![You prepare your standardization solution by weighing out 0.0800 grams of standard ascorbic acid (176.12 g/mol) and dissolving t in 50.00 mL of water. You then add ten drops of starch indicator. How many milliliters of 0.014M triiodide soluton will you need to reach the endpoint of the titration? [(enter only the number]](http://img.homeworklib.com/questions/69e62300-c977-11eb-8b05-eb1fdcf49d9d.png?x-oss-process=image/resize,w_560)
Homework Answers
Mass of Ascorbic acid = 0.0800g
Molar mass of Ascorbic acid = 176.12g/mol
Volume ofAscorbic acid (V1) = 50 mL =0.050 L
Molarity of ascorbic acid(M1) = moles/volume = (0.0800g/176.12g/mol)/ 0.050L
= 0.00908 M
Ascorbic acid + triiodide 
dehydroascorbic + 2H+ + 3I-
From reaction:
One mole of ascorbic acid reacts with one mole of triiodide.
molarity of triiodide(M2) = 0.014 M
Let V2 be the volume of triiodide needed.
Using molarity equation:
M1V1=M2V2
0.00908M * 50 mL = 0.014 M * V2
V2 = 32.45 mL
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