A) \[8.2798\]
B) \[10.6990\]
C) \[12.7854\]
D) \[13.3344\]
Correct Answer: B
Solution :
Given, \[[{{H}_{3}}{{O}^{+}}]=5\times {{10}^{-4}}g.eq/.L\] \[[{{H}_{3}}{{O}^{+}}]\times [O{{H}^{-}}]={{10}^{-14}}\] \[[O{{H}^{-}}]=\frac{{{10}^{-14}}}{5\times {{10}^{-4}}}\] \[=0.2\times {{10}^{-10}}\] \[=2\times {{10}^{-11}}\] \[pH=-\log [O{{H}^{-}}]\] \[=-\log (2\times {{10}^{-11}})\] \[=-(\log 2+\log {{10}^{-11}})\] (\[\because \]\[{{\log }_{10}}2=0.3010\]) \[=11-0.3010\] \[pOH=10.6990\]
You need to login to perform this action.
You will be redirected in
3 sec
