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Degree of Dissociation

BOH is a weak...

Question

$B O H$ is a weak base.Molar concentration of BOH that provides a $[O H^{-}] of 1.5 \times 10^{- 3} M [K_{b} [B O H] = 1.5 \times 10^{- 5} M]$ is:

0.15 M

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0.01515 M

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0.0015 M

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0.5 \times 10^{- 5} M

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Solution

The correct option is A $0.15 M$
$B O H ⇌ B^{+} + O H^{-}$
By Ostwald's dilution law,
$[O H^{-}] = \sqrt{K_{b} C}$
$1.5 \times 10^{- 3} = \sqrt{1.5 \times 10^{- 5} \times C}$
$∴ C = 0.15 M$

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$B O H$ is a weak base. Molar concentration of $B O H$ that provides a $[O H^{-}]$ of $1.5 \times 10^{- 3} M$ is:

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