Question

What is the ${OH}^{-}$ concentration of a $0.08M$ solution of ${CH}_{3}COONa$.

$\left[K_a\right({CH}_{3}COOH)=1.8\times {10}^{-5}]$

• A. $\left[{OH}^{-}\right]=6.664\times {10}^{-6}$
• B. $\left[{OH}^{-}\right]=6.664\times {10}^{-4}$
• C. $\left[{OH}^{-}\right]=6.664\times {10}^{-8}$
• D. $\left[{OH}^{-}\right]=6.664\times {10}^{-2}$

Answer 1

Answer: (A)

$\left[{OH}^{-}\right]=6.664\times {10}^{-6}$

${CH}_{3}CO{O}^{\ominus }+H_{2}O\rightleftharpoons {CH}_{3}COOH+{OH}^{\rightleftharpoons }$
$0.08$ $K_b=\frac{{10}^{-14}}{1.8\times {10}^{-5}}$
$\frac{x^2}{0.08}=\frac{10}{1.8}\times {10}^{-10}$
$x^2=0.44\times {10}^{-10}$
$x=0.66\times {10}^{-5}$