A) \[13\]
B) \[13.3\]
C) \[12.3\]
D) \[12.0\]
Correct Answer: B
Solution :
\[CaO+{{H}_{2}}O\xrightarrow{{}}Ca{{(OH)}_{2}}\] \[[Ca{{(OH)}_{2}}]=0,.01\] mole/\[100C.C=0.1\,mol\,\,{{L}^{-1}}=0.1M\] Assuming complete ionization \[[O{{H}^{-}}]=0.2M\] \[[{{H}^{+}}]=\frac{{{10}^{-14}}}{0.2}=5\times {{10}^{-14}}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log \,\,(5\times {{10}^{-14}})\] \[=14-\log 5=14-0.699=13.3\]You need to login to perform this action.
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