Question

stion 2 of 18 ) Calculate the pH for each case in the titration of 50.0 mL of 0.140 M HCIO(aq) with 0.140 M KOH(aq). Use the ionization constant for HCIO What is the pH before addition of any KOH? pH = What is the pH after addition of 250 mL KOH? What is the pH after addition of 35.0 mL KOH? S15 PM BadC TE RT Y U o P DF CV B NM <

Answer 1

Solution :- Given thout :- = 500ml Helo = 0.140 colo kolt cas pH before additan of Kott. obtain pH of 0.140 m Holo. ka - [Hot ] [lo] . [Helo ] - Hioxio-8 ---- HO X10 2.57048 x com. [#30t] =746 x 105 m - pH log (A me x 10" H012. . . . Initial moles of Hclo 5000 to 3 Lix 0-100 md Halo 5.00 x 10 mol Holo. Initial concentration of KOH (OY) OH is o.1o.my

(b) moles of ot in 250 ml of 0-40 m Koh 0.100 molo 2510 x L xol - =0.0025 mol OH co + H₂O obtain Ewilibrium moles Helo tot Inital mol) 0.00500 ou oodso change (mol) -0.00250 -0.002so Fruilibrium (mol) <0.00250 20 -40.00 250 0.00250 pH = pka + log + [clo [Helo] cc1c9(Hierto-8) + log 6-60350 (0-025 0.003s0 /0.075 stog 7:39 + log Contato - 440) HO

C) moles of ot in 3510 ml opro 140 M Kolt 35104103 mx 0.100 mol ott - -0.00350 mol oH obtain cauilibrium molee. I tido t ott Intial moly 0.00560 - 0.0356 clot H2o 0 -0.00350 -0.00350 +0.00350 0.00150 o zuilibrium (moly 0.00350 pH = pta + log ] (clo [tico] E-log Chee xro 8 + log 0.00350 0-085 0.0015 0 0-085 Enloq (4:39) tlog (O.olit 0.0176 & 7.76 1777

(o) moles of oth in 50.00 ml of 0.140 m kott : 50.00 x10 3 L 0.100 mot ot . - - 0.005OD mol obtain equilibrium moless- co + H₂. fclot of Temol) Ovesoo 0.00500 clmol) - 0.00500 -0.00500 E (mod) ao ao +0.00500 0.00500. 0.00500 mol 0050 M. 0.00 L Holo tot (doj obrain pH- . co + H2O (1) Im 0.0500 ссм) x x EM) 0-0500 x kb 2x² be . 00.500-2 40x10-14 4.0x108 . ott = 112 x 10h M

pot = log [ot ] =3.95 pH = 14.00 -pott - 10.05 . e) moles of ot PR Coomb of 0.140 M koht: 60.0 xlo + x 0.100 mol op = 0.00600. mol OH - + H₂O obtain Equilibrium moles : Hiclo + ott - Initial (mol) 0.00500 0.00600 Ċ (mol) -0.00500 E (mol) ao 0.00190 -0.00500 O 00500 0.00500 Tot 7= 0.00100 mol - 0.110 L = 0.00909 M pott pf a-log (OH) = 2.04 itkop - port = 1196