An artificial fruit beverage contains 12.0 g of tartaric acid (diprotic acid), H2C4H4O6

An artificial fruit beverage contains 12.0 g of tartaric acid (diprotic acid), H2C4H4O6, to achieve the tartness. It is titrated with a basic solution that has a density of 1.045 g/cm3 and contains 5.00 mass percent KOH. What volume of the basic solution is required?

studying for an exam and am stumped...
the answer should be 172 ml.....
thanks!!

1) Find moles of acid needed to be titrated
Molar mass of tartaric acid (calculated from molecular weights of each element) = 150 g/mol
12 g divided by 150g/mol = 0.08 moles tartaric acid
Because its a diprotic acid, that means two protons dissociate, so multiply molar mass times two (each mole of compound creates two moles of H+) so that makes 0.16 moles acid

2) That means you need 0.16 moles KOH to neutralize/titrate. Find the number of grams 0.16 moles refers to: 0.16 moles * 56g/mol = 8.96 grams KOH

3) If solution contains 5 percent by mass KOH, you need 20 times the grams of liquid solution (100%/5%): 8.96 times 20 =179.2 g solution

4) If density is 1.045 g/cm3, you need 179.2g divided by density 1.045 g/cm3 = ~172 cm3 or 172 mL

Tartaric acid's molecular weight is about 150, so 12.0 g is about 0.08 moles. Since the acid is diprotic, you need two equivalents of base to neutralize it, or 0.16 moles of KOH. A 5 % KOH solution contains 50 g KOH per liter and is thus about 0.89 molar in KOH. Therefore you need 0.16/0.89 x 1000 mL to neutralize it. That's 180 mL. Since I approximated some of my calculations, that could account for the 8 mL discrepancy in my answer.

Hi, I found a link where u can try.
http://zhun.info/139523/fruits-beverage