Determine the pH during the titration of 22.2 mL of 0.283 M perchloric acid by 0.184...

Question

Question

Determine the pH during the titration of 22.2 mL of 0.283 M perchloric acid by 0.184 M potassium hydroxide at the following p

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Answer 1

Answer #1

Answer:-

This question is is answered by using the simple concept of titration of strong acid and strong base. Using the simple stoichiometry.

The answer is given in the image,

Anscells (HUON] = 0.283M [KOU] = 0.184 M H404 +KOH + KAO4 +H20 1) [H+] = CHU04) = 0.283 pH=- log (H+) =- loy(0-283) PH= 0.55

Excess mores q koh added = 7.8752x10^-6-2826X10-3 = 1.5926 X10-3 mol. con] = 1:5926 X10-3x1000 - 6500 [04] = 0.0 245 POH= -lo

Answer 2

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