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Question: A weak monobasic acid is \(15\%\) dissociated in \(0.01moldm^{- 3}\) solution. Limiting molar conduc...

A weak monobasic acid is 15%15\% dissociated in 0.01moldm−30.01moldm^{- 3} solution. Limiting molar conductivity of acid at infinite dilution is 4×10−2ohm−1m2mol−14 \times 10^{- 2}ohm^{- 1}m^{2}mol^{- 1}. What will be the conductivity of 0.05moldm−30.05moldm^{- 3} solution of the acid?

A

8.94×10−6ohm−1cm2mol−18.94 \times 10^{- 6}ohm^{- 1}cm^{2}mol^{- 1}

B

8.94×10−4ohm−1cm2mol−18.94 \times 10^{- 4}ohm^{- 1}cm^{2}mol^{- 1}

C

4.46×10−6ohm−1cm2mol−14.46 \times 10^{- 6}ohm^{- 1}cm^{2}mol^{- 1}

D

2.23×10−5ohm−1cm2mol−12.23 \times 10^{- 5}ohm^{- 1}cm^{2}mol^{- 1}

Answer

8.94×10−4ohm−1cm2mol−18.94 \times 10^{- 4}ohm^{- 1}cm^{2}mol^{- 1}

Explanation

Solution

Ka=Cα2=0.01×(0.05)2=2.5×10−5K_{a} = C\alpha^{2} = 0.01 \times (0.05)^{2} = 2.5 \times 10^{- 5}

Ka=Cα2K_{a} = C\alpha^{2}

2.5×10−5=0.05×α22.5 \times 10^{- 5} = 0.05 \times \alpha^{2}

α=0.0233\alpha = 0.0233

α=ΔmcΔm∞\alpha = \frac{\Delta_{m}^{c}}{\Delta_{m}^{\infty}}

Δmc=0.0223×4×10−2=8.94×10−4ohm−1cm2mol−1\Delta_{m}^{c} = 0.0223 \times 4 \times 10^{- 2} = 8.94 \times 10^{- 4}ohm^{- 1}cm^{2}mol^{- 1}