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NCERT Class XII Chemistry
Chapter - Electrochemistry
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Question : 11 of 32
Marks: +1, -0
Conductivity of 0.00241 M acetic acid is 7.896 × 10−5^{-5} S cm−1^{-1}. Calculate its molar conductivity. If Λm0\Lambda^{0}_{m} m for acetic acid is 390.5 S cm2^{2}mol−1^{-1}, what is its dissociation constant?
Solution:  
Λmc=κ×1000M\Lambda^{c}_{m} = \frac{\kappa \times 1000}{M}
=(7.896×10−5)×10000.00241= \frac{(7.896 \times 10^{-5}) \times 1000}{0.00241}
=32.76 S cm2 mol−1=32.76 \, \mathrm{S} \, \mathrm{cm}^{2} \, \mathrm{mol}^{-1}
α=ΛmcΛm0\alpha = \frac{\Lambda^{c}_{m}}{\Lambda^{0}_{m}}
=32.76390.5=8.4×10−2= \frac{32.76}{390.5} = 8.4 \times 10^{-2}
Ka=cα21−αK_{a} = \frac{c \alpha^{2}}{1 - \alpha}
=0.00241×(8.4×10−2)21−0.084= \frac{0.00241 \times (8.4 \times 10^{-2})^{2}}{1 - 0.084} =1.86×10−5=1.86 \times 10^{-5}
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