A) between 7 and 8
B) between 5 and 6
C) between 6 and 7
D) between 10 and 11
Correct Answer: A
Solution :
\[[O{{H}^{-}}]\]in the diluted base\[=\frac{{{10}^{-6}}}{{{10}^{2}}}={{10}^{-8}}\] Total\[[O{{H}^{-}}]={{10}^{-8}}+[O{{H}^{-}}]\]of water \[=({{10}^{-8}}+{{10}^{-7}})M\] \[={{10}^{-8}}[1+10]M\] \[=11\times {{10}^{-8}}M\] \[pOH=-\log 11\times {{10}^{-8}}\] \[=-\log 11+8\log 10\] \[=6.9586\] \[pH=14-6.9586\] \[=7.0414\]
You need to login to perform this action.
You will be redirected in
3 sec
