A) \[1.8\times {{10}^{-5}}\]
B) \[1.8\times {{10}^{5}}\]
C) \[1.8\times {{10}^{-9}}\]
D) \[5.59\times {{10}^{-10}}\]
Correct Answer: A
Solution :
\[N{{H}_{4}}OH+{{H}^{+}}NH_{4}^{+}+{{H}_{2}}O\] \[\therefore \] \[K=\frac{[NH_{4}^{+}]\,[{{H}_{2}}O]}{[N{{H}_{4}}OH]\,[{{H}^{+}}]}=1.8\times {{10}^{9}}\] Again, \[N{{H}_{3}}+{{H}_{2}}O\xrightarrow{{}}N{{H}_{4}}OHNH_{4}^{+}+O{{H}^{-}}\] \[\therefore \] \[K=\frac{[NH_{4}^{+}]\,[O{{H}^{-}}]}{[N{{H}_{4}}OH]}\] Now \[\frac{K}{K}=\frac{[NH_{4}^{+}]\,[O{{H}^{-}}]}{[N{{H}_{4}}OH]}\times \frac{[N{{H}_{4}}OH]\,[{{H}^{+}}]}{[NH_{4}^{+}]\,[{{H}_{2}}O]}\] \[=\,[O{{H}^{-}}]\,\,[{{H}^{+}}]\] (\[\because {{H}_{2}}O\] is in excess) \[={{K}_{w}}=1\times {{10}^{-14}}\] \[\therefore \] \[K=K\times 1\times {{10}^{-14}}\] \[=1.8\times {{10}^{9}}\times {{10}^{-14}}=1.8\times {{10}^{-5}}\]
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