I was given a question that I got wrong, then was given the answer...
The question is as follows:
"In a titration experiment you used 30 mL of vinegar and 70 mL of NaOH. If the molarity of the base is .500 M calculate the concentration of acetic acid in the vinegar in % w/v."
Apparently the correct answer was 7%, but I am stumped as to how that is....
Please help me understand. Best step by step explanation will get best answer.
The question is as follows:
"In a titration experiment you used 30 mL of vinegar and 70 mL of NaOH. If the molarity of the base is .500 M calculate the concentration of acetic acid in the vinegar in % w/v."
Apparently the correct answer was 7%, but I am stumped as to how that is....
Please help me understand. Best step by step explanation will get best answer.
-
moles of base = (0.500 mol/L) (0.070 L) = 0.035 mol
the acid and base react in a 1;1 molar ratio, so 0.035 mol of acid present.
grams of acid = (0.035 mol) (60.0516 g/mol) = 2.102 g
w/v percent = (2.102 g / 30 mL) (100) = 7.0%
the acid and base react in a 1;1 molar ratio, so 0.035 mol of acid present.
grams of acid = (0.035 mol) (60.0516 g/mol) = 2.102 g
w/v percent = (2.102 g / 30 mL) (100) = 7.0%