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 \[=({{10}^{-8}}+{{10}^{-7}})M\] \[={{10}^{-8}}[1+10]M\] \[=11\times {{10}^{-8}}M\] \[pOH=-\log 11\times {{10}^{-8}}\] \[=-\log 11+8\log 10\] \[=6.9586\] \[pH=14-6.9586\] \[=7.0414\]
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