Working out moles of vitamin C (ascorbate) from DCPIP titration

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ismadi25
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Hi! Please can someone help with the question below, I have the solution but I don't understand the steps for getting to the answer.

Q7. If a solution (10 mL) of a vegetable extract requires 3.5 ml of a 0.1 M solution of DCPIP to produce a persistent pink colour, how many moles of ascorbate did the 10 mL solution contain?

A. 3.5 x 10-4
B. 3.5 X 10-3
C. 3.5 x 10-5
D. 0.35
E. 0.035

Answer: since the stoichiometry of this reaction is 1 to 1, in effect, the answer is provided by the amount of DCPIP used. 3.5 mL of the DCPIP solution contains 3.5/10 (= 0.35) of what is in 10 mL of the 0.1 M solution and that is 0.1x (10/1000), which is 0.001, so finally, the DCPIP used = 0.35 x 0.001, which is 0.00035, or 3.5 x 10-4 Moles.

I thought the number of moles of DCPIP would equal those of ascorbate because of the 1:1 ratio. If moles = conc. x vol. then Moles (DCPIP) = 0.1M x 3.5ml = 0.35 moles... Why is this wrong? Thank you!
Last edited by ismadi25; 1 month ago
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charco
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(Original post by ismadi25)
Hi! Please can someone help with the question below, I have the solution but I don't understand the steps for getting to the answer.

Q7. If a solution (10 mL) of a vegetable extract requires 3.5 ml of a 0.1 M solution of DCPIP to produce a persistent pink colour, how many moles of ascorbate did the 10 mL solution contain?

A. 3.5 x 10-4
B. 3.5 X 10-3
C. 3.5 x 10-5
D. 0.35
E. 0.035

Answer: since the stoichiometry of this reaction is 1 to 1, in effect, the answer is provided by the amount of DCPIP used. 3.5 mL of the DCPIP solution contains 3.5/10 (= 0.35) of what is in 10 mL of the 0.1 M solution and that is 0.1x (10/1000), which is 0.001, so finally, the DCPIP used = 0.35 x 0.001, which is 0.00035, or 3.5 x 10-4 Moles.

I thought the number of moles of DCPIP would equal those of ascorbate because of the 1:1 ratio. If moles = conc. x vol. then Moles (DCPIP) = 0.1M x 3.5ml = 0.35 moles... Why is this wrong? Thank you!
moles = molarity x volume IN LITRES
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tony_dolby
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(Original post by ismadi25)
Hi! Please can someone help with the question below, I have the solution but I don't understand the steps for getting to the answer.

Q7. If a solution (10 mL) of a vegetable extract requires 3.5 ml of a 0.1 M solution of DCPIP to produce a persistent pink colour, how many moles of ascorbate did the 10 mL solution contain?

A. 3.5 x 10-4
B. 3.5 X 10-3
C. 3.5 x 10-5
D. 0.35
E. 0.035

Answer: since the stoichiometry of this reaction is 1 to 1, in effect, the answer is provided by the amount of DCPIP used. 3.5 mL of the DCPIP solution contains 3.5/10 (= 0.35) of what is in 10 mL of the 0.1 M solution and that is 0.1x (10/1000), which is 0.001, so finally, the DCPIP used = 0.35 x 0.001, which is 0.00035, or 3.5 x 10-4 Moles.

I thought the number of moles of DCPIP would equal those of ascorbate because of the 1:1 ratio. If moles = conc. x vol. then Moles (DCPIP) = 0.1M x 3.5ml = 0.35 moles... Why is this wrong? Thank you!
You need to convert the ml into dm3 by dividing it by 1000. In the lab, almost every piece of glassware is graduated in cm3 or ml. Calculations are done in dm3 . 0.1M stands for 0.1 moles per dm3 which is why your volume must be in dm3 too.
Last edited by tony_dolby; 1 month ago
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ismadi25
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(Original post by charco)
moles = molarity x volume IN LITRES
Thank you very much for the quick reply!
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ismadi25
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(Original post by tony_dolby)
You need to convert the ml into dm3 by dividing it by 1000. In the lab, almost every piece of glassware is graduated in cm3 or ml. Calculations are done in dm3 . 0.1M stands for 0.1 moles per dm3 which is why your volume must be in dm3 too.
Ahh ok I see, that makes sense! Thank you so much for the quick response, much appreciated!
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