Vinegar is an aqueous solution of acetic acid, ch3cooh. a 5.00 ml sample of a particular vinegar requires 26.90 ml of 0.175 m naoh for its titration. what is the molarity of acetic acid in the vinegar?

The molarity of acetic acid in the vinegar is 0.94 M

calculation

Step 1: write the balanced equation between CH3COOH + NaOH

that is CH3COOH + NaOH → CH3COONa + H2O

step 2 : find the moles of NaOH

moles = molarity x volume in L

volume in liters = 26.90/1000=0.0269 l

moles = 0.175 mol / L x 0.0269 L = 0.0047 moles of NaOH

Step 3: use the mole ratio to find moles of CH3COOH

that is the mole ratio of CH3COOH: NaOH is 1:1 therefore the moles of CH3COOH is = 0.0047 moles

Step 4: find the molarity of CH3COOH

molarity = moles/volume in liters

volume in liter = 5.00/1000 = 0.005 l

molarity is therefore=0.0047 moles / 0.005 l = 0.94 M