Question #70ea1

1 Answer
May 29, 2016

${K}_{a} \left(H O A c\right) = 1.82 \times {10}^{- 5}$

Explanation:

${H}_{3} C - C \left(= O\right) O H + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} C - C \left(= O\right) {O}^{-} + {H}_{3} {O}^{+}$

${K}_{a}$ $=$ $\frac{\left[{H}_{3} {O}^{+}\right] \left[{H}_{3} C - C \left(= O\right) {O}^{-}\right]}{\left[{H}_{3} C - C \left(= O\right) O H\right]}$

Now it is a given that $\left[{H}_{3} {O}^{+}\right]$ $=$ $0.0019 \cdot m o l \cdot {L}^{-} 1$. Given the stoichiometry of the equation, $\left[H {O}^{-}\right]$ $=$ $0.0019 \cdot m o l \cdot {L}^{-} 1$ as well, and $\left[{H}_{3} C - C \left(= O\right) O H\right] = \left(0.200 - 0.0019\right) \cdot m o l \cdot {L}^{-} 1$.

Thus ${K}_{a}$ $=$ ${\left(0.0019\right)}^{2} / \left(0.200 - 0.0019\right)$ $=$ $1.82 \times {10}^{- 5}$.