A) Basic
B) Acid
C) Neutral
D) Both A and B
Correct Answer: A
Solution :
\[[O{{H}^{-}}]\]ion conc.\[=0.05\frac{mol}{l}\] \[=5\times {{10}^{-2}}\frac{mol}{l}\] \[pOH=-\log \,\,[O{{H}^{-}}]\]\[=-\log \,\,[5\times {{10}^{-2}}]\] \[pOH=1.30\];\[pH+pOH=14\] \[pH=14-pOH\]\[=14-1.30=12.7\]You need to login to perform this action.
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