A) \[3\times {{10}^{-6}}\]
B) \[3\times {{10}^{-5}}\]
C) \[2\times {{10}^{-5}}\]
D) \[5\times {{10}^{-5}}\]
Correct Answer: C
Solution :
Key Idea \[pH=-\log \,[{{H}^{+}}]\] Here, \[pH=4.7\] \[\therefore \] \[4.7=-\log \,[{{H}^{+}}]\] \[4.7=\log \,{{[{{H}^{+}}]}^{-1}}\] Antilog \[4.700={{[{{H}^{+}}]}^{-1}}\] \[50120.0=\frac{1}{[{{H}^{+}}]}\] \[[{{H}^{+}}]=\frac{1}{50120}\] \[[{{H}^{+}}]=0.00002\] \[[{{H}^{+}}]=2\times {{10}^{-5}}\]
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