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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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BOH is a weak base. Molar concentration of BOH provides a [OH^-] of $1.5 \times 10^{-3}$ M is [Given $K_b$ (BOH) = $1.5 \times 10^{-3}$]

(a) $1.5 \times 10^{-5}$ M

(b) 0.015 M

(c) 0.0015 M

(d) 0.15 M

Can you answer this question?
 
 

1 Answer

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Answer: 0.15 M
 
$BOH \rightleftharpoons B^+ + OH^-$
 
$K_b = \frac{[B^+][OH^-]}{[BOH]}$
Since, $[B^+] = [OH^-]$
$\Rightarrow K_b = \frac{[OH^-]^2}{[BOH]}$
$\Rightarrow [BOH] = \frac{[OH^-]^2}{K_b} = \frac{(1.5 \times 10^{-3})^2}{1.5 \times 10^{-5}} = 1.5 \times 10^{-1} M$ = 0.15 M
 

 

answered Nov 29, 2013 by mosymeow_1
 

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