Question

Part A:
The electrical conductivities of the following 0.100 M solutions were measured in an apparatus that contained a light bulb as the indicator of conductivity. Rank the solutions in order of decreasing intensity (brightest to dimmest) of the light bulb.
Rank from brightest to dimmest bulb. To rank items as equivalent, overlap them.


HF,CH3OH, KI, Al(NO3)3

Brightest Dimmest






Part B:

A 0.100 M solution of K2SO4 would contain the same total ion concentration as which of the following solutions?

0.0800
0.100
0.0750
0.0500

Answer 1

Concepts and reason

The concept used to solve this problem is based upon the nature of electrolytes.

Substances are classified into strong electrolytes and weak electrolytes on the basis of their behavior in solution.

Fundamentals

Strong electrolytes are the substances which get dissociated completely in solution and conduct electricity. Weak electrolytes are the substances which do not get dissociated completely in solution. Hence, their electricity conduction is weaker.

The conductivity of salts in the solution is determined on the basis of their number of ions present in the solution. Higher is the number of ions formed in the solution, more will be the conductivity and this provides bright light to the bulb connected to the circuit.

The ionic concentration of an electrolytic solution is calculated by the formula given as follows:

${\rm{Ionic}}\;{\rm{concentration}} = {\rm{Number}}\;{\rm{of}}\;{\rm{ions}}\;{\rm{in}}\;{\rm{the}}\;{\rm{solution}} \times {\rm{Concentration}}\;{\rm{of}}\;{\rm{solution}}$ …… (1)

(A)

The conductivity of ${\rm{Al}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{3}}}$ is higher than ${\rm{KI}}$ and ${\rm{HF}}$ because of the formation of highest number of ions in the solution. The conductivity of ${\rm{KI}}$ is higher than ${\rm{HF}}$ because of less covalent character in ${\rm{KI}}$ and for methanol it is lowest because of highest covalent character.

The molecules are arranged in the order of decreasing conductivities as follows:

${\rm{Al}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{3}}} > {\rm{KI}} > {\rm{HF}} > {\rm{C}}{{\rm{H}}_3}{\rm{OH}}$

(A)

The brightness of the bulb increases with an increase in the conductivity of the solution.

Therefore, on the basis of the conductivities, the order from brightest to dimmest light will be as follows:

${\rm{Al}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{3}}} > {\rm{KI}} > {\rm{HF}} > {\rm{C}}{{\rm{H}}_3}{\rm{OH}}$

(B)

Total ions in ${{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4}$ are $3$ . Hence, total ionic concentration of $0.100{\rm{ M}}$ ${{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4}$ is calculated from the equation (1) as follows:

$\begin{array}{c}\\{\rm{Ionic}}\;{\rm{concentration}} = 3 \times 0.100{\rm{ M}}\\\\ = 0.300{\rm{ M}}\\\end{array}$

Total ions in ${\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}$ are $3$ . Hence, total ionic concentration of $0.080{\rm{ M}}$ ${\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}$ is calculated from the equation (1) as follows:

$\begin{array}{c}\\{\rm{Ionic}}\;{\rm{concentration}} = 3 \times 0.080{\rm{ M}}\\\\ = 0.240{\rm{ M}}\\\end{array}$

Total ions in ${\rm{NaCl}}$ are $2$ . Hence, total ionic concentration of $0.100{\rm{ M NaCl}}$ is calculated from the equation (1) as follows:

$\begin{array}{c}\\{\rm{Ionic}}\;{\rm{concentration}} = 2 \times 0.100{\rm{ M}}\\\\ = 0.200{\rm{ M}}\\\end{array}$

Total ions in ${\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4}$ are $4$ . Hence, total ionic concentration of $0.075{\rm{ M}}$ ${\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4}$ is calculated from the equation (1) as follows:

$\begin{array}{c}\\{\rm{Ionic}}\;{\rm{concentration}} = 4 \times 0.075{\rm{ M}}\\\\ = 0.300{\rm{ M}}\\\end{array}$

Total ions in ${\rm{NaOH}}$ are $2$ . Hence, total ionic concentration of $0.050{\rm{ M}}$ ${\rm{NaOH}}$ is calculated from the equation (1) as follows:

$\begin{array}{c}\\{\rm{Ionic}}\;{\rm{concentration}} = 2 \times 0.050{\rm{ M}}\\\\ = 0.100{\rm{ M}}\\\end{array}$

As the number of moles of ions present in the ${{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4}$ and ${\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4}$ solution are equal, therefore, the total ion concentration of $0.100{\rm{ M}}$ ${{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4}$ is same as that of $0.075{\rm{ M}}$ ${\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4}$ .

Ans: Part A

Correct order from brightest to dimmest light will be as follows:

${\rm{Al}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{3}}} > {\rm{KI}} > {\rm{HF}} > {\rm{C}}{{\rm{H}}_3}{\rm{OH}}$

Part B

The total ion concentration of $0.100{\rm{ M}}$ ${{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4}$ solution is same as that of $0.075{\rm{ M}}$ ${\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4}$ solution.