Question

use the following information to answer the next two questions. the concentration of vitamin c, c6h8o6(aq), in a sample of grapefruit juice can be measured by titration with an aqueous iodine solution, using starch as an indicator. when unreacted iodine is present, the starch forms a purple complex. the titration reaction is represented by the following equation. c6h8o6(aq) + i2(aq) → c6h6o6(aq) + 2 h+(aq) + 2 i-(aq) in an experiment to determine the vitamin c concentration in grapefruit juice, a student titrates 10.00 ml samples of grapefruit juice with a 0.100 mol/l i2(aq) solution. the student records the following data. volume of i2(aq) used during the titration trial initial burette reading (ml) final burette reading (ml) i 10.17 29.69 ii 29.69 49.12 iii 10.00 29.51 iv 29.51 48.93 11. the concentration of vitamin c in the grapefruit juice is a. 5.14×10 - 2 mol/l b. 9.74×10 - 2 mol/l c. 1.05×10 - 1 mol/l d. 2.89×10 - 1 mol/l

Explanation:

Step1: Calculate volume of iodine used in each trial

Trial I: $V_1=29.69 - 10.17=19.52$ mL.
Trial II: $V_2=49.12 - 29.69 = 19.43$ mL.
Trial III: $V_3=29.51 - 10.00=19.51$ mL.
Trial IV: $V_4=48.93 - 29.51 = 19.42$ mL.

Step2: Calculate average volume of iodine used

$\bar{V}=\frac{19.52 + 19.43+19.51+19.42}{4}=\frac{77.88}{4}=19.47$ mL.

Step3: Determine moles of iodine used

$n_{I_2}=C\times V$, where $C = 0.100$ mol/L and $V=19.47\times10^{- 3}$ L. So $n_{I_2}=0.100\times19.47\times10^{-3}=1.947\times10^{-3}$ mol.

Step4: Use mole - ratio from reaction equation

From the reaction $C_6H_8O_6(aq)+I_2(aq)
ightarrow C_6H_6O_6(aq)+2H^+(aq)+2I^-(aq)$, the mole - ratio of $C_6H_8O_6$ to $I_2$ is 1:1. So $n_{C_6H_8O_6}=n_{I_2}=1.947\times10^{-3}$ mol.

Step5: Calculate concentration of vitamin C

The volume of the grapefruit juice sample is $V_{sample}=10.00$ mL = $10.00\times10^{-3}$ L. $C_{C_6H_8O_6}=\frac{n_{C_6H_8O_6}}{V_{sample}}=\frac{1.947\times10^{-3}\text{ mol}}{10.00\times10^{-3}\text{ L}} = 0.1947$ mol/L = $1.95\times10^{-1}$ mol/L.

Answer:

B. $1.95\times10^{-1}$ mol/L