HC_9H_7O_4 (aq) + H_2O(l) H_3O^+ (aq) + C_9H_7O_4^- (aq) The molecular formula o

Question

HC_9H_7O_4 (aq) + H_2O(l) H_3O^+ (aq) + C_9H_7O_4^- (aq) The molecular formula of acetylsalecylic acid, also know as aspirin, is HC_9H_7O_4. The dissociation of HC_9_H_7O_4(aq) is represented by the equation above. The pH of 0.0100 M HC_9H_7O_4(aq) is measured to be 2.78. (a) Write the expression for the equilibrium constant, K_a, for the reaction above. (b) Calculate the value of K_a for acetylaslicylic acid. (c) An aqueous solution of aspirin is buffered to have equal concetrations of HC_9H_7O_4 (aq) and C_9H_7O_4^-(aq) Calculate the pH of the solution.

Explanation / Answer

1. ka = [ C9H7O4-] [H3O+] [HC9H7O4]

2.   HC9H7O4 + H2O <--> H3O+ + C9H7O4- { molarity of HC9H7O4 = 0.01 M}

initial 0.01 0 0

final 0.01-x x x

ka = [ C9H7O4-] [H3O+] [HC9H7O4] = x2 (0.01-x)

since; pH = - log [H+]

2.78= -log [H+]

[H+] = 0.0025 = X { since    [H3O+]= [H+]= X }

hence, by formula: ka = [ C9H7O4-] [H3O+] [HC9H7O4] = x2 (0.01-x) = (0.0025)2 (0.01- 0.0025)

=8.33* 10-4

3. ka = [ C9H7O4-] [H3O+] [HC9H7O4]

taking logarithm we get,

pH = pKa - log  [HC9H7O4] [ C9H7O4-]

when   [HC9H7O4] = [ C9H7O4-]

then, pH = pKa = 8.33* 10-4

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.