Homework Answers
Answer:-
This question is is solved by using the simple concept of reaction of strong acid and strong base which ionizes completely and give neutral solution at equivalance point.
The answer is given in the image,
![Answers (4404) = 0.198 M Thox] = 0.157 HClO4 + KOH - ka04 + H20 cas 0130 +3= [1404) = 0.198 Pus-log[H30+3 =- lo, (0.198) PH=](http://img.homeworklib.com/questions/384ae3e0-77ae-11ea-abb0-2bed245c8503.png?x-oss-process=image/resize,w_560)
![= 1600m2 Excess kou added-20-6-16-0 = 4.6mi [on) = 0.157x4.6 (12.7+2006) [OH-] = 0.0217 M pons-log con] =- log (0.0217) POH](http://img.homeworklib.com/questions/38b84210-77ae-11ea-911b-094cc73960e4.png?x-oss-process=image/resize,w_560)
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