9. Vitamin C tablets contain ascorbic acid (CH.o, 176.1232 g/mol) and a st fille

Question

9. Vitamin C tablets contain ascorbic acid (CH.o, 176.1232 g/mol) and a st filler" c which holds them together.Your goal today is to determine how much vitamin C or ascorbeith calcium hydrox present in tablet. This can be determined by titrating ascorbic acid, HCl Ok, 1s HO, with calcium hydroxide solution, Ca(OH)(ag) (74.0918 g/mol) according to the following equa 2 HCHo. (aq) + Ca(OH)2 (aq) Ca(CHah (aq) + 2HO (1) First you must make 500 mL a 0.1339 M standard solution of Ca(OHh of Ca(OH)2(density 1.0634 g/mL). How many mL of the 10.45% mass solution of use to prepare this solution? Describe How to prepare this solution. (5 pts.) from a 10.45% mass solution a. Ca(OH)2would you How to prepare: in anErlenmeyer flask. Thelascorbic acid is titrated with the 0.1339 M Ca(OH): solution. The initial reading on the buret was 5.68 mL and final reading on the buret was 20.02 mL. Determine the number of moles of Ca(OH) added to the solution. (3 pts.) b. 1 tablet (0.5702 g) of vitamin C/ascorbic acid and filler is dissolved in 100.0 mL of water and placed

Explanation / Answer

Q1) First calculate the molarity of stock Ca(OH)2 solution with 10.45% masssolution and density 1.0634g/mL

Let the stock be 1000g

then mass of Ca(OH)2 = 104.5g in the solution

and volume of solution = mass of solution/ density

= 1000g /1.0634

= 940.38 mL

Thus the molarity of tock solution = (mass of solute/molar mass) (1000/volume )

= (104.5/74)(1000/940.38)

= 1.5017 M

[Alternatively it can be calculated directly using the formula

Molarity = (%by mass x 10 x density) /molar mass]

Now we need to prepare 0.1339 M solution

we know V1M1 = V2M2 for dilutions

0.1339M x 500mL = 1.5017 M x V ml

Thus V = 44.58mL

Q2) When the acid is titrtaed with Ca(OH)2 , the base is taken in burette and acid is in Erlenmayer FlasK(conical flask)

Thus volume of Ca(OH)2 used = final -initial burette reading

=20.02 - 5.68

= 14.34 mL

Thus moles of Ca(OH)2 used = volume in L x molarity

= 14.34x 10-3 L x 0.13339M

= 1.92 x10-3 moles

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