A) 2.89
B) 3.89
C) 4.89
D) 0.89
Correct Answer: A
Solution :
Amount of dissociation\[(\alpha )=1.3%=\frac{1.3}{100}\] \[[{{H}^{+}}]=\alpha .C=\frac{1.3}{100}\times 0.1\] \[=1.3\times {{10}^{-3}}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log [1.3\times {{10}^{-3}}]\] \[=2.89\]
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