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BOH is a weak base. Molar concentration of BOH that provides a [OH^-] of 1 .5 x 10^-3 M
[K_b(BOH) = 1.5x 10^-5M] is

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0.15 M

0.0015 M

$1 .5 \times 10^{- 5} M$

0.1515 M

answer is A.

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Detailed Solution

$BOH ⇌ B^{+} + {OH}^{-}$

By Ostwald's dilution law,

$[{OH}^{-}] = \sqrt{K_{b} C}$

$1 .5 \times 10^{- 3} = \sqrt{1 .5 \times 10^{- 5} \times C}$

C = 0.15 M

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