Question

$BOH$ is a weak base. Molar concentration of, BOH that provides a $[OH^{-}]$ of $1.5\times 10^{-3}M$ is $[Kb(BOH)=1.5\times 10^{-5}]$

• A. $1.5\times 10^{-5}M$
• B. 0.015 M
• C. 0.0015 M
• D. 0.15 M

Answer 1

Answer: (D)

0.15 M

Answer 2

$BOH \rightleftharpoons B ^{+}+ OH ^{-}$ $K_{b} =\frac{\left[B^{+}\right]\left[ OH ^{-}\right]}{[ BOH ]}$ $\left[ B ^{+}\right] =\left[ OH ^{-}\right]$ $K_{b} =\frac{\left[ OH ^{-}\right]^{2}}{[ BOH ]}$ $\because [ BOH ] =\frac{\left[ OH ^{-}\right]^{2}}{K_{b}}$ $=\frac{\left(1.5 \times 10^{-3}\right)^{2}}{1.5 \times 10^{-5}}$ $=1.5 \times 10^{-1} M$ $=0.15\, M$