A) 14.00
B) 13.70
C) 13.00
D) 12.70
Correct Answer: D
Solution :
Given \[[O{{H}^{-}}]=5\times {{10}^{-2}}\] \[\therefore \]\[pOH=-\log 5\times {{10}^{-2}}\] \[=-\log 5+2\log 10=1,30\] \[\because \]\[pH+pOH=14\] \[\because \]\[pH=14-pOH\] \[=14-1.30=12.70\]You need to login to perform this action.
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