A) four times
B) \[5.55\times {{10}^{4}}\]times
C) \[5.55\times {{10}^{6}}\]times
D) \[{{10}^{-2}}\]times
Correct Answer: B
Solution :
[b] \[pH=\frac{1}{2}[p{{K}_{a}}-\log 1]=\frac{p{{K}_{a}}}{2}\] pH' (twice of pH)\[=p{{K}_{a}}\] \[\therefore \]\[p{{K}_{a}}=\frac{1}{2}[p{{K}_{a}}-\log C]\] \[-\log C=p{{K}_{a}}=-\log {{K}_{a}}\] \[C={{K}_{a}}=1.8\times {{10}^{-5}}M\] dilution\[=\frac{1}{C}=5.55\times {{10}^{4}}times\]
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