Assuming 90 mg of vitamin C present in 29.57 mL of the cranberry juice, how many

Question

Assuming 90 mg of vitamin C present in 29.57 mL of the cranberry juice, how many milliliters of 0.0102 W KIO_3 would be required to reach the stiochiometric point? A 64.11-g sample of a citrus juice was prepared for the analysis procedure described in this experiment. An 8.95-mL volume of 0.0111 M KIO_3 titrated the sample to the deep-blue I_3^-, starch endpoint. How many moles of I_3^- were generated for its reaction with the ascorbic acid? How many moles of ascorbic acid are in the sample? How many milligrams of ascorbic acid (molar mass = 176.1 g/mol) are in the sample? Calculate the number of milligrams of ascorbic acid per 100 g of citrus juice. Compare the result to the values listed in Table 30.1.

Explanation / Answer

(b) Let us write down the step by reaction between KIO3 and ascorbic acid (Vitamin C)

IO3- + 5 I- + 6 H+ ---------> 3 I2 + 3 H2O …..(1)

C6H8O6 + I2 ---------> C6H6O6 + 2 I- + 2 H+ ……(2)

Multiplying (2) by 3 and adding (1) and (2), we get the overall reaction as

IO3- + 3 C6H8O6 ------> 3 C6H6O6 + I- + 6H2O

Molar mass of ascorbic acid is 176.12 gm/mol.

As per the balanced equation, IO3- and ascorbic acid have a 1:3 ratio, i.e., 1 mole of IO3- neutralizes 3 moles of ascorbic acid.

(a) The sample of cranberry juice contains 90 mg vitamin C.

Therefore, moles of ascorbic acid present is (90/1000 gm)/(176.12 gm/mol) = 5.11*10-4 mole (we need to convert mg to gm).

Molarity of ascorbic acid solution (volume is 29.75 mL = 0.02975 L) is

5.11*10-4 mole/0.02975 L = 0.01718 M.

Let x mL of IO3- solution be required for the titration. As per the last step, 0.01718 M ascorbic acid solution will be neutralized by (0.01718/3) = 5.7267*10-3 M IO3- (since they have a 1:3 molar ratio). But, the given concentration of IO3- is 0.0102 M.

Therefore,

0.0102 mol/ L = x.(5.7267*10-3 mol/L)

or, x = 1.78

The volume of IO3- required for titration is 1.78 mL (ans)

6(a) Let us again write down the equations for the reactions. We have equations (1) and (2) and add a third one as

IO3- + 8I- + 6H+ ------> 3I3- + 3H2O

The molar ratio of IO3- and I3- IS 1:3.

We start with 8.95 mL of 0.0111 M IO3- solution. Therefore, moles of IO3- in the solution is (8.95 mL/1000 mL)(1 L).(0.0111 mol/L) = 9.9345*10-5 mole.

Since IO3- and I3- have 1:3 molar ratio, hence the moles of I3- generated is (9.9345*10-5 x 3) = 2.98035*10-4 2.98*10-4 (corrected upto 2 decimal places) (ans).

(b) From equation (2) above, we see that IO3- and ascorbic acid have 1:3 molar ratio.

Hence, amount of ascorbic acid in the solution is (9.9345*10-3 x 3) = 2.98035*10-4 2.98*10-4 mol (corrected to 2 decimal places) (ans).

(c) Molar mass of ascorbic acid is 176.1 gm/mol; hence, amount of ascorbic acid present in the solution = (2.98*10-4 mol)(176.1 gm/mol) = 0.05248 gm = 52.48 mg (since 1 gm =1000 mg) (ans)

(d) The said amount of ascorbic acid, i.e, 52.48 mg is present in 64.11 gm sample of citrus juice.

Therefore, amount of ascorbic acid per 100 gm of the citrus juice is (52.48 mg ascorbic acid)(100 gm citrus juice)/(64.11 gm citrus juice) = 81.86 mg (ans)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.