Read the passage carefully and answer the Questions.
Molar conductivity (\( Λ_m \)) of a solution at a given concentration (c) is the conductance of volume, V of solution, containing one mole of electrolyte kept between the two electrodes with area of cross-section A and at a distance of unit length. It increases with the decrease in concentration and when the concentration approaches zero, the molar conductivity is called limiting molar conductivity (\( Λ^0_m \)). For a strong electrolyte, \( Λ_m \) increases linearly with dilution and is given by \( Λ_m = Λ^0_m -Ac^{1/2} \). The value of the constant A for a given solvent depends on the type of electrolyte along with temperature. According to Kohlrausch law, the value of \( Λ^0_m \) for an electrolyte is \( Λ^0_m=v_+λ^0_+ + v_-λ^0_- \), where \( v_+ \) and \( v_- \) are the number of cations and anions, respectively, per molecule of the electrolyte and \( λ^0_+ \) and \( λ^0_- \) are limiting molar conductivities of cation and anion, respectively. Kohlrausch law finds many applications, like determining the solubility of a sparingly soluble salt, determining the degree of dissociation (\( Λ_m/Λ^0_m \)) and the dissociation constant of a weak electrolyte.
If the equivalent conductivity of 1 M Phthalic acid is \( 12.8\, S\, cm^2\, eq^{-1} \) and the limiting equivalent conductivities of the hydrogen and Phthalate ions are 42 and \( 288.42\, S\, cm^2\, eq^{-1} \), respectively, the degree of dissociation of Phthalic acid will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.0387
A weak electrolyte is only partly dissociated in solution, while at infinite dilution it is taken as fully dissociated. The degree of dissociation is therefore the ratio of the conductivity at the given concentration to the limiting conductivity:
\( \displaystyle \alpha = \frac{\Lambda}{\Lambda^{0}} \)
By Kohlrausch’s law, the limiting conductivity of the acid is the sum of the limiting conductivities of its ions. Here the conductivities are given per equivalent, so both the hydrogen ion and the phthalate ion are counted once:
\( \displaystyle \Lambda^{0} = \lambda^{0}(\ce{H+}) + \lambda^{0}(\text{phthalate}) \)
\( \displaystyle \Lambda^{0} = 42 + 288.42 = 330.42\ \mathrm{S\,cm^{2}\,eq^{-1}} \)
Then:
\( \displaystyle \alpha = \frac{12.8}{330.42} \)
\( \displaystyle \alpha = 0.0387 \)
So the degree of dissociation of phthalic acid is 0.0387, that is, only about 3.9% of the acid is dissociated.
Shortcut: \( \alpha \) must be less than 1, and 12.8 is a little less than one-twenty-fifth of 330, so \( \alpha \) is a little less than 0.04. Only 0.0387 fits; 0.387 is ten times too large.
Note: The question is from beyond the NCERT textbook.