Answer to Question #134349 in General Chemistry for neyola

Question

2. Tartaric acid, H2C4H4O6, is often present in wines and precipitates from solution as the wine ages. A titration was carried out to determine the concentration of the acid solution. Tartaric acid reacts with sodium hydroxide in the following reaction:
H2C4H4O6(aq) + 2 NaOH(aq) Na2C4H4O6(aq) + 2 H2O(l)
a) If 20.00 mL of the tartaric acid solution required 22.62 mL of 0.1185 M NaOH to neutralize, calculate the molarity of the tartaric acid solution.
b) Using the answer from part (a), calculate how many grams of tartaric acid are present in a 750.0 mL bottle of wine?

Answer 1

 a)

Tartaric acid reacts with sodium hydroxide in the following reaction:

H2C4H4O6(aq) + 2 NaOH(aq) Na2C4H4O6(aq) + 2 H2O(l)

From the Above reaction Tatric Acid have two lossing H⁺ .

Let Molarity of  tartaric acid solution is y M .

So , moles of  tartaric acid in solution  = 20 * y / 1000 .

So , moles of NaOH Required = 2 * 20 * y / 1000 = 0.04y .

So , 0.04y = 22.62 * 0.1185 / 1000 = 0.0026 .

So , y = 0.0026 / 0.04 = 0.065 M . Answer .

b)

volume of Solution = 750 ml .

Molarity of Solution = 0.065 M .

Moles of  tartaric acid = 750 * 0.065 / 1000 = 0.04875 moles .

Molar mass of  tartaric acid = 150.087 g / mol .

Mass of  tartaric acid = moles of  tartaric acid * molar mass of  tartaric acid .

= 0.04875 * 150 .087 = 7.316 g answer .