A 25 mL sample of juice is diluted to 250 mL. A 50 mL aliquot was treated with 20 mL of 0.023M I2; back-titration of the excess I2 required 6.71mL of 0.045M Na2S2O3. M of vit.C in original juice?

Question

C6H8O6 + I2  C6H6O6 + 2H+ + 2I

I2 + 2S2O3 2-  2I- + S4O6 2-

Answer 1

C6H8O6 + I2 → C6H6O6 + 2H⁺ + 2I^-

I2 + 2S2O3^2- → 2I^- + S4O6^2-

Given:

1. 50 mL of juice mixed with 20 mL of 0.023 M I₂
2. _{Of the 20mL 0.023 M I2, only 6.71mL of 0.045 M Na2S2O3 is needed}

_Desired:

1. _{Molarity of original juice, C6H8O6}

_Strategy:

1. _{Find moles of S2O32-}
2. _{Find moles of C6H8O6}
3. _{Find Molarity}

_{1) 6.71mL of 0.045 M Na2S2O3 is needed}

_{mL to mol:}

(6.71mL / 1) (0.045 mol / 1000mL)=0.00030195 mol Na2S2O3

Na2S2O3 → 2Na⁺ + S2O3^2-

0.00030195 mol.....0........................................0

-x...........................+2x....................................+x

0..........................2(0.0030195).............0.0030195

So, there's 0.00030195 mol S2O32-

2) Convert mol S2O32- to mol C6H8O6

from the second equation 2 mol S2O3^2- = 2 mol I^-

from the first equation 2 mol I^- = 1 mol C6H8O6

Using the second equation

0.00030195 mol S2O32- (2 mol I- / 2 mol S2O32-)=0.00030195 mol I-

Using the first equation

0.00030195 mol I- (1 mol C6H8O6 / 2 mol I-) = 0.000150975 mol C6H8O6

3) Find molarity. Divide moles by liters

M=?

n=0.000150975 mol

Volume = 50 mL

M = 0.000150975 / (50/1000 L)

0.0030195 M for the 50 mL sample

What's the molarity of the 25 mL sample?

Well, 50 mL contained 50/250=0.2 of the moles in the 250 mL sample:

(50mL/1) (0.0030195 mol / 1000 mL)=0.000150975 mol in 50 mL

So, in the 250 mL sample, there's 5 times as many moles (250/50=5)

So, 5(0.000150975)=0.000754875 mol in 250 mL

The 25 mL sample was diluted to 250 mL

So, the volume changed due to addition of water, but the moles of C6H8O6 solute never changed. So, there is also 0.000754875 mol in 25 mL

Find molarity:

M= n/V

M=?

n=0.000754875 mol

V=25 mL = 25/1000=0.025 L

M=0.030195 M

original vitamin C sample is a 25 mL sample of 0.030195 M vitamin C