A) \[11.80\]
B) \[10.69\]
C) \[7.22\]
D) \[12.24\]
Correct Answer: B
Solution :
\[[O{{H}^{-}}]=C.\,\alpha =0.01\times \frac{5}{100}\] \[=5\times {{10}^{-4}}\] \[pOH=-\log [O{{H}^{-}}]\] \[=-[\log 5\times {{10}^{-4}}]\] \[=4\log 10-\log 5\] \[=4-0.06989\] \[=3.3010\] \[pH=14-pOH\] \[=14-3.3010=10.6990\]
You need to login to perform this action.
You will be redirected in
3 sec
