Question

If equilibrium constant of ${CH}_{3}COOH+H_{2}O\rightleftharpoons {CH}_3{COO}^{-}+H_3{O}^{+}$ is $1.8\times {10}^{-5}$, equilibrium constant for ${CH}_{3}COOH+{OH}^{-}\rightleftharpoons {CH}_3{COO}^{-}+H_{2}O$ is?

• A. $1.8\times {10}^{-9}$
• B. $1.8\times {10}^9$
• C. $5.55\times {10}^{-9}$
• D. $5.55\times {10}^{10}$

Answer 1

The correct option is B $1.8\times {10}^9$

$C{H}_{3}COOH+H_{2}O\equiv C{H}_{3}CO{O}^{-}+H_3{O}^{+}$

$K_1=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}COOH\right]}---\left(i\right)$

$C{H}_{3}COOH+O{H}^{-}\equiv C{H}_{3}CO{O}^{-}+H_{2}O$

$K_2=\frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}---\left(ii\right)$

$H_{2}O+H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$

$K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]---\left(iii\right)$

comparing all three quation we get

$\frac{K_1}{K_2}=K_w$

$K_2=\frac{K_1}{K_w}=\frac{1.8\times O^{-5}}{1\times {10}^{-19}}=1.8\times {10}^{-9}$