Question

Calculate the pH for each case in the titration of 50,0 mL of 0.180 M HCIO (aq) with 0.180 M KOH(aq). Useioa constant for HCIO What is the pH before addition of any KOH? pH= | 4.07 What is the pH after addition of 25.0 mlL KOH? 7.40 pH = Question Source: MRG-General ChemistryPublishe University Science 415 PM A d ENG 11/5/201

Answer 1

NOTE- Ka of HClO is here taken as $2.95\times 10^{-8}$ because it is the right value. And all the answers are according to it.

If u take Ka of HClO as $4\times 10^{-8}$ ,then you will get exactly same answer value as written in answer box.

I have done according to the actual value. Every calculations is right and procedure is correct .

ka otHdo = a-95 x/o Befre the additian of koH o Find the p of oIs0M Hdo meane we have As Huo i a Weau auid t dissouales. Hdo initial O-18 chamge At ulbri is-x х - х Ka 2-31a Hene 2 2-45 X10 We can assue that x ii swall relaire h Hhe Small inihal on ceuhaha of Hdo because ka it Ve 2 a-a5 x lo= Thue fore O18 a.4s XI0° 0. 0000 72,8 H](uo] = o-0000728 M pH r -lalo.0008] 413 PH- 43

Th minchu will be buffer au kot i Story base mel ttlo i weak aud Vohume ot tlo = 50 0 mL -0.05L Molau of elo 0-1s0 M Vohmu of koH a5 0mL 0.025 L Molasdy of kon 0.180 M Equatian, Hdo t KoH Kou t o (ag) laa) Moles of Ho O 05LXO 1DM 0.009 moles Moles of KoH O.0a5 L X0 120M = o.0045 moles Henu Hod w be left unreaute Urreatee tou = 0.00q- 0 0045=0.0045 moles Now total Vohm 50 +95 -15 ml = 0 .075 L Hou= Woles of Hal rema Volume hota welesity of The = 0.06M O 075 Molar of clo" = moles of uo (kou) femed Total Vohume O.0045 1 : 0.06M

As this i a buffer, We witt use Homdesr son -Hasselbalch equahan to find pt. pt- pka t Casid Ka =a.g5 xI6 pka 7-53 6.06) pt 53 +log [90.0 pH-53