A) 10
B) 8
C) 9
D) 12
Correct Answer: D
Solution :
\[2Na+2{{H}_{2}}O\xrightarrow{{}}2NaOH+{{H}_{2}}\] \[\underset{0.023}{\mathop{(2\times 23=46)}}\,\] \[\underset{0.04}{\mathop{(2\times 40=80)}}\,\] \[\therefore \] \[[O{{H}^{-}}]=\frac{0.04}{40}\times 10={{10}^{-2}}\] \[\because \] \[pOH=-\log \,\,[O{{H}^{-}}]\] \[pOH=2\] \[\because \] \[pH+pOH=14\] \[pH=14-pOH\] \[pH=14-2\] \[pOH=12\]
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