Question

Conductivity of 0.00241 M acetic acid is \[7.896 \times {10^{ - 5}}Scm\], calculate its molar conductivity. If \[{\lambda ^ \wedge }\{ 0\} m\]​ for acetic acid is \[390.5Sc{m^2}mo{l^{ - 1}}\], what is its dissociation constant?

Answer 1

Hint: Conductivity refers to the ability or capability of a substance to allow passage of electricity through it. Molar conductivity refers to the conducting power of the ions constituting in an electrolyte solution, when an unit mole of electrolyte is dissolved in a solution. It is denoted by the unit “\[S{m^2}mo{l^{ - 1}}\]”.

Formula used: There are 3 formulas used to find the solution they are –
1) ∧m = \[\dfrac{{(K \times 1000)}}{M}\]
 2) \[\alpha = \dfrac{{\lambda ^\circ m}}{{\lambda m}}\]
 3) \[K = 1 - \alpha C\alpha \]

Complete step by step solution:
Here, Given: K=\[7.896 \times {10^{ - 5}}Scm\] C=M=\[0.00241mol{L^{ - 1}}\]
 It is asked to find the Molar conductivity:-
∧m=\[\dfrac{{(K \times 1000)}}{M} = \dfrac{{(7.896 \times 10 - 5 \times 1000)}}{{0.00241}}\]
       =\[32.76Sc{m^2}mo{l^{ - 1}}\]
So, the Degree of dissociation is:-
\[\alpha = \dfrac{{\lambda ^\circ m}}{{\lambda m}}\]​= \[\dfrac{{390.5}}{{32.76}}\]​= 0.084
Now, we need to find the dissociation constant (K) :-
\[
   \Rightarrow K = 1 - \alpha C\alpha \\
   = (1 - 0.084)0.00241 \times 0.084 \\
   = 1.86 \times {10^{ - 5}}mol{L^{ - 1}} \\
 \]
So, the molar conductivity of 0.00241 M acetic acid is \[32.76Sc{m^2}mo{l^{ - 1}}\]. Hence, the value of the dissociation constant (K) is \[1.86 \times {10^{ - 5}}mol{L^{ - 1}}\].

Additional Information :
Generally, the molar conductivity of an electrolytic solution is defined as the conductance of the volume of the solution contained in an unit mole of electrolyte solution. This solution is placed in between two electrodes of an unit area of cross-section(i.e,1 cm apart from each other).
Thus, more the conductivity, more efficiently the charges are carried easily from one electrode to the other by overcoming the unit distance that separates them.
Compounds having ionic bonds susceptible to breaking down in molten or liquid phase are good candidates for electricity conduction.
Note: 1. Always be careful of the units used, while solving problems like these. Do not confuse the negative and positive powers of the constants.
2. Remember the values if needed and read the question carefully before attempting it.