A 0.0100 M solution of NH4OH has molar conductivity 11.27 S cm^2 mol^-1. The limiting molar conductivity of NH4OH is 271.4 S cm^2 mol^-1 and K_w = 1.0x 10^-14. Which of the following statement(s) is/are correct? (Use K_b = (calpha^2)/(1-alpha).)

The degree of dissociation is about 0.0415

K_b of NH4OH is about 1.8x 10^-5

The pH of the solution is about 10.62

The degree of dissociation is about 0.0415

K_b of NH4OH is about 1.8x 10^-5

The pH of the solution is about 10.62

On diluting the solution to 0.00100 M, the degree of dissociation becomes ten times larger

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ec206

A $0.0100 M$ solution of $NH X_{4} OH$ has molar conductivity $11.27 S c m^{2} mo l^{- 1}$ . The limiting molar conductivity of $NH X_{4} OH$ is $271.4 S c m^{2} mo l^{- 1}$ and $K_{w} = 1.0 \times 1 0^{- 14}$ . Which of the following statement(s) is/are correct? (Use $K_{b} = \frac{c α ^{2}}{1 - α}$ .)

AThe degree of dissociation is about $0.0415$
B $K_{b}$ of $NH X_{4} OH$ is about $1.8 \times 1 0^{- 5}$
CThe pH of the solution is about $10.62$
DOn diluting the solution to $0.00100 M$ , the degree of dissociation becomes ten times larger

α = 11.27/271.4 = 0.0415

, so (A) is correct.

K_{b} = \frac{0.0100 \times ( 0.0415 ) ^{2}}{1 - 0.0415} = \frac{1.724 \times 1 0 ^{- 5}}{0.9585} = 1.80 \times 1 0^{- 5}

, so (B) is correct.

[OH X^{-}] = c α = 4.15 \times 1 0^{- 4}

pOH = 3.38

pH = 14 - 3.38 = 10.62

, so (C) is correct. On ten-fold dilution

α

grows roughly as

10 \approx 3.2

times (to about 0.125, from the quadratic), not ten times, so (D) is wrong.