Chapter 4, Problem 90E

(a)

Interpretation Introduction

Interpretation:

The following equation should be balanced in acidic medium.

  ${\text{IO}}_3^{-}(\text{aq})+I^{-}\text{(aq})\to {\text{I}}_3^{-}(\text{aq)}$

Concept Introduction:

The chemical reaction in which both oxidation and reduction process takes place is known as redox reaction. In this reaction, transfer of electrons takes place among the elements.

A reduction or an oxidation reaction is known as half reaction.

Balance all atoms including oxygen and hydrogen atoms are carried out by addition of water molecule (to balance oxygen) and hydrogen ion (to balance hydrogen) in the half reactions. Number of electrons and charge should be balanced after that makes the number of electrons equal in both oxidation and reduction reactions by multiplying with an integer. The last step is to add both half reactions.

Answer to Problem 90E

The balanced equation is:

  ${\text{IO}}_3^{-}(\text{aq})+6{\text{H}}^{\text{+}}{\text{(aq)+8I}}^{-}\text{(aq})\to 3{\text{I}}_3^{-}({\text{aq)+3H}}_{\text{2}}\text{O(l)}$

Explanation of Solution

The given reaction is:

  ${\text{IO}}_3^{-}(\text{aq})+{\text{I}}^{-}\text{(aq})\to {\text{I}}_3^{-}(\text{aq)}$

The above reaction is separated (oxidation-reduction reaction) as:

  ${\text{IO}}_3^{-}(\text{aq})\to {\text{I}}_3^{-}(\text{aq)}$

  ${\text{I}}^{-}\text{(aq})\to {\text{I}}_3^{-}(\text{aq)}$

Balance the atoms other than hydrogen and oxygen.

  ${\text{3IO}}_3^{-}(\text{aq})\to {\text{I}}_3^{-}(\text{aq)}$

  ${\text{3I}}^{-}\text{(aq})\to {\text{I}}_3^{-}(\text{aq)}$

Balance oxygen atoms.

  ${\text{3IO}}_3^{-}(\text{aq})\to {\text{I}}_3^{-}({\text{aq)+9H}}_{\text{2}}\text{O}$

  ${\text{3I}}^{-}\text{(aq})\to {\text{I}}_3^{-}(\text{aq)}$

Balance hydrogen atoms.

  ${\text{3IO}}_3^{-}(\text{aq})+18{\text{H}}^{\text{+}}\to {\text{I}}_3^{-}({\text{aq)+9H}}_{\text{2}}\text{O}$

  ${\text{3I}}^{-}\text{(aq})\to {\text{I}}_3^{-}(\text{aq)}$

Balance charge and number of electrons.

  ${\text{3IO}}_3^{-}(\text{aq})+18{\text{H}}^{\text{+}}+16e^{-}\to {\text{I}}_3^{-}({\text{aq)+9H}}_{\text{2}}\text{O}$ (1)

  ${\text{3I}}^{-}\text{(aq})\to {\text{I}}_3^{-}(\text{aq)+2}{e}^{-}$ (2)

Multiply equation (2) by 8

  ${\text{3IO}}_3^{-}(\text{aq})+18{\text{H}}^{\text{+}}+16e^{-}\to {\text{I}}_3^{-}({\text{aq)+9H}}_{\text{2}}\text{O}$

  ${\text{24I}}^{-}\text{(aq})\to 8{\text{I}}_3^{-}(\text{aq)+16}{e}^{-}$

Add both equations.

  ${\text{3IO}}_3^{-}(\text{aq})+18{\text{H}}^{\text{+}}+16e^{-}+{\text{24I}}^{-}\text{(aq})\to {\text{I}}_3^{-}({\text{aq)+9H}}_{\text{2}}\text{O+}8{\text{I}}_3^{-}(\text{aq)+16}{e}^{-}$

The balanced equation is written as:

  ${\text{3IO}}_3^{-}(\text{aq})+18{\text{H}}^{\text{+}}{\text{(aq)+24I}}^{-}\text{(aq})\to 9{\text{I}}_3^{-}({\text{aq)+9H}}_{\text{2}}\text{O(l)}$

Simplify the equation as:

  ${\text{IO}}_3^{-}(\text{aq})+6{\text{H}}^{\text{+}}{\text{(aq)+8I}}^{-}\text{(aq})\to 3{\text{I}}_3^{-}({\text{aq)+3H}}_{\text{2}}\text{O(l)}$

(b)

Interpretation Introduction

Interpretation:

The minimum mass of solid potassium iodide and the minimum volume of 3.00 M $\text{HCl}$ needed to convert all of the ${\text{IO}}_3^{-}$ ions to ${\text{I}}^{-}$ ions.

