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# QUANTITATIVE ANALYSIS CHEM 315

Texas A&M

GPA 3.63

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This 3 page Class Notes was uploaded by Dahlia Douglas on Wednesday October 21, 2015. The Class Notes belongs to CHEM 315 at Texas A&M University taught by Manuel Soriaga in Fall. Since its upload, it has received 16 views. For similar materials see /class/226234/chem-315-texas-a-m-university in Chemistry at Texas A&M University.

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Date Created: 10/21/15

CHM 235 Quantitative Analysis Spring 2007 Dr SA Skrabal SOLUTIONS TO PROBLEM SET 10 Monoprotic buffers and acidbase equilbria 10 April 2007 Please refer to Appendix G in your textbook for required values of Ka Please refer to Appendix G in your textbook for required values of Ka Note I have expressed values of pH and pOH to only two decimal places for convenience 1 A What is the pH of a 0075 M NH3 ammonia solution B What is the pH of a 0025 M NH4C ammonium chloride solution C What is the pH of 50 mL of a solution consisting of 0075 M NH3 and 0025 M NH4CI A Weak base Kb KMKa 100 x 1044 569 x 10quotquot 1 76 x 10395 x2 F x x2 0075 x Multiply out to obtain x2 1 76 x 105x 132 x 10396 0 Use quadratic formula to obtain X OH 114 X 10393 M pOH log1 14 x 103 294 pH 14oo 294 1106 Or assume X ltlt F to obtain X 115 X 103 M and pH 1106 Good assumption since 114 X 10393 lt 0075 B Weak acid K XZF X 569X103910 XZF x x20025 x Multiply out to obtain x2 569 x 1U x 142 x 103911 0 Use quadratic formula to obtain X H307 3 77 X 106 M pH log3 77x 105 Or assume X ltlt F to obtain X 3 77 X 106 M and pH 542 Good assumption since 3 77 X 10396 ltlt 0 025 C Buffer pH pKa logNH3NH4 924 log00750 025 9 72 2 A What is the pH of 250 mL of 0040 M benzoic acid B A solution is prepared by adding 250 mL of 20 M KOH to 250 mL of 0040 M benzoic acid What is the pH of the solution after mixing A Weak acid K 628 x 105 XZF x x2 0040 x Multiply out to obtain x2 628 x 105x 251 x 10396 0 Use quadratic formula to obtain X H 30 155 X 103 M pH log155 x 103 w Or assume X ltlt F to obtain X 158 X 103 M and pH 280 Good assumption since 158 X 10393 ltlt 0 040 B OH C5H5C02H C5H5C0239 H20 Strong base reacts With weak acid completely Initial mol C5H5C02H 0 250 L0040 molL 10 X 10392 mol C5H5C02H neglecting dissociation lnitialmol OH 2 50 X 10393 L20 molL 50 X 10393 mol OH Initial mol C5H5COZ39 0 neglecting dissociation Final mol C5H5C02H 10 x 102 50 x 103 50 x 103 mol Final mol C5H5C0239 50 X 10393 mol Buffer pH 420 log50 x 1039350 x 103 M 3 A What is the pH of 200 mL of a 00250 M sodium hypochlorite NaOCl solution This is the active ingredient in most bleaches B Suppose 100 mL of 014 M HNO3 is added to this solution What would be the pH C The solution in B is then diluted to a total volume of 500 mL What would be the pH A Weak base K mm 100 x 101430 x 108 33 x 107 XZF x xZo 0250 x Multiply out to get x2 33 x 10397x 825 x 109 o X OH 907X10395 M pOH 404 pH 996 Or assume K X2F to get same pH B Add 100 X 10393 L0 14 molH30L 14 X 10393 mol H30 HN03 is a strong acid Strong acid reacts with weak base completely pKa of HOCI 753 Initial mol OCI 200 X 10393 L0 0250 moVL 500 X 103 mol neglecting association Initial mol HOCI 0 neglecting association InitialmoIH30 14 x 103 mol Final mol ocr 500 x 103 14 x 103 36 x 103 mol Final mol HOCI 14 X 10393 mol Buffer pH 753 log36 x 10314 x 103 M C Moderate dilution will have no effect on pH because volume cancels out in numerator and denominator of term after the log in HH equation 4 A What is the pH of 2500 mL of 100 M hydroxyacetic acid also known as glycolic acid B What volume of 200 M NaOH must be added to 2500 mL of 100 M hydroxyacetic acid to produce a buffer solution