A) \[1.0\times {{10}^{4}}\]
B) \[1.0\times {{10}^{-4}}\]
C) \[1.0\times {{10}^{-10}}\]
D) \[2.5\times {{10}^{-1}}\]
Correct Answer: A
Solution :
[a] \[HA+BOH\rightleftharpoons BA+{{H}_{2}}O\] \[\therefore {{K}_{H}}=\frac{{{K}_{w}}}{{{K}_{a}}\times {{K}_{b}}}.\] The inverse of \[{{K}_{H}}\] is\[{{K}_{c}}=\frac{{{K}_{a}}\times {{K}_{b}}}{{{K}_{w}}}.\] \[=\frac{2\times {{10}^{-5}}\times 5\times {{10}^{-6}}}{1\times {{10}^{-14}}}=1.0\times {{10}^{4}}\]
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