A) 0.1 RT
B) 0.11 RT
C) 1.1 RT
D) 0.01 RT
Correct Answer: B
Solution :
Dissociation of acid occurs as \[\operatorname{HA} \,\,{{H}^{+}}\,+{{A}^{-}}\] Initially, c 0 0 At equilibrium, \[c-c\,\alpha \] \[c\alpha \] \[c\,\alpha \] \[[{{H}^{+}}]=c\,\alpha \] \[{{10}^{-\,2}}=c\,\alpha \] \[\alpha =\frac{{{10}^{-\,2}}}{c}=\frac{{{10}^{-\,2}}}{0.1}\] \[=\,\,\,{{10}^{-\,1}}=0.1\] We know that \[\alpha =\frac{i-1}{n-1}\,\,\,\Rightarrow \,0.1=\frac{i-1}{2-1}\] \[i=1.1\] \[\pi =i\,cRT\] \[\pi =1.1\times 0.1\times RT\,\,=\,\,0.11\,\,RT\]
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