A 40.00 mL wine sample is titrated with a 1.04 times 10^-1M solution of sodium h

Question

A 40.00 mL wine sample is titrated with a 1.04 times 10^-1M solution of sodium hydroxide. The equivalence point is determined to be 29.267 mL. What is the acidity of the wine expressed as (g tartaric acid/L, MW tartaric acid 3150.085 g/mole)? 5.71 11.4 11.42 3.38 5.710 The acidity of a wine sample is determined to be 14.21 g tartaric acid/L What is the mass of tartaric acid needed to prepare an equivalent concentration solution in a 50 mL volumetric flask? (MW tartaric acid =150.085 g/mole.) 710.5 0.7105 3.70 0.710 0710

Explanation / Answer

M1V1 = M2V2

M1 = 1.04 * 10-1 = 0.104 M ; V1 = 29.267 mL

M2 =? V2 = 40 mL

M2 = M1V1 / V2 = (0.104 * 29.267) / 40 = 0.0761 M

M = moles / vol (L) ; moles = M * V = 0.0761* 0.04 = 0.00304 moles od tartaric acid present

Weight = molecular weight * moles = 150*0.00304 = 0.4569

If 40 mL wine contains 0.4569 g of tartaric acid, then 1000 mL or 1L will contain 11.42 g. So the concentration expressed in g/L is 11.42.

15) 14.21 g/L of tartaric acid means 14.21 g of tartaric acid in 1000 mL of water. SO per 50 mL itequals to 0.7105 g of tartaric acid.

1000 mL = 14.21 g

50 mL = ?

50*14.21/1000 =0.7105 g o

So 0.7105 g of tartaric acid in 50 mL equal to that concentratoon

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