A sample of white wine is to be analyzed for its tartaric acid content by titrat

Question

A sample of white wine is to be analyzed for its tartaric acid content by titration with a standard solution of NaOH. A 30mL sample is diluted to 75mL and 23.82mL if this diluted solution is titration with 22.22mL of .0522N NaOH. Assume both ionizable hydrogen atoms react with the base and calculate the percent tartaric acid (H2C6H4O6) in white wine. (The density of the wine is 1.00) A sample of white wine is to be analyzed for its tartaric acid content by titration with a standard solution of NaOH. A 30mL sample is diluted to 75mL and 23.82mL if this diluted solution is titration with 22.22mL of .0522N NaOH. Assume both ionizable hydrogen atoms react with the base and calculate the percent tartaric acid (H2C6H4O6) in white wine. (The density of the wine is 1.00)

Explanation / Answer

From the given data -

The number of equivalents of NaOH reacted with tartaric acid = (0.02222 L) (0.0522 mol/L)

                                                                                        = 0.0011598 eq

Same number of equivalents must be present in the sample of tartaric acid of diluted solution.

Thus Normality of tartaric acid = 0.0011598 eq / 0.02382 L = 0.0486 N

Thus, number of equivalents present in 75 mL sample = 0.0486 eq/L * 0.075 L = 0.00365 eq

Thus, same number of equivalents are there in the original sample.

Therefore number of mols of tartaric acid present = 0.00365 eq * 2 = 0.0073 mol = (0.0073 mol)(174 g/mol)

                                                                                                                                = 1.27 g

There are 30 mL(30 g of wine) , thus percent of tartaric acid present = (1.27 g/ 30 g) *100 = 4.23 %

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