Example 16 3 A 1 0L buffer solution 0 100mol HC2H3O2 0 100mol NaC2H3O2 Ka HC2H3O2 1 8x10 5 a pH of this buffer solution pka logka log 1 8x10 5 4 74 pH pka log base acid pH 4 74 log 0 100 0 100 b pH after of 0 010 mol of NaOH is added to the buffer solution NaOH HC2H3O2 H2O NaC2H3O2 NaOH aq HC2H3O2 aq H2O l NaC2H3O2 aq Before 0 00mol 0 100 mol Addition Addition 0 010mol After Addition 0 00mol 0 090 mol 0 100 mol 0 110 mol 0 010 0 010 0 100 0 010 0 100 0 010 after completion of the reaction HC2H3O2 aq H2O H3O C2H3O2 0 090 mol 0 110 mol pH pka log base c how much would the pH be if 0 010 mol of NaOH was added to 1 0L of pure acid pH 4 74 log 0 110mol V 0 090mol V 4 74 0 087 pH 4 83 water NaOH 0 010mol 0 010M 1 0L OH 0 010M pOH log 0 010 2 00 pH 14 00 2 00 pH 12 00 November 14 2012 Buffer range pH pKa 1 lowest pH for effective buffer pH pKa 1 highest pH for effective buffer suppose you want a pH 3 00 pKa HA 5 00 pH pKa log base acid pH 5 00 2 pH 3 16 3 buffer effectiveness how much acid and base should a buffer contain Buffer effectiveness depends on 2 factors 1 base should be in the range of 0 10 to 10 2 a buffer with a greatest amount of acid and conjugate base is most acid resistant to pH changes and most effective Dilution makes the buffer less effective this requires base 0 01 this requires ratio of base acid to be outside the effective buffer range acid which is 0 10 10 not a good buffer in this case Buffer capacity the amount of acid or base that can be added to a buffer without destroying its effectiveness 16 4 titrations and pH curves equivalence point is the point in the titration when the number of moles of base is stoichiometrically equal to the number of moles of acid Titration of a strong acid with a strong base generates a pH 7 00 at equivalence point Same for titration of a strong base with a strong acid November 16 2012 1 titration of a strong acid with a strong base being added calculations euquivalence point volume pH before equivalence point pH at equivalence point pH after equivalence point consider the titration of 25 0 mL of 0 100M HCL with 0 100 M NaOH 1 calculation of the volume of NaOH required to reach equivalence point HCL 25 0mL 0 025 L 0 100M moles of HCL 0 100M 0 025L 0 00250mol HCl NaOH 0 100M Vequivalence completely disocaties because it is strong NaOH aq HCl aq H2O l NaCl aq Since HCl and NaOH are 1 1 they both have the same number of moles at equivalence nNaOH 0 00250mol now we know it because it s the same as HCl Vequivalence 0 00250mol 0 100M 0 0250L or 25 0 mL 2 find initial pH of solution before any NaOH is added what is present in the solution before NaOH HCl 0 100M in H20 HCl is a strong acid pH log 0 100M 1 00 pH 1 00 3 since you are adding a base it should increase the pH pH after adding 5 00 mL of NaOH HCL NaOH NaCl H2O Before NaOH 0 00250 mol 0 Addition After 0 0 5x10 3 mol 0 5x10 3 mol 0 00250 0 5x10 3 0 0 00200mol HCl after reaction and addition of NaOH 0 00200mol 0 0250 0 0050L 0 0667M pH log 0 0667M 1 18 4 pH at 25 0mL of NaOH pH at equivalence point Only species present are NaCl and H2O NaCl is pH neutral and H2O is 7 00 Therefore the pH at equivalence point 7 00 5 pH after addition of 30 0 mL of NaOH moles of NaOH 0 100M 30 0x10 3L 3 00x10 3 mol moles of NaOH remaining after equivalence 3 00x10 3 2 5x10 3 0 5x10 3 mol NaOH 0 5x10 3 mol 0 0250 0 030 L 0 00909 M H3O 1 0x10 14 0 00909 pH log 1 10x10 12 pH 11 96 November 19 2012 Section 16 7 16 8 will not be on the exam 4 The titration of a weak acid with a strong base Consider the titration of 25 0 mL of 0 100M HCHO2 with 0 100M NaOH Ka HCHO2 1 8x10 4 1 volume of NaOH required to reach equivalence point HCHO2 25 0mL 1000L 025L 0 100M 0025 moles HCHO2 HCHO2 aq NaOH aq CHO2Na aq H2O l Number of moles of NaOH 0 0025 moles NaOH Volume of equivalence point 0 0250 L 2 initial pH of HCHO2 before addition of NaOH this is a WEAK ACID use ice tabe HCHO2 aq H2O CHO2 aq H3O aq 0 00 x x 0 x x 1 1 ratio of HCHO2 NaOH NaOH 100 M and 0 0025 mol 0 100 0 00250mol V M n V I 0 100 M C E 0 100 x x Ka x2 0 100 x 1 8x10 4 x2 0 100 x can neglect x solve for x H3O 4 24x10 3 M pH log 4 24x10 3 pH 2 37 3 pH after addition of 5 00 mL of NaOH 5 mL of NaOH correspond to NaOH 0 100 M and 5 0 mL mol of NaOH added 0 100M 5 0x10 3 L NaOH added 0 50x10 3 mol HCHO2 NaOH NaCHO2 H2O NaOH Before addition 0 0025 mol Addition After 0 0025 0 50x10 3 mol 0 0 0 0020 mol 0 00 0 50x10 3 mol 0 50x10 3 mol To find pH you can use the Henderson hachalbach equation to determine pH The region of the curve before equivalence point in the titration curve of a weak acid or a weak base is called the buffer range pH pKa log NaCHO2 HCHO2 pH log 1 8x10 4 log 0 50x10 3 0 0020 pH 3 74 0 60 pH 3 14 4 pH after addition of 12 5 mL half of volume of NaOH half equivalence point V 12 5 mL equivalence point Vequivalence 25 mL At equivalence point half of the initial quantitiy of HCHO2 has reacted and formed the NaCHO2 HCO2 NaCHO2 pH pKa log NaCHO2 HCHO2 pH pKa pH 3 74 5 pH after addition of 25 0 mL of NaOH this will be a weak base at equivalence point all of HCHO2 has disappeared and formed NaCHO2 …
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