Question

If the equilibrium constant of $BOH\rightleftharpoons B^{\oplus }+\stackrel{⊝}{O}H$ at ${25}^{o}C$ is $2.5\times {10}^{-6}$, then equilibrium constant for $BOH+H^{\oplus }\rightleftharpoons B^{\oplus }+H_{2}O$ at the same temperature is

• A. $4.0\times {10}^{-9}$
• B. $4.0\times {10}^5$
• C. $2.5\times {10}^8$
• D. $2.5\times {10}^{-6}$

Answer 1

The correct option is C $2.5\times {10}^8$

If the equilibrium constant of $BOH\rightleftharpoons B^{\oplus }+\stackrel{⊝}{O}H$ at ${25}^{o}C$ is $2.5\times {10}^{-6}$, then equilibrium constant for $BOH+H^{\oplus }\rightleftharpoons B^{\oplus }+H_{2}O$ at the same temperature is $2.5\times {10}^8$
$BOH\rightleftharpoons B^{\oplus }+\stackrel{⊝}{O}H$. $K=2.5\times {10}^{-6}$.....(1)
$H^{+}+O{H}^{-}\rightleftharpoons H_{2}O$ $\frac{1}{K_w}={10}^{14}$ ......(2)
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$BOH+H^{\oplus }\rightleftharpoons B^{\oplus }+H_{2}O$ $K^{\prime }=?$......(3)
The reaction (3) is obtained by adding reactions (1) and (2),
$K^{\prime }=K\times \frac{1}{K_w}=2.5\times {10}^{-6}\times {10}^{14}=2.5\times {10}^8$