A) 1
B) 12.87
C) 11.13
D) 1.35
Correct Answer: C
Solution :
\[\underset{0.1-x}{\mathop{N{{H}_{3}}}}\,+{{H}_{2}}O\,\overset{+}{\mathop{N}}\,{{H}_{4}}+\overset{-}{\mathop{O}}\,H\] \[{{K}_{b}}=\frac{\left[ NH_{4}^{+} \right]\left[ O{{H}^{-}} \right]}{\left[ N{{H}_{3}} \right]}\] Neglecting \[x\] in \[0.1\,-x,\] we get \[1.8\times {{10}^{-5}}=\frac{{{x}^{2}}}{0.1}\] \[{{x}^{2}}=1.8\times {{10}^{-6}}\] \[\Rightarrow \,\,x=1.35\times {{10}^{-3}}\] \[x=\,\left[ O{{H}^{-}} \right]=1.35\times {{10}^{-3}}\] \[pOH=3-\log 1.35=3-0.13=2.87\] \[pH=14-2.87=11.13\]
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