15.00 mL of a 1:5 diluted wine sample plus ~20 mL DI water and 3 drops of phenol

Question

15.00 mL of a 1:5 diluted wine sample plus ~20 mL DI water and 3 drops of phenolphthalein were mixed and titraited to completion using 23.86 mL of 0.0784 M NaOH solution. The density of the original wine sample was 1.095 g/mL. Molarity of the acid in the original wine sample is 0.267 M. A) what is the percent by mass of acid in the original wine sample? ton f Tartaric Acid Concentration in Wine General Chemistry I, CHEM 111AF V06192018 This color change will signify the endpoint, which is assumed to be close to the equivalence point (when the stoichiometric ratio of reactants has been reached). The titrant will only react with the indicator after the entire analyte sample has reacted with the titrant. After the NaOH has been standardized with KHP, a sample of wine (the analyte) will be titrated with a strong base (the titrant) to determine the concentration of tartaric acid in a wine sample (equation 4.2). S0 mi buret Buret clamp GH06 (aq) +(2NaOH (aq) 2 H,OU2Na (aq) C4H,O, (ag) (4.2) Erlenmeyer flask Tartaric acid is assumed to be the only acid present in the wine. A quantitative amount of analyte will be placed in an Erlenmeyer flask with a few drops of indicator (Figure 3). The titrant will be placed in a buret and slowly delivered to the analyte until a permanent pale pink Ring stand Figure 3. Set-up for titration.owy Note the relative positions of the buret t color is obtained. and Erlenmeyer flask Safety Sodium hydroxide is corrosive, if you come in contact with this strong base, wash immediately with water for 15 minutes. Waste All waste produced in this experiment should be placed in the appropriately labeled waste container. Materials 50 mL buret wine ring stand buret clamp phenolphthalein 100 mL volumetric flask 25 ml volumetric pipet 5 ml volumetric pipet 0.1 M NaOH pipet bulbs 10 ml weighing bottles 125 mL Erlenmeyer flask KHP

Explanation / Answer

Total volume= 15+20 ml= 35 ml(diluted acid sample)

Using the formula, M1V1 = M2V2 as the equivalent must be equal to the equivalence point of titration.

So, M1× 35= 0.0784× 23.86

M1= 0.0784× 23.86/ 35

M1= 0.0534

Given, Density of original wine sample= Mass of wine/ Volume of wine.

Given, 1:5 diluted wine sample means One part of solute in 5 parts of solution.

Mass of wine = 20 g

Volume of wine= 20/1.095 = 18.264

Molality of acid= no.of moles/ volume of solution

0.267= no.of moles/ 18.26

No. Of moles= 4.875

Mass % is the ratio of mass of acid to the mass of mixture

Mass% = Mass of acid/ Mass of mixture.

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