A weak monobasic acid HA is 10.0% dissociated in a $0.0100 M$ solution. Its limiting molar conductivity is $400 S c m^{2} mo l^{- 1}$ . Without using the approximation $1 - α \approx 1$ , the molar conductivity of a $0.00100 M$ solution of HA, in $S c m^{2} mo l^{- 1}$ , is closest to

Answer: A

From the 0.0100 M data:

K_{a} = \frac{c α ^{2}}{1 - α} = \frac{0.0100 \times ( 0.100 ) ^{2}}{0.900} = 1.111 \times 1 0^{- 4}

. At

c = 0.00100

c α^{2} + K_{a} α - K_{a} = 0

, i.e.

0.00100 α^{2} + 1.111 \times 1 0^{- 4} α - 1.111 \times 1 0^{- 4} = 0

, giving

α = \frac{- 1.111 \times 1 0 ^{- 4} + ( 1.111 \times 1 0 ^{- 4} ) ^{2} + 4 ( 0.00100 ) ( 1.111 \times 1 0 ^{- 4} )}{2 ( 0.00100 )} = 0.282

Λ_{m} = α Λ_{m}^{\circ} = 0.282 \times 400 = 113 S c m^{2} mo l^{- 1}

. The approximation

α = K_{a} / c = 0.333

gives

133

; keeping the same

α

gives

40