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

(a) H₂SO₄ + 2YOH = Y₂SO₄ + 2H₂O

(b) C(YOH) = C(H₂SO₄)*V(H₂SO₄)*2/V(YOH) = 0.012*23.5*2/(50-23.5) = 0.0213 mol/L

(c) $\Lambda(Y_2SO_4)=0.147Ω^{-1}m^{-1} = 0.00147S*cm^{-1}$

C(Y₂SO₄) = 0.012*23.5/50 = 0.00564 mol/L = 0.00000564 mol/cm³

$\Lambda_m^0(SO_4^{2-})=160 S*cm^2/mol$

As concetration of Y₂SO₄ is small:

$\Lambda(Y_2SO_4) = \Lambda_m^0(Y^+)*C(Y_2SO_4)*2+\Lambda_m^0(SO_4^{2-})*C(Y_2SO_4)$

0.00147 = $\Lambda_m^0(Y^+)$*0.00000564*2+160*0.00000564

0.0005676 = $\Lambda_m^0(Y^+)$*0.00000564*2

$\Lambda_m^0(Y^+)$ = 50.3 S*cm²/mol