The following procedure was used to determine the percentage of acetic acid, CH3COOH, in a commercial sample of vinegar:
(i) Sodium hydroxide solution of unknown concentration was standardized using KHP (potassium hydrogen phthalate, molar mass=202.2g.) 13.5 mL of NaOH was required to neutralize 2.60 g of KHP.
(ii) The standardized NaOH solution was then used to titrate a 10.0 mL sample of commercial vinegar. 8.50 mL of NaOH was required to reach the end point.
Answer the following:
a) What is the molarity of the NaOH solution
b) What is the percentage of acetic acid in commercial vinegar? Assume that the density of commercial vinegar is 1.00 g/mL.
I did part a and got .95 M. Did I get part a correct and how on earth do I do part b?? I really need a step by step answer so that I can understand how to solve not only this problem but also similar problems.
Thanks so much and 10 points to a best answer!!Moles KHP = 2.60 g / 202.2 = 0.0129 = moles NaOH
Concentration NaOH = 0.0129 / 0.0135 =0.952 M
Moles NaOH in 8.50 mL ( = 0.0085 L) = 0.0085 L x 0.952 mol/L= 0.00809 = moles CH3COOH in 10.0 mL
Mass CH3COOH = 0.00809 mol x 60.05 g/mol = 0.486 g
% = 0.486 x 100 / 10 .0 = 48.6