A) \[1.5\times {{10}^{-5}}M\]
B) 0.015 M
C) 0.0015 M
D) 0.15 M
Correct Answer: D
Solution :
\[BOH{{B}^{+}}+O{{H}^{-}}\] \[{{K}_{b}}=\frac{[{{B}^{+}}][O{{H}^{-}}]}{[BOH]}\] \[[{{B}^{+}}]=[O{{H}^{-}}]\] \[{{K}_{b}}=\frac{{{[O{{H}^{-}}]}^{2}}}{[BOH]}\] \[\therefore \] \[[BOH]=\frac{{{[O{{H}^{-}}]}^{2}}}{{{K}_{b}}}=\frac{{{(1.5\times {{10}^{-3}})}^{2}}}{1.5\times {{10}^{-5}}}\] \[=1.5\times {{10}^{-1}}M\] \[=0.15M\]
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