The molar conductivity vs √C curve for Na₂SO₄, H₂SO₄, and NH₄OH are shown below in random order…
(a)[3]
The molar conductivity vs $\sqrt{C}$ curve for $\text{Na}_2\text{SO}_4$, $\text{H}_2\text{SO}_4$, and $\text{NH}_4\text{OH}$ are shown below in random order.
Identify the curve that corresponds to $\text{Na}_2\text{SO}_4$, $\text{H}_2\text{SO}_4$, and $\text{NH}_4\text{OH}$. Justify your answer.
(b)[2]
The molar conductivity ($\Lambda_m$) of a dilute solution of methanoic acid is $34\cdot1\text{ S cm}^2/\text{mol}$. Calculate its degree of dissociation.
(Given $\lambda^0(\text{H}^+) = 349\cdot6\text{ S cm}^2/\text{mol}$ and $\lambda^0(\text{HCOO}^-) = 54\cdot6\text{ S cm}^2/\text{mol}$)
Curve 1 = $\mathrm{H_2SO_4}$; curve 2 = $\mathrm{Na_2SO_4}$; curve 3 = $\mathrm{NH_4OH}$.
1. $\mathrm{NH_4OH}$ is a weak electrolyte: its $\Lambda_m$ falls steeply at higher concentration and rises sharply on dilution because the degree of ionisation increases, so it is curve 3.
2. $\mathrm{H_2SO_4}$ and $\mathrm{Na_2SO_4}$ are strong electrolytes, whose $\Lambda_m$ falls linearly with $\sqrt{C}$ (Debye-Huckel-Onsager).
3. $\mathrm{H_2SO_4}$ has the higher $\Lambda_m^\circ$ because of the very mobile $\mathrm{H^+}$ ion, so it is the top curve 1, and $\mathrm{Na_2SO_4}$ is curve 2.
Answer (b)
Formula used: Degree of dissociation from molar conductivities: alpha = Lm/Lm0 with Lm = 34.1 S cm^2 mol^-1, Lm0 = 404.2 S cm^2 mol^-1, so alpha = 0.0844
