Question #70ea1

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Question #70ea1

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Answer 1 · 2016-05-29T17:47:34

$K_{a} (H O A c) = 1.82 \times 10^{- 5}$ $K_a(HOAc)=1.82xx10^(-5)$

Explanation:

$H_{3} C - C (= O) O H + H_{2} O ⇌ H_{3} C - C (= O) O^{-} + H_{3} O^{+}$ $H_3C-C(=O)OH +H_2O rightleftharpoonsH_3C-C(=O)O^(-) + H_3O^+$

$K_{a}$ $K_a$ $=$ $=$ $\frac{[H_{3} O^{+}] [H_{3} C - C (= O) O^{-}]}{[H_{3} C - C (= O) O H]}$ $([H_3O^+][H_3C-C(=O)O^(-)])/([H_3C-C(=O)OH])$

Now it is a given that $[H_{3} O^{+}]$ $[H_3O^+]$ $=$ $=$ $0.0019 \cdot m o l \cdot L^{- 1}$ $0.0019*mol*L^-1$ . Given the stoichiometry of the equation, $[H O^{-}]$ $[HO^-]$ $=$ $=$ $0.0019 \cdot m o l \cdot L^{- 1}$ $0.0019*mol*L^-1$ as well, and $[H_{3} C - C (= O) O H] = (0.200 - 0.0019) \cdot m o l \cdot L^{- 1}$ $[H_3C-C(=O)OH]=(0.200-0.0019)*mol*L^-1$ .

Thus $K_{a}$ $K_a$ $=$ $=$ $\frac{{(0.0019)}^{2}}{0.200 - 0.0019}$ $(0.0019)^2/(0.200-0.0019)$ $=$ $=$ $1.82 \times 10^{- 5}$ $1.82xx10^(-5)$ .

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Question #70ea1

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