# Equivalent conductivity of barium chloride at infinite dilution

Find out the equivalent conductivity of barium chloride at infinite dilution given that the ionic conductivities of $$\ce{Ba^{2+}, Cl-}$$ are $$\pu{127 ohm^-1 cm^2 equivalent^-1}, \pu{76 ohm^-1 cm^2 equivalent^{-1}}$$ respectively.

Equivalent conductivity is defined as.

The equivalent conductance of an electrolyte is defined as the conductance of a volume of solution containing one equivalent weight of dissolved substance when placed between two parallel electrodes 1 cm apart, and large enough to contain between them all of the solution.

One gram equivalent weight would be simply half of the entire compound so
$$\text{Equivalent Conductivity} = \dfrac{1}{2} \times 127 + \dfrac{1}{2} \times 2 \times 76$$

This comes out to be 139.5 however the answer is 203.

• No, equivalents don't work like this. – Mithoron Dec 5 '18 at 23:40
• @Mithoron how do they work – Avnish Kabaj Dec 6 '18 at 4:11
• – Mithoron Dec 6 '18 at 22:48
• @Mithoron Isn't that what I've done or am I hopelessly confused? – Avnish Kabaj Dec 7 '18 at 7:59

One must be really careful with the units while dealing with conductivity, conductance problems.

If you read the question carefully, the equivalent conductivity of $$\ce{Ba^{2+}}$$ and $$\ce{Cl^{-1}}$$ are provided to you.

So the molar conductivity of $$\ce{Ba^2+}$$ is: $$2 \times 127 ~\pu{ohm^-1 mole^-1} = 254 ~\pu{ohm^-1 mole^-1}$$

and that of $$\ce{Cl^{-1}} = 76~\pu{ohm^-1 cm^2 mole^-1}$$

Now apply Kohlrausch's law of molar conductivity of solution at infinite dilution:

$$\lambda^o_{\ce{BaCl2}}= \lambda^o _{\ce{Ba^{2+}}} + 2\lambda^o_{\ce{Cl^{-1}}} \\ \implies \lambda^o_{\ce{BaCl2}} = 254 + 2\times 76 = 406 ~\pu{ohm^-1 cm^2 mole^-1}$$

Now, as you say, equivalent conductivity is $$\dfrac{1}{2}$$ times the molar conductivity for $$\ce{BaCl_2}$$ so equivalent conductivity of $$\ce{BaCl_2} = \dfrac{1}{2}\times 406 = 203 ~ \pu{ohm^-1 cm^2 equivalent^-1}$$ which is the correct answer.

Alternatively, observe that $$203 = 127 + 76$$ so Kohlrausch's law for equivalent conductivity of strong electrolyte at infinite dilution may be stated as:

The equivalent conductivity of a strong electrolyte at infinite dilution is equal to the sum of the equivalent conductivities of the anions and cations.