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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