A) \[4.8\text{ }basic\]
B) \[4.8\text{ }neutral\]
C) \[4.8\text{ }acidic\]
D) \[1.6\text{ }acidic\]
Correct Answer: D
Solution :
\[[{{H}^{+}}]=3.3\times {{10}^{-2}}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log [3.3\times {{10}^{-2}}]\] \[=-\{\log 3.3+(-2\log 10)\}\] \[=-\{0.5185-2\}\] \[\therefore \] \[pH=1.4815acidic\] \[\therefore \] s\[pH\approx 1.6\] acidic (approximately)
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