Asked Oct 30, 2019

Expert Answer

Step 1

Given that,

Mass of acetic acid = 0.1 M

Volume of acetic acid= 25mL

Mass of KOH = 0.1 M

Step 2

A: when 0.00 mL of KOH is added

Firstly we calculate the mole


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Mole of CH,COOH-0.1M 25.0mL 2.5 mole Mole of KOH 0.1M x 0.00mL -0 mole remaining mole of CH,COOH-2.5 mole Total volume 25.0mL mole of acid remaining Н" volume 2.5 mole 25.0mL pH-log [H -log (0.1) н" pH 1.0

Step 3

B: When 12.5mL of KOH added

Mole of acetic acid = 2.5 mole

Mole of KOH = 1.25 mol

1.25 moles of both w...


Image Transcriptionclose

1.25 mole н]- 37.5mL 33.333х10°м PH-1og (3.333x 10*м) 1.48


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