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The molar conductivity of acetic acid at infinite dilution is 390.7 and for 0.01 M acetic acid is 3.907 S cm2 mol-1. Calculate pH of solution.

Option: 1

4


Option: 2

6


Option: 3

9


Option: 4

2


Answers (1)

best_answer

As we have learnt,

\mathrm{\alpha=\frac{\Lambda_{m}^{c}}{\Lambda_{m}^{o}}=\frac{3.907}{390.7}=0.01}


                  \mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{\: \: H}^{+}
Initial conc          c                              0                       0
final conc           c(1 - \alpha)                   c\alpha                     c\alpha

\begin{array}{l}{\therefore\left[\mathrm{H}^{+}\right]=c \alpha=0.01 \times 0.01=10^{-4} \mathrm{\: M}} \\ {\mathrm{pH}=-\log \left(10^{-4}\right)=4}\end{array}

Posted by

Ritika Jonwal

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