A) 4
B) 10
C) 7
D) 9
Correct Answer: B
Solution :
\[Ba{{(OH)}_{2}}\] ⇌ \[B{{a}^{2+}}+2O{{H}^{-}}\] One molecule on dissociation furnishes \[2O{{H}^{-}}\] ions. So, \[[O{{H}^{-}}]=2\times {{10}^{-4}}N\] \[N=M\times 2\] ; \[M=\frac{N}{2}=\frac{2\times {{10}^{-4}}}{2}={{10}^{-4}}\] pOH \[=-\log [O{{H}^{-}}]=-\log (1\times {{10}^{-4}})=-4\] \[pH+pOH=14\]; \[pH=14-4=10\].You need to login to perform this action.
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