Question

Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.160 M KOH(aq), with 0.160 M HI(aq). Number Note: Enter your answers with two decimal places. (a) before addition of any HI Number (b) after addition of 13.5 mL of HI Number (c) after addition of 23.5 mL of HI Number (d) after the addition of 35.0 mL of HI Number (e) after the addition of 40.5 mL of HI Number (1) after the addition of 50.0 mL of HI

Answer 1

molarity Volume of kon of KOH (m) = 0.160m (W = .350 mL Fon KOH solution MV = 0.160X3550 nmol concentration of kom = 506 mmol molarity of HI = 0.160. Ansorer la before addition of any HI. Pon of the ROH solution = - dog [ou] = -dog10160) = -(-0.7958) pol = 0.7958 Hence pH = 14- pon. = 14 - 0.7958 TpH = 13.20 Answer (b). After addition of 13.5 mL of 0.160M HI. for HI solution mv = 13.5*0.160 mmol = 2.16. mmol after addition of HI Premaning commount of ROH in = (506 – 8.16 ) mmol 5 3.44 mmol Tatal volume of the solution = 3550 mil + 13.5 mL = 48.5 mL concentration of remaining :ROH = 3.44 mmol = 0.0709m 48.5 mL

Hence pon of the solution - - log [on-] = -log 10.07092) =-(-401492) = 11492 Hence pH = 14-POH = 14-111492 PM = 12.85] Answer (C) af for addition of 93.5 mL of HI: Then milimples of HI 93.5 x 0.76 mmol . = 3.76 mmol remaining mmoles of KDH affer addition of HI is = 1516 – 3.76 ) mmol = 1.84 mmol dutal volume of solution = (23.5 + 35:0) mL = 58.5 mL Hence remaining concentration of kom = 1.84 M - 58.5 M so pon of the solution = 0.03145 = -log[on] = -(-4:50-23) = 105023 Henie ; pH = 14- POH = 14-105023 pH = 12.49]

Ansivos (d) after addition of 350 mL of HT for HI milimoles = 35.0% 0.160 mmol I sob mmol Antal volume of solution = 35.0 + 35.0 = 70 mL remaining concentration in mmol of KOH = 506 - 5.6. = 0 Here OH concentratim in 2010 meams have affrey addition of HI the kon solution is completly metredized. I pH =7000 Answer (e often addition afhen addition of 40.5 mL of HI means milimoles of HI = 40.5 x 0.160mmol = 6.48 mmol Here mmoles of HI is more than the KOH minder hence the remaining mmol of HI = 6.48 - 516 = 0.88 mmol Total volume of solution = 4015+3500 = 7500 ml concentration of remaining HI Solution = GD = 0011737 Hence pH = - до Гн11 = -log(0.01173) = - (1993) TpH = 1.937 1 Answer (f) after addition of 5000 mL of HI milimoles of HI = 50.0 X 0.160 - 8 mmol Mere mmol of HI is more than KoH mmol Hence the

remaining mmol Tatal volume of = of MI = solution = 8-5.6 5000 + 35.0 2.4 mmol = 85 mL Remaining concentration of HI = = 0.02823 Hence PM = -Jog [+] = -log(0.02823) = -61.549) py = 1.55