A) 8
B) 10
C) 12
D) 6
Correct Answer: C
Solution :
0.01 M \[Ba{{(OH)}_{2}}=0.02N\,\,Ba{{(OH)}_{2}}\] \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[[0.02N]\times [50\,\,ml]={{N}_{2}}\times 100\,\,ml\] \[{{N}_{2}}=\frac{0.02\times 50}{100}\]\[={{10}^{-2}}N\]; \[[O{{H}^{-}}]={{10}^{-2}}\]N \[pOH=2\] or \[pH=12\]You need to login to perform this action.
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