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