A) 14.00
B) 13.70
C) 13.00
D) 12.70
Correct Answer: D
Solution :
[d] Given \[\left[ O{{H}^{-}} \right]=5\times {{10}^{-2}}\] \[\therefore \text{ }pOH=-log5\times {{10}^{-2}}\] \[=-log5+2log10=1.30\] \[\therefore \text{ }pH+pOH=14\] \[\therefore \text{ }pH=14-pOH\] \[=14-1.30=12.70\]You need to login to perform this action.
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