A) \[1.96\times {{10}^{-2}}\,\,mol/l\]
B) \[1.96\times {{10}^{-3}}\,\,mol/l\]
C) \[1.5\times {{10}^{-4}}\,\,mol/l\]
D) \[1.96\times {{10}^{-1}}\,\,mol/l\]
Correct Answer: C
Solution :
\[pH=3.82\] \[=-\log \,[{{H}^{+}}]\] \[\therefore [{{H}^{+}}]=1.5\times {{10}^{-4}}\] mole/litre.
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