Question

A student has conducted an experiment to measure the conductivity of electrolyte

Y₂SO₄ and was found to be 0.147 Ω^-1 m^-1. This solution was prepared by reacting 23.50

mL of 0.012 mol dm^-3 of H₂SO₄ and YOH that gives the total volume of reaction

mixture is 50 mL.

(a) Write the chemical equation for the reaction.

(b) Find the concentration of YOH

(c) Find the limiting molar conductivity of Y ion

Answer 1

The chemical equation for the reaction is:

2YOH + H₂SO₄ $\leftrightarrow$ Y₂SO₄ + 2H₂O.

The total volume of reaction mixture is 50 mL, therefore the volume of the YOH reacted is:

$V(YOH) = 50 - 23.50 = 26.50$ mL.

As it can be seen from the reaction equation, the volume number of the moles of YOH that react is twice the number of the moles of H₂SO₄:

$n(YOH) = 2n(H_2SO_4)$ .

Therefore, the concentration of YOH is:

$c(YOH) = \frac{n(YOH)}{V(YOH)} = \frac{2n(H_2SO_4)}{V(YOH)}$

$c(YOH) = \frac{2c(H_2SO_4)V(H_2SO_4)}{V(YOH)} = \frac{2·0.012·23.50·10^{-3}}{26.50·10^{-3}}$

$c(YOH) = 0.0213$ mol dm^-3.

Then, the concentration of Y₂SO₄ is:

$c(Y_2SO_4) = \frac{c(H_2SO_4)·V(H_2SO_4)}{V_r} = \frac{0.012·23.50·10^{-3}}{50·10^{-3}}$

$c(Y_2SO_4) = 0.00564$ mol dm^-3.

The molar conductivity of the reaction solution is:

$\lambda_m = \frac{\kappa}{c} = \frac{0.147}{0.00564·10^3} = 0.026$ Ω^-1 m² mol^-1, or 261 Ω^-1 cm² mol^-1.

As the concentrations are small, we can consider this molar conductivity as a limiting molar conductivity.

$\lambda_m\approx\lambda_m^0$ .

The limiting conductance of SO₄^2- is 160 Ω^-1 cm² mol^-1. Finally, the limiting molar conductivity of Y ion is:

$\lambda_m^0(Y^{+}) =\frac{\lambda_m^0 - \lambda_m^0(SO_4^{2-})}{2} = \frac{261- 160}{2} = 51$ Ω^-1 cm² mol^-1.

The value resembles to that of sodium ion, Na⁺.