-4. A XXXXXXXXXXg sample of pure ascorbic acid is weighed out and dissolved in 50 mL of distilled...

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-4. A XXXXXXXXXXg sample of pure ascorbic acid is weighed out and dissolved in 50 mL of distilled water in an Erlenmeyer flask. One week later, when one third of the water in the flask has evaporated, this solution is titrated against XXXXXXXXXXM KI03 solution and requires 18.75 mL of it to reach the end point. a. What mass of ascorbic acid has been oxidized by one of the components of air over the weeklong waiting period? • b. What has happened to the concentration of ascorbic acid in the solution in the flask over this waiting period (before the titration.)? Show your calculations.
5. An old container of KOH has been found in the chemistry lab. A 4.00 g sample of this impure KOH is dissolved in 30.0 mL of 2.500 M HCI, transferred to a 250 0 mL volumetric flask and diluted to the mark with distilled water. A 15.00 mL sample of the resulting solution is pipetted into an Erlenmeyer flask and titrated with 4.32 mL of standardized 0.120 MNaOH. Determine the percent purity of the sample of KOH found in the lab (Molar mass of KOH = 56.1 g/mol).

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David · Accepted Answer

Solution:TTs171112_51130_31
236612_1_question-1 
4. a.             0.1967 g of pure ascorbic acid = 0.1967/176.12 moles  
                                                                       = 0.001117 moles of ascorbic acid. 
     [ Moles of solute = volume of solution used (in Liter) x molarity of the solution] 
Moles of KIO3 present in 18.75 mL(= 0.01875L) of 0.01735 M KIO3 solution  
=   0.01875L x 0.01735 moles/L = 0.0003253. 
The overall reaction (oxidation of ascorbic acid by potassium iodate) can be 
represented by the following equation: 
3 HC6H7O6(aq) + KIO3(aq)            3 C6H6O6(aq) +  KI(aq)  +  3 H2O 
The stoichiometric relationship between ascorbic acid and potassium iodate can 
be seen from the above equation. 
                     1 mole of KIO3 reacts with 3 moles of HC6H7O6 
Therefore, 0.0003253 moles of KIO3 reacts with 3 x 0.0003253 moles of HC6H7O6 
                                                                                   = 0.0009759 moles of HC6H7O6. 
                                                                                   = 0.0009759 x 176.12 g of HC6H7O6 
                                                                                                                                = 0.1719 g of HC6H7O6
So, the mass of ascorbic acid oxidized =  (0.1967 - 0.1719) g = 0.02480 g
b.  There are two consequences of waiting period  which have affected the 
concentration of the ascorbic acid solutions: 
1. Due to oxidation, the amount of ascorbic is reduced from 0.1967 g  to 0.1719 g. 
2.  One third of water in the flask have evaporated i.

-4. A XXXXXXXXXXg sample of pure ascorbic acid is weighed out and dissolved in 50 mL of distilled water in an Erlenmeyer flask. One week later, when one third of the water in the flask has evaporated,...

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