Concept Introduction:

Mole is SI unit which is used to measure the quantity of the substance. It is the quantity of a substance which contains same number of atoms as present in accurately 12.00 g of carbon-12 is known as mole.

Number of moles of a compound is defined as the ratio of given mass of the compound to the molar or molecular mass of the compound.

The mathematical expression is given by:

Number of moles = $\frac{\text{ mass of the compound}}{\text{molar mass of the compound}}$

Molarity is defined as the ratio of number of moles to the volume of solution in L.

The mathematical expression is:

  $\text{Molarity =}\frac{\text{number of moles}}{\text{volume of solution in L}}$

Answer to Problem 90E

Minimum mass of $\text{KI}$ required = 3.732 g Minimum volume of $\text{HCl}$ = $\text{0}\text{.00749 L}$

Explanation of Solution

Given information:

Mass of potassium iodate = 0.6013 g

Molar mass of potassium iodate = 214.0 g/mole

Number of moles = $\frac{\text{ mass of the compound}}{\text{molar mass of the compound}}$

Put the values,

Number of moles of potassium iodate = $\frac{0.6013g}{214.0g/mole}$

Number of moles of potassium iodate = $0.002810\text{ mole}$

The balanced equation is:

  ${\text{IO}}_3^{-}(\text{aq})+6{\text{H}}^{\text{+}}{\text{(aq)+8I}}^{-}\text{(aq})\to 3{\text{I}}_3^{-}({\text{aq)+3H}}_{\text{2}}\text{O(l)}$

According to the reaction, ratio between ${\text{IO}}_3^{-}$ and ${\text{I}}^{-}$ is 1:8.

Thus, number of moles of ${\mathrm{I}}^{-}$ = $0.002810\text{ mole}\times \text{8}$

  = $0.02248\text{ mole}$

Molar mass of $\text{KI}$ = 166.0 g/mole

Minimum mass of $\text{KI}$ required = $0.02248\text{ mole}\times \text{166}\text{.0 g/mole}$

  = 3.732 g

Number of moles of $\text{KI}$ is equal to the number of moles of $\text{HCl}$ . Thus, number of moles of $\text{HCl}$ is $0.02248\text{ mole}$ .

Molarity of $\text{HCl}$ = 3 M

Volume = $\frac{\text{number of moles}}{\text{molarity}}$

Put the values,

Minimum volume of $\text{HCl}$ = $\frac{\text{0}\text{.02248 mole}}{\text{3 M}}$

  = $\text{0}\text{.00749 L}$

(c)

Interpretation Introduction

Interpretation:

The balance equation for the reaction between ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}$ and ${\text{I}}_3^{-}$ in acidic solution should be written.

Concept Introduction:

The chemical reaction in which both oxidation and reduction process takes place is known as redox reaction. In this reaction, transfer of electrons takes place among the elements.

A reduction or an oxidation reaction is known as half reaction.

Balance all atoms including oxygen and hydrogen atoms are carried out by addition of water molecule (to balance oxygen) and hydrogen ion (to balance hydrogen) in the half reactions. Number of electrons and charge should be balanced after that makes the number of electrons equal in both oxidation and reduction reactions by multiplying with an integer. The last step is to add both half reactions.

