A) 10
B) 11
C) 9
D) 12
Correct Answer: D
Solution :
\[\underset{2\,\,mol}{\mathop{2Na}}\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\underset{2\,\,mol}{\mathop{2{{H}_{2}}O}}\,\xrightarrow{{}}\underset{2\,\,mol}{\mathop{2NaOH}}\,+{{H}_{2}}\] Given, \[=\frac{0.023}{23}mol\,\,\,\,\frac{100}{22400}mol\] \[=1\times {{10}^{-3}}mol\,\,\,\,=4.46\times {{10}^{-3}}mol\] Thus, Na is the limiting reagent and decide the amount of \[NaOH\] formed. \[\because \] mole Na give \[NaOH=1\,\,mol\] \[\therefore \] \[1\times {{10}^{-3}}\] mole Na will give \[NaOH\] \[=1\times {{10}^{-3}}mol\] Concentration of \[[O{{H}^{-}}]=\frac{1\times {{10}^{-3}}\times 1000}{1000}=1\times {{10}^{-2}}\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log (1\times {{10}^{-2}})\] \[=2\] \[pH=14-2=12\]
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