A) \[12\]
B) \[13\]
C) \[1\]
D) \[12.96\]
Correct Answer: B
Solution :
\[0.05\,\,M\,\,Ba{{(OH)}_{2}}\]Solution \[\cong 2\times 0.05\,\,N\,\,Ba{{(OH)}_{2}}\] \[\cong 0.10\,\,N\,\,Ba{{(OH)}_{2}}\] \[\therefore \] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log (0.10)=1\] \[pH=14-pOH=14-1=13\]You need to login to perform this action.
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