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