Answer to Problem 90E

The balanced equation is written as:

  ${\text{2S}}_{\text{2}}{\text{O}}_3^{2-}(\text{aq})+{\text{I}}_3^{-}(\text{aq)}\to {\text{S}}_4{\text{O}}_6^{2-}\text{(aq})+3{\text{I}}^{-}\text{(aq})$

Explanation of Solution

The reaction between ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}$ and ${\text{I}}_3^{-}$ is:

  ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}\text{(aq})+{\text{I}}_3^{-}\text{(aq)}\to {\text{S}}_4{\text{O}}_6^{2-}\text{(aq})+{\text{I}}^{-}\text{(aq)}$

The above reaction is separated (oxidation-reduction reaction) as:

  ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}(\text{aq})\to {\text{S}}_4{\text{O}}_6^{2-}\text{(aq})$

  ${\text{I}}_3^{-}(\text{aq)}\to {\text{I}}^{-}\text{(aq})$

Balance all the atoms other than hydrogen and oxygen.

  ${\text{2S}}_{\text{2}}{\text{O}}_3^{2-}(\text{aq})\to {\text{S}}_4{\text{O}}_6^{2-}\text{(aq})$

  ${\text{I}}_3^{-}(\text{aq)}\to 3{\text{I}}^{-}\text{(aq})$

Balance oxygen atoms.

  ${\text{2S}}_{\text{2}}{\text{O}}_3^{2-}(\text{aq})\to {\text{S}}_4{\text{O}}_6^{2-}\text{(aq})$

  ${\text{I}}_3^{-}(\text{aq)}\to 3{\text{I}}^{-}\text{(aq})$

Balance hydrogen atoms.

  ${\text{2S}}_{\text{2}}{\text{O}}_3^{2-}(\text{aq})\to {\text{S}}_4{\text{O}}_6^{2-}\text{(aq})$

  ${\text{I}}_3^{-}(\text{aq)}\to 3{\text{I}}^{-}\text{(aq})$

Balance charge and number of electrons.

  ${\text{2S}}_{\text{2}}{\text{O}}_3^{2-}(\text{aq})\to {\text{S}}_4{\text{O}}_6^{2-}\text{(aq})+2e^{-}$ (1)

  ${\text{I}}_3^{-}(\text{aq)+2}{e}^{-}\to 3{\text{I}}^{-}\text{(aq})$ (2)

Add both equations.

  ${\text{2S}}_{\text{2}}{\text{O}}_3^{2-}(\text{aq})+{\text{I}}_3^{-}(\text{aq)+2}{e}^{-}\to {\text{S}}_4{\text{O}}_6^{2-}\text{(aq})+2e^{-}+3{\text{I}}^{-}\text{(aq})$

The balanced equation is written as:

  ${\text{2S}}_{\text{2}}{\text{O}}_3^{2-}(\text{aq})+{\text{I}}_3^{-}(\text{aq)}\to {\text{S}}_4{\text{O}}_6^{2-}\text{(aq})+3{\text{I}}^{-}\text{(aq})$

(d)

Interpretation Introduction

Interpretation:

The molarity of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ should be calculated.

Concept Introduction:

Mole is SI unit which is used to measure the quantity of the substance. It is the quantity of a substance which contains same number of atoms as present in accurately 12.00 g of carbon-12 is known as mole.

Number of moles of a compound is defined as the ratio of given mass of the compound to the molar or molecular mass of the compound.

The mathematical expression is given by:

Number of moles = $\frac{\text{ mass of the compound}}{\text{molar mass of the compound}}$

Molarity is defined as the ratio of number of moles to the volume of solution in L.

The mathematical expression is:

  $\text{Molarity =}\frac{\text{number of moles}}{\text{volume of solution in L}}$

Answer to Problem 90E

Molarity of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ = $\text{0}\text{.0468 M}$

Explanation of Solution

Given information:

Molarity of ${\text{KIO}}_3$ = 0.0100 M

Volume of ${\text{KIO}}_3$ = 25.00 mL

Volume of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ = 32.04 mL

  $\text{KI}$ present in excess

The mathematical expression for calculating molarity is:

  $\text{Molarity =}\frac{\text{number of moles}}{\text{volume of solution in L}}$

Rearrange the above formula in terms of number of moles:

  $\text{Number of moles}=\text{Molarity }\times \text{volume of solution in L}$

Convert the given volume in mL to L.

