Question

From the equations

$$\begin{gathered} \mathbf{HOCl}\mathbf{"}\mathbf{}\mathbf{\rightleftharpoons }{\mathbf{H}}^{\mathbf{+}}\mathbf{+}{\mathbf{OCl}}^{\mathbf{-}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{K}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}} \\ \mathbf{HOCl}\mathbf{}\mathbf{+}{\mathbf{OBr}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{HOBr}\mathbf{}\mathbf{+}{\mathbf{OCl}}^{\mathbf{-}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{K}\mathbf{=}\mathbf{15}\mathbf{} \end{gathered}$$

find the value of K for the reaction$\mathbf{HOBr}\mathbf{\rightleftharpoons }{\mathbf{H}}^{\mathbf{+}}\mathbf{+}{\mathbf{OBr}}^{\mathbf{-}}$.

Short answer

For a given reaction, the value of K is $2.0\times {10}^{-9}$.

Step-by-step solution

The equilibrium constant.

Consider a weak-acid equilibrium reaction:

$\mathbf{HA}\mathbf{\rightleftharpoons }{\mathbf{K}}_{\mathbf{a}}^{\mathbf{\rightleftharpoons }}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}{\mathbf{A}}^{\mathbf{-}}$

The acid dissociation constant can be calculated as follows:

${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[\mathrm{HA}\right]}$

Consider a reaction of a weak base:

$\mathit{B}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\stackrel{{\mathbf{K}}_{\mathbf{b}}}{\mathbf{\rightleftharpoons }}\mathit{B}{\mathit{H}}^{\mathbf{+}}\mathbf{+}\mathit{O}{\mathit{H}}^{\mathbf{-}}$

The value of ${\mathbf{K}}_{\mathbf{b}}$can be calculated as:

${\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\left[{\mathrm{BH}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[B\right]}$

Here , ${\mathbf{K}}_{\mathbf{b}}$ is called thebase dissociation constant.

The value of ${\mathbf{K}}_{\mathbf{w}}$is calculated by the formula:

${\mathbf{K}}_{\mathbf{w}}\mathbf{=}{\mathbf{K}}_{\mathbf{a}}\mathbf{\times }{\mathbf{K}}_{\mathbf{b}}$

The equilibrium constant for the given reaction.

The given equation is:

$\mathrm{HOCl}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{OCl}}^{-}\hspace{1em}K=3.0\times {10}^{-8}$

$\mathrm{HOCl}+\mathrm{OBr}\rightleftharpoons \mathrm{HOBr}+{\mathrm{OCl}}^{-}\hspace{1em}K=15$

The equilibrium constant for the given reaction:

$\mathrm{HOBr}+{\mathrm{OCl}}^{-}\rightleftharpoons \mathrm{HOCl}+{\mathrm{OBr}}^{-}\hspace{1em}{K}_1=1/15$

role="math" localid="1663330678422" $\mathrm{HOCL}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{OCl}}^{-}{\mathrm{K}}_2=3.0\times {10}^{-8}$

$\mathrm{HOBr}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{OBr}}^{-}{\mathrm{K}}_3={\mathrm{K}}_1{\mathrm{K}}_2$

$$\begin{gathered} K_3=K_1{K}_2 \\ =\left(1/15\right)\times 3.0\times {10}^{-8} \\ =\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}} \end{gathered}$$

The third equation is obtained by reversing the first reaction and adding it to the second reaction. The value of the equilibrium constant of the final equation is obtained by multiplying the first two reaction's equilibrium constants.