A) \[3.98\times {{10}^{-6}}\]
B) \[3.68\times {{10}^{-6}}\]
C) \[3.88\times {{10}^{6}}\]
D) \[3.98\times {{10}^{8}}\]
Correct Answer: A
Solution :
Antilog of 0.6 is \[[{{H}^{+}}]\approx 4\times {{10}^{-6}}M\] Alternate Solution \[[{{H}^{+}}]={{10}^{-pH}}\] Taking log both sides \[\log [{{H}^{+}}]=-pH\log 10\] \[\log [{{H}^{+}}]=-5.4\times 1=-5.4\] \[[{{H}^{+}}]=\]antilog\[(-5.4)=\frac{1}{anti\log (5.4)}\] \[[{{H}^{+}}]=3.98\times {{10}^{-6}}M\]You need to login to perform this action.
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