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if 40.70ml of standard 0.3173 m KOH reacts with 65.50ml of acetic acid what is the...

if 40.70ml of standard 0.3173 m KOH reacts with 65.50ml of acetic acid what is the molarity of this acid solution

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Answer #1

a. Convert molality of KOH to Molarity:

0.3173 m means 0.3173 moles /kg solvent (water).

0.3173 X 56.1 g/mole = 17.8 grams KOH.

Total mass of solution = 1000 grams solvent (water) + 17.8 grams HF = 1017.8 grams solution

Total Volume of solution = 1017.8 grams X 1mL/2.12 grams (density) = 480.09 mL = 0.48009 L

Molarity of KOH = 0.3173 moles solute / 0.48009 L = 0.6609 M KOH

M1V1 = M2V2

M1 = Molarity of KOH = 0.6609 M

V1 = Volume of KOH = 40.70 mL

M2 = Molarity of Acetic acid = ?

V2 = Volume of Acetic acid = 65.50 mL

M1V1 = M2V2

0.6609 * 40.70 = M2 * 65.50

M2 = 0.4107 M acetic acid solution

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