# chemistry

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Titration of 25.00 mL of KOH required 200.00 mL of 0.0050 M acetic acid, CH3COOH. What is the molarity (M) of the KOH solution? Please show calcutions.

• chemistry -

HAc = CH3COOH = acetic acid
KOH = potassium hydroxide
mL KOH x M KOH = mL HAc x M HAc
Substitute and solve for M KOH

• chemistry -

0.04

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