Question

IF molar conductance at infinite dilution of (NH4)2SO4 , NaOH , AND Na2SO4 SOLUTION are X1 , X2 AND X3 respectively , then molar conductance of NH4OH SOLUTION IS : ​

Answer 1

Answer:

x1+x2-x3

Explanation:

Answer 2

The molar conductance of ammonium hydroxide$NH_4OH$ is $\frac{X_1 + 2X_2 - X_3}{2}$

Explanation:

Given: The molar conductance of

• $(NH_4)_2SO_4=X_1$
• $NaOH = X_2$
• $Na_2SO_4 =X_3$

Using Kohlrausch's law, the above can be written as:

1. $\Lambda_{(NH_4)_2SO_4} = 2\lambda_{(NH_4^+)} +\lambda_{(SO_4^{2-})} = X_1$
2. $\Lambda_{NaOH} = \lambda_{(Na^+)} +\lambda_{(OH^-)} = X_2$
3. $\Lambda_{Na_2SO_4} = 2\lambda_{(Na^+)} +\lambda_{(SO_4^{2-})} = X_3$

To find: The molar conductance of $NH_4OH (\Lambda_{NH_4OH} = \lambda_{(NH_4^+)} +\lambda_{(OH^-)})$

Solution:

• Multiplying equation 2 by 2

$(\lambda_{(Na^+)} +\lambda_{(OH^-)} = X_2)\times 2$

$2\lambda_{(Na^+)} +2\lambda_{(OH^-)} = 2X_2$

• Now adding with equation 1

$2\lambda_{(Na^+)} +2\lambda_{(OH^-)} = 2X_2$  + $2\lambda_{(NH_4^+)} +\lambda_{(SO_4^{2-})} = X_1$

$2\lambda_{(Na^+)} +2\lambda_{(OH^-)} + 2\lambda_{(NH_4^+)} +\lambda_{(SO_4^{2-})} = X_1 + 2X_2$

• Now subtracting equation 3 from the sum, we get,

$(2\lambda_{(Na^+)} +2\lambda_{(OH^-)} + 2\lambda_{(NH_4^+)} +\lambda_{(SO_4^{2-})} = X_1 + 2X_2) - (2\lambda_{(Na^+)} +\lambda_{(SO_4^{2-})} = X_3)$

$2\lambda_{(Na^+)} +2\lambda_{(OH^-)} + 2\lambda_{(NH_4^+)} +\lambda_{(SO_4^{2-})} - 2\lambda_{(Na^+)} -\lambda_{(SO_4^{2-})} = X_1 + 2X_2 - X_3$

Canceling out the common terms

$2\lambda_{(OH^-)} + 2\lambda_{(NH_4^+)} = X_1 + 2X_2 - X_3$

Taking out 2

$2(\lambda_{(OH^-)} + \lambda_{(NH_4^+)}) = X_1 + 2X_2 - X_3$

Thus, the molar conductance for ammonium hydroxide$, NH_4OH$ becomes

$\lambda_{(OH^-)} + \lambda_{(NH_4^+)} = \frac{X_1 + 2X_2 - X_3}{2}$

Hence, the molar conductance of ammonium hydroxide is $\frac{X_1 + 2X_2 - X_3}{2}$