Calculate the ph for each case in the titration of 50.0 ml of 0.150 m hclo(aq) with 0.150 m koh(aq). use the ionization constant for hclo

PH = 7.58

KOH will interact with HClO to provide KClO. you may have an answer that contains unreacted HClO and KOH. this is often a solution.

HClO + KOH → KClO + H2O

HClO reacts with KOH in a very 1: 1 molar ratio

Mole of acid in 50 ml of an answer of 0.150 m = 50/1000 * 0,150 = 0.0075 mole of acid

Mole of KOH in 30 ml of 150 M solution = 30/1000 * 0,150 = 0.0045 mole of KOH

This reacts to provide 0.0045 moles of KClO and there are 0.0030 moles of HClO unreacted

The volume of the answer = 50 ml + 30 ml = 80 ml = 0.080 l

Deposition of acid in solution = 0.0030 / 0.080 = 0.0375M

KClO deposit in solution = 0.0045 / 0.080 = 0.0562 m

pKa HClO = -log (4.0 * 10 ^ -8) = 7.40

PH = pKa + log ([KClO] / [HClO])

PH = 7.40 + notes (0.0562 / 0.0375)

PH = 7.40 + 1.50 notes

PH = 7.40+ 0.18

PH = 7.58

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Mole of Acid  

HClO  

Details

Grade: College

Subject: Chemistry

Keyword: mole, acid, HClO

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