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

beta decay is the answer