A) 8.279
B) 12.285
C) 10.699
D) 13.335
Correct Answer: C
Solution :
\[[{{H}^{+}}]=0.0005=5\times {{10}^{-4}}\] \[[O{{H}^{-}}][{{H}^{+}}]={{10}^{-14}}\] \[[O{{H}^{-}}]\times 5\times {{10}^{-4}}={{10}^{-14}}\] \[[O{{H}^{-}}]=\frac{{{10}^{-14}}}{5\times {{10}^{-4}}}=2\times {{10}^{-11}}\] \[pOH=-\log (2\times {{10}^{-11}})\] \[=10.699\]
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