Question

15.1
A 39.8 mL sample of a 0.448 M aqueous hypochlorous acid solution is titrated with a 0.234 M aqueous potassium hydroxide solution. What is the pH after 50.5 mL of base have been added?

15.2

What is the pH at the equivalence point in the titration of a 24.9 mL sample of a 0.313 M aqueous hypochlorous acid solution with a 0.479 M aqueous barium hydroxide solution?

12.1

How many grams of solid ammonium bromide should be added to 1.00 L of a 0.101 M ammonia solution to prepare a buffer with a pH of 9.990 ?

12.2

How many grams of solid sodium cyanide should be added to 1.00 L of a 0.132 M hydrocyanic acid solution to prepare a buffer with a pH of 8.401 ?

Answer 1

is. I ... pla of Hocd = 7:53 initial moles of Holl = 39.8 ML X 0.448m = 39.8x163x *0.448 molly = 17-8304x16%mo? Imitial moles of KOH = 0.234 mol x 898 50.5 mL 3 11.817x1ős mol Hodt. KOH + H20+ Koch

Hou and Kou react in til ratio . moles of kod tormed = moles of KOH moc = 11.817x1ó? mol mole of Hod left in solution nuoa = = 17.8304 x16?-11.877xlo?mos = 6:0134 xromol Henderson Hallalbalch equation is uned for pH of buffer solution. HOCI KOU РЧ, 109 * NHOCO 57-53+ log 11.8178163 6.0134x103

pH = 7.53+0.2934 = 7.8234 15.? At the equivalence point all Hod is converted to · Roch . .. moles of kod = initial moles of Hoch = 24.9 mL x 0.313M -707937x10 Simol Volume of BacoH 2 required = 707937x183 mol . 2 x Bacollz.cone. 7.7937X163 mol 2X0.479 mol/L = 8.135 ml Total volume = 24.9 mL +8.135 mL = 33.035 mL

conci of kold at . molel of kod equivalence point - Total volume = 7.7937X163 mol 33.035 mL. . . koci].0.23592.m pkb of koll = 14-pka = 14-7.53 = 6.47 .. pkb log kh Kb = 186.47 = 3.38x107 Kod undergoes hycholy na at equivalence point kod + H2O HOCI + oHS I 0.23592 C -2 E 0.23592- da +x . +n Kb = HOCIAZIOHJE * KOCIJE 0.23592-X

Neglect a in denominator because it is small 3:38x167 = x2 => x=023592x3:38 x1o7 . 0.23592 32= 07.974 x16 8 =) *= 17.974 x104 [OH-] = x= 2.824.x107m pOH = 109HTE 34092-894x1õ4 = 3.549 pH= 14-pol = 14-3.549 PM - 10.451

plate of NH3 = 4.75 , p. 14-24=14-9.99 Henderson Hamalbalch equation in poM = pkb + log moles of NHA BV moles of NH3 46 4.5 4 103 moles oÝ NHA RY IX x 0.101 molly log moles of NMA BY 0.101 mol = 4.01-4.75 = -0.74 moles of NHA BY 0.101 mol. 150.74 - 3 = 0.18197 58. moles of NHA BY = 0.101X0.18197 mol = 0.018379 moi

magn of NH BV should be added = 0.018379 mol x molar mash of NHq BV = 0.018379 mol x 97.94 g/mol @ 1.8 ġ 12.2 pra pka + log moles of NaGN I moles of HCN , pea of H(N = 9.21 / - na ľ 8.401-9-21 = 109 moles of the CN 0.132 mol x 14 IX 1X moles of NaCN = 0.132 x 160.807= 0.0204915 mol mau of Nach . should be added . molar man = 0.0204915 mol X Of NACN - = 0.0204915 9/61 x 49.00729 maól = 1.0049 (007 v