Chapter 8, Problem 8.124QA

Interpretation Introduction

To find:

a) Write a balanced chemical equation based on the reaction (i).

(i) $I{O}_3^{-}\left(aq\right)+I^{-}\left(aq\right)\to I_2\left(aq\right)$

(ii) $C_6{H}_8{O}_6\left(aq\right)+I_2\left(aq\right)\to C_6{H}_6{O}_6\left(aq\right)+2I^{-}\left(aq\right)+2H^{+}\left(aq\right)$

b) Combine equations (i) and (ii) to show the overall reaction in the titration.

c) Find the concentration of ascorbic acid in $mg/L$ in lemon juice.

Answer to Problem 8.124QA

Solution:

a) The balanced equation based on the reaction described in (i) is

$I{O}_3^{-}\left(aq\right)+5I^{-}\left(aq\right)+6\mathrm{ }{H}^{+}\left(aq\right)\to 3I_2\left(aq\right)+3 H_{2}O\left(l\right)$

b) The overall reaction in the titration by combining equations (i) and (ii) is

$I{O}_3^{-}\left(aq\right)+3C_6{H}_8{O}_6\left(aq\right)\to {3C}_6{H}_6{O}_6\left(aq\right)+I^{-}\left(aq\right)+3 H_{2}O\left(l\right)$

c) Concentration of ascorbic acid in lemon juice is $543.2 mg/L$.

Explanation of Solution

1) Concept:

To balance a redox reaction, first check the $∆O.N$. Balance the reaction, and add appropriate coefficient to balance the $∆O.N$. Add the appropriate number of $H_{2}O$ to balance $O$ atoms and $H^{+}$ to balance $H$ atoms.

2) Formula:

i) $M=\frac{n}{V}$

$n=M \times V$

ii) $1L=1000 mL$

iii) $1 g=1000 mg$

3) Given:

i) Volume of lemon juice = $25 mL$

ii) Total volume = $100.0 mL$

iii) Concentration of $KI{O}_3=0.002455 M$

iv) Volume of $KI{O}_3=10.47 mL$

4) Calculation:

a) To write the balanced equation based on the reaction described in (i):

$+5$ $-1$ $0$

$I{O}_3^{-}\left(aq\right)+I^{-}\left(aq\right)\to I_2\left(aq\right)$

Assign oxidation number to $I$ and $O$ atoms as

In $I{O}_3^{-}$, the oxidation number of iodine changes from $+5 to 0$ $(i.e. \Delta O.N. = +5)$

The oxidation number for $I^{-}$ is $-1,$ which changes to $0$ in $I_2$ $(i.e. \Delta O.N. = -1)$

So, to balance the charge, we need to add a coefficient $5$ in front of $I^{-}$ and $3$ in front of $I_2$ to balance the $I$ atoms:

$I{O}_3^{-}\left(aq\right)+5I^{-}\left(aq\right)\to 3I_2\left(aq\right)$

Here $I$ atoms are balanced.

The $O$ atoms are balanced by adding $3 H_{2}O$ on right side. The equation changes to

$I{O}_3^{-}\left(aq\right)+5I^{-}\left(aq\right)\to 3I_2\left(aq\right)+3 H_{2}O\left(l\right)$

We need to balance the $H$ atoms by adding $6H^{+}$ on the left side. The equation changes to

$I{O}_3^{-}\left(aq\right)+5I^{-}\left(aq\right)+6\mathrm{ }{H}^{+}\left(aq\right)\to 3I_2\left(aq\right)+3 H_{2}O\left(l\right)$

All atoms as well as the charge is balanced for the above reaction. So, this is a balanced equation.

b) To write the overall reaction equation by combining equations (i) and (ii):

Combining equations (i) and (ii) to show the overall reaction in the titration:

i. $I{O}_3^{-}\left(aq\right)+I^{-}\left(aq\right)\to I_2\left(aq\right)$

ii. $C_6{H}_8{O}_6\left(aq\right)+I_2\left(aq\right)\to C_6{H}_6{O}_6\left(aq\right)+2I^{-}\left(aq\right)+2H^{+}\left(aq\right)$

We have balanced equation (i) in part (a).

iii. $I{O}_3^{-}\left(aq\right)+5I^{-}\left(aq\right)+6\mathrm{ }{H}^{+}\left(aq\right)\to 3I_2\left(aq\right)+3 H_{2}O\left(l\right)$

We need to balance equation (ii) in order to combine them:

To match the $H^{+}$ ion and $I_2$ molecules, multiply the equation (ii) by $3$.

iv. $3C_6{H}_8{O}_6\left(aq\right)+3I_2\left(aq\right)\to {3C}_6{H}_6{O}_6\left(aq\right)+6I^{-}\left(aq\right)+6H^{+}\left(aq\right)$

Now add equation (iii) and (iv). We get

$I{O}_3^{-}\left(aq\right)+5I^{-}\left(aq\right)+6\mathrm{ }{H}^{+}\left(aq\right)\to 3I_2\left(aq\right)+3 H_{2}O\left(l\right)$

$3C_6{H}_8{O}_6\left(aq\right)+3I_2\left(aq\right)\to {3C}_6{H}_6{O}_6\left(aq\right)+6I^{-}\left(aq\right)+6H^{+}\left(aq\right)$

$I{O}_3^{-}\left(aq\right)+3C_6{H}_8{O}_6\left(aq\right)\to {3C}_6{H}_6{O}_6\left(aq\right)+I^{-}\left(aq\right)+3 H_{2}O\left(l\right)$

So, the overall reaction in the titration is

$I{O}_3^{-}\left(aq\right)+3C_6{H}_8{O}_6\left(aq\right)\to {3C}_6{H}_6{O}_6\left(aq\right)+I^{-}\left(aq\right)+3 H_{2}O\left(l\right)$

c) Finding concentration of ascorbic acid in lemon juice:

From the given molarity and volume of $I{O}_3$, we find number of moles as

$n_{KI{O}_3}=M_{KI{O}_3}\times V_{KI{O}_3}$

$10.47 mL\times \frac{1 L}{1000 mL}=0.01047 L KI{O}_3$

$0.002455 M KI{O}_3\times 0.01047 L KI{O}_3=2.57 \times {10}^{-5}mol I{O}_3^{-}$

Calculate the moles of $C_6{H}_8{O}_6$ (ascorbic acid):

$I{O}_3^{-}\left(aq\right)+3C_6{H}_8{O}_6\left(aq\right)\to {3C}_6{H}_6{O}_6\left(aq\right)+I^{-}\left(aq\right)+3 H_{2}O\left(l\right)$

From the above reaction, $1 mol I{O}_3^{-}$ reacts with $3 mol C_6{H}_8{O}_6$, so, calculating moles of acid as

$0.0000257 mol I{O}_3^{-}\times \frac{3 mol C_6{H}_8{O}_6 }{1 mol I{O}_3^{-}}=7.711 \times {10}^{-5} mol C_6{H}_8{O}_6$

Convert the moles to grams and then to milligrams as

$0.0000771 mol C_6{H}_8{O}_6\times \frac{176.12 g}{1 mol}\times \frac{1000 mg}{1 g}=13.58 mg C_6{H}_8{O}_6$

Calculate the concentration in $mg/L$:

$\frac{13.58 mg}{0.025 L}=543.2 mg/L C_6{H}_8{O}_6$

Concentration of ascorbic acid in lemon juice is $543.2 mg/L$

Conclusion:

Redox reactions are balanced by calculating $∆O.N.$ values of the elements that are reduced and oxidized.

The stoichiometry in the balanced equation is used to find the concentration of the analyte.