of pH 400 A Weak acid K 148 x 10 XZF x xZ1oo x Multiply out to obtain x2 148 x 104x 148 x 10quot 0 Use quadratic formula to obtain X H307 121 X 102 M pH Iog121 x 102 Or assume X ltlt F to obtain X 122 X 102 M pH 191 Good assumption since X ltlt F B Strong base reacts with weak acid completely Initial mol HA 2500 X 10393 L100 moVL 0250 mol HA neglecting dissociation Initial mol OH from NaOH X Final mols of A39 after reaction X neglecting dissociation Final mols of HA after reaction 0 250 X neglecting dissociation A buffer has been formed so use HH equation pH pKa logmolA39mol HA pKa of hydroxyacetic acid 383 400 383 log mol A39mol HA 4 00 3 83 log molA39mol HA 017 logmolA39mol HA 10077 mol A39moI HA xo25o x 148 Multiply out to get 0370 148X X 248X 0370 0 X 0149 mol of OH needed 0149 mol OHL2 00 mol OH 745 X 10392 L or 745 mL ofNaOH needed Check work with HH eqn39 pH 3 83 log0 1490 250 0149 4 00 5 What mass of sodium formate HC02 Na FW 6801 must be added to 5000 mL of 100 M formic acid to produce a buffer solution that has a pH of 365 Remember that in a buffer initial concentrations of A39 and HA do not change signi cantly from what you start with pKa of formic acid 3 74 lnitialmols HA 5000 X 10393 L100 molL 0500 mol HA Use HH eqn39 365 374 logmol A390 500 mol HA 0090 logmol A390 500 mol HA 10009 mol A390 500 mol HA 0813 moIA39 08130 500 mol 0406moIA39 needed 0406mol HCOZNa6801 g HCOZNamolHCOZNa 276 g HCOZNa needed Refer to Table 92 in your textbook Harris 2007 to answer the following two questions 6 A mass of 1000 g of ACES is dissolved in H20 and brought to a volume of 500 mL A What is the pH of this solution B Suppose 120 mL of 120 M HCI is added to this solution What would be the pH A ACES is a weak acid K 1039 5 141 x 10quot F 1000 g ACES1 molACES182200 g ACES10 500 L 101 molL K XZF x 141x10397 x21o1 x Assume x ltlt 101 so K 141x107 x21o1 X H307 3 77X 10quot M assumption was good since 377 X 104 ltlt 101 pH 342 B Adding strong acid to weak acid SA dominates H3 0 concentration 120 X 10393 L120 mol H30L 0144 mol H30 this dwarfs the H30 contributed by the WA 0144 molH30512 X 10393 L 281 X 101 molH30L pH 055 7 Which of the substances in Table 92 would be suitable for preparing buffers with the following pH ranges A 67 i 02 B 70 i 01 C 80 i 02 Looking for pH z pKa A ADA BIS TRIS propane PIPES ACES B MOPSO imidazole HCI C HEPPS TRICINE glycine amide hydrochloride TRIS HCI 8 A What is the pH of 250 mL of a solution consisting of 015 M acetic acid and 026 M sodium acetate CH3C02N3 B Suppose 375 mL of 50 M NaOH was added to the solution in A What would be the pH C Suppose another 375 mL of 50 M NaOH was added to the solution in B What would be the pH A Buffer mols CH3C02H 0250 L0 15 molL 38 x 10392 mol HA mols CH3C0239 0250 L0 25 moVL 55 x 102 molA39 pH 4 75 log65 x 1039238 x 102 g B Add strong base OIf Initial mol HA 3 8 X 10392 mol HA neglecting dissociation Initial molA39 65 X 10392 me A39 neglecting dissociation Initial mol OH 3 75 x 10393 L50 moVL 19 x 10392 mol OH Final mol HA 38X 10392 19x10392 19x 10392 mol HA Final molA39 55 x 10392 19 x 10392 moi 84 x 10392 moi Still a buffer pH 4 75 log84 x 10219 x 102 M C Add more strong base Initial mol HA 19 X 10392 mol HA Initial molA39 84 x 10392 mol A39 Initial mol OH 3 75 x 10393 L50 moVL 19 x 10392 mol OH Final mol HA 19 X 102 19 X 102 0 mol HA neglecting association Final mol A39 8 4 x 10392 1 9 x 10392 10 x 10391 mol A39 neglecting association This is no longer a buffer it is a solution of A39 weak base CH 3C0 239 Kb KMK 100 x 10441 75 x10395 5 71 x 10quotquot XZF x x21o x10391 x Multiply out to getx2 571 x 103910x 5 71 x 103911 o x OH 755 x 10396 M pOH 512 pH 8 88 Or assume Kb X2F to get same answer

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