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rewona [7]
2 years ago
7

Please help 25 points!!!​

Chemistry
1 answer:
scoundrel [369]2 years ago
3 0

Answer:

solve this and get 25 points.

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When 0.040 mol of propionic acid, c2h5co2h, is dissolved in 750 ml of water, the equilibrium concentration of h3o+ ions is measu
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<span>Answer is: Ka for propinoic acid is 6,57·10</span>⁻⁵.<span>
Chemical reaction: C</span>₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).<span>
n(C</span>₂H₅COOH) = 0,04 mol.<span>
V(C</span>₂H₅COOH) = 750 mL = 0,75 L.<span>
c(C</span>₂H₅COOH) = 0,04 mol ÷ 0,75 L.<span>
c(C</span>₂H₅COOH) = 0,053 mol/L = 0,053 M.<span>
[C</span>₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [C</span>₂H₅COO⁻] · [H₃O⁺] / [C₂H₅COOH].<span>
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10</span>⁻⁵.

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