Since, 1L=1000 mL

Thus, volume in L = $25\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}$

  = 0.025 L

Now,

  ${\text{Number of moles of KIO}}_3=\text{0}\text{.0100 M }\times 0.025\text{ L}$

  = $\text{0}\text{.00025 mole}$

The balanced equation is:

  ${\text{IO}}_3^{-}(\text{aq})+6{\text{H}}^{\text{+}}{\text{(aq)+8I}}^{-}\text{(aq})\to 3{\text{I}}_3^{-}({\text{aq)+3H}}_{\text{2}}\text{O(l)}$

According to the reaction, ratio between ${\text{IO}}_3^{-}$ and ${\text{I}}_3^{-}$ is 1:3.

Thus, number of moles of ${\mathrm{I}}_3^{-}$ = $\text{0}\text{.00025 mole}\times 3$

  = $0.00075\text{ mole}$

The balanced equation between ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}$ and ${\text{I}}_3^{-}$ is written as:

  ${\text{2S}}_{\text{2}}{\text{O}}_3^{2-}(\text{aq})+{\text{I}}_3^{-}(\text{aq)}\to {\text{S}}_4{\text{O}}_6^{2-}\text{(aq})+3{\text{I}}^{-}\text{(aq})$

According to the reaction, ratio between ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}$ and ${\text{I}}_3^{-}$ is 2:1.

Thus, number of moles of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ = $0.00075\text{ mole}\times 2$

Number of moles of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ = $0.0015\text{ mole}$

  $\text{Molarity =}\frac{\text{number of moles}}{\text{volume of solution in L}}$

Put the value,

  $\text{Molarity =}\frac{0.0015\text{ mole}}{32.04\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}}$

Molarity of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ = $\text{0}\text{.0468 M}$

(e)

Interpretation Introduction

Interpretation:

The preparation of 500.0 mL ${\text{KIO}}_{\text{3}}$ solution from part (d) should be determined.

Concept Introduction:

Mole is SI unit which is used to measure the quantity of the substance. It is the quantity of a substance which contains same number of atoms as present in accurately 12.00 g of carbon-12 is known as mole.

Number of moles of a compound is defined as the ratio of given mass of the compound to the molar or molecular mass of the compound.

The mathematical expression is given by:

Number of moles = $\frac{\text{ mass of the compound}}{\text{molar mass of the compound}}$

Molarity is defined as the ratio of number of moles to the volume of solution in L.

The mathematical expression is:

  $\text{Molarity =}\frac{\text{number of moles}}{\text{volume of solution in L}}$

Answer to Problem 90E

Solution of ${\text{KIO}}_3$ is prepared by dissolving 1.07 g of ${\text{KIO}}_3$ in 500.0 mL of water in volumetric flask.

Explanation of Solution

From part (d) molarity of ${\text{KIO}}_{\text{3}}$ = 0.0100 M

Volume = 500.0 mL

The mathematical expression for calculating molarity is:

  $\text{Molarity =}\frac{\text{number of moles}}{\text{volume of solution in L}}$

Rearrange the above formula in terms of number of moles:

  $\text{Number of moles}=\text{Molarity }\times \text{volume of solution in L}$

Convert the given volume in mL to L.

Since, 1L=1000 mL

Thus, volume in L = $500\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}$

  = 0.5 L

Now,

  ${\text{Number of moles of KIO}}_3=\text{0}\text{.0100 M }\times 0.5\text{ L}$

  = $\text{0}\text{.005 mole}$

Molar mass of ${\text{KIO}}_3$ = 214.0 g/mole

Number of moles = $\frac{\text{ mass of the compound}}{\text{molar mass of the compound}}$

  $\text{0}\text{.005 mole}$ = $\frac{\text{ mass of the compound}}{\text{214}\text{.0 g/mole}}$

Mass of ${\text{KIO}}_3$ = $\text{214}\text{.0 g/mole}\times 0.005\text{ mole}$

  $\text{=1}\text{.07 g}$

Thus, solution of ${\text{KIO}}_3$ is prepared by dissolving 1.07 g of ${\text{KIO}}_3$ in 500.0 mL of water in volumetric flask.