Question

When a 21.1 mlL sample of a 0.415 M aqueous hypochlorous acid solution is titrated with a 0.394 M aqueous barium hydroxide solution, what is the pH after 16.7 ml. of barium hydroxide have been added? pH=

Answer 1

2 HOCI + BaCoth —> 240 + Balcios Amoles of Hoc reach with Imole Bacond HOCI = 21.1m1, 0.415M = 0.415 molleit number of moles = Molarity x volume Lin L) Given Ăn: 6.His molit) x(21.1x10-3L) = 8.7565X10-3 mois 58.76 X163 mol Ba(OH), M = 0.194 M. V=16.7 m 316.7x16?L na (0.394 x16PX16.7) moll = 6.5748 x16? mol = 6.58X10? mol q HOCI + Ba(ohh ~ Initially: 8.76X103 6.58x10? allo + Ba(CIO) change - ax ax +x Finally: 8-76815222 6.58x16?- In this reaction Imole Ba(OH), reach with a mol of HOC ; 8.76X16-3 8.76X10- -= 4.38X1omol Bacow) Yeacted.

Finally: 8.76X169-8-76* : 8.76X16-7-8-76x103. HOCI 7 Q2 = 8.76X10? = x= 4.38x10-3 Ba(OH) = 6:58x163X = 6·58x16?- 4. 38x16? = 2-2x16?mo) :: In the solution 2.2X16 "mol of BacoH) is present. Ba(OH) is strong acid. BaCOH Concentration 5 moles = 2:2X163 mol Total volume of solution 21.1m1 +167mi = 37.8 mi Ms Wines = a.2x16 *ml = 37.88162 L 37-8X10-31 M= 0.058 molle Ba(OH), Bata? +2014 . [on] = 2 (0.058) = 0.116 M polt, -log [or] = -109 (0.116) = -(-0.94) POH = 0.94 we know that pH + por = 14 .. pH = 14 - pol = 14-0.94 = 13.06 .: pH = 13.06