Question

Molar conductances of $BaC{l}_2,H_{2}S{O}_4$ and HCl at infinite dilutions are $x_1,x_2$ and $x_3$ respectively. Equivalent conductance of $BaS{O}_4$ at infinite dilution will be :

• A. $\frac{(x_1+x_2-x_3)}{2}$
• B. $x_1+x_2-2x_3$
• C. $\frac{(x_1-x_2-x_3)}{2}$
• D. $\frac{(x_1+x_2-2x_3)}{2}$

Answer 1

The correct option is (C)

$\frac{(x_1+x_2-2x_3)}{2}$

According to Kohlraush's law, the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of equivalent conductivity of cation and anions at infinite dilution.

The molar conductance of BaCl2BaCl_2BaCl2​ at infinite dilution is:

λBaCl2∞=λBa2+∞+2λCl−∞=X1.....................(1)\lambda _{BaCl_{2}}^{\infty}=\lambda _{Ba^{2+}}^{\infty}+2\lambda _{Cl^{-}}^{\infty}=X_{1}.....................(1)λBaCl2​∞​=λBa2+∞​+2λCl−∞​=X1​.....................(1)

The molar conductance of H2SO4H_2SO_4H2​SO4​ at infinite dilution is:

λH2SO4∞=2λH+∞+λSO42−∞=X2.....................(2)\lambda _{H_{2}SO_{4}}^{\infty}=2\lambda _{H^{+}}^{\infty}+\lambda _{SO_4^{2-}}^{\infty}=X_{2}.....................(2)λH2​SO4​∞​=2λH+∞​+λSO42−​∞​=X2​.....................(2)

The molar conductance of $HCl$ at infinite dilution is:

λHCl∞=λH+∞+λCl−∞=X3............................(3)\lambda _{HCl}^{\infty}=\lambda _{H^{+}}^{\infty}+\lambda _{Cl^{-}}^{\infty}=X_{3}............................(3)λHCl∞​=λH+∞​+λCl−∞​=X3​............................(3)

The molar conductance of BaSO4BaSO_4BaSO4​ at infinite dilution is:

λBaSO4∞=λBa2+∞+λSO42−∞=X4......................(4)\lambda _{BaSO_{4}}^{\infty}=\lambda _{Ba^{2+}}^{\infty}+\lambda _{SO_4^{2-}}^{\infty}=X_{4}......................(4)λBaSO4​∞​=λBa2+∞​+λSO42−​∞​=X4​......................(4)

Adding the equation (1) and (2)

λBaCl2∞+λH2SO4∞=λBa2+∞+2λCl−∞+2λH+∞+λSO42−∞=X1+X2..............(5)\lambda _{BaCl_{2}}^{\infty}+\lambda_{H_2SO_4}^{\infty}=\lambda _{Ba^{2+}}^{\infty}+2\lambda _{Cl^{-}}^{\infty}+2\lambda_{H^+}^{\infty}+\lambda_{SO_4^{2-}}^{\infty}=X_{1}+X_{2}..............(5)λBaCl2​∞​+λH2​SO4​∞​=λBa2+∞​+2λCl−∞​+2λH+∞​+λSO42−​∞​=X1​+X2​..............(5)

Multiply the equation with (2)

$2{\lambda }_{HCl}^{\infty }=2{\lambda }_{H^{+}}^{\infty }+2{\lambda }_{C{l}^{-}}^{\infty }=2X_3............................\left(6\right)$

Substract equation (6) from equation (5)

λBaCl2∞+λH2SO4∞−2λHCl∞=λBa2+∞+λSO42−∞=X1+X2−2X3\lambda _{BaCl_{2}}^{\infty}+\lambda _{H_2SO_4}^{\infty}-2\lambda_{HCl}^{\infty}=\lambda _{Ba^{2+}}^{\infty}+\lambda _{SO_4^{2-}}^{\infty}=X_{1}+X_{2}-2X_{3}λBaCl2​∞​+λH2​SO4​∞​−2λHCl∞​=λBa2+∞​+λSO42−​∞​=X1​+X2​−2X3​

λBaSO4∞=X1+X2−2X3\lambda _{BaSO_4}^{\infty}= X_{1}+X_{2}-2X_{3}λBaSO4​∞​=X1​+X2​−2X3​