What is the definition of K_a? Write out the equilibrium reaction of a generic a

Question

What is the definition of K_a? Write out the equilibrium reaction of a generic acid HA, that defines the equilibrium constant K_a. Write the general expression you would use to calculate K_a from concentration(s). How are K_a and pK_a related? How would you calculate one from the other? How does the k_a value of an acid relate to the pH of its aqueous solution? What is the definition of K_b? Write out the equilibrium reaction of a generic base. B. that defines the equilibrium constant K_b. Write the general expression you would use to calculate K_b from concentration(s). How are K, and pk_b related? How would you calculate one from the other? How does the K_b value of a base relate to the pH of its aqueous solution? How would you change the reaction you wrote in part (a) if you were instead asked about K_b for the conjugate base of the acid HA? When and how are the K_a, of an acid and the K_b of a base related? The pH of a solution of acetylsalicylic acid (commonly known as aspirin. "HA" = HC_9H_7O_4) was determined to be 1.74. If the K_a value for this weak acid is 3.3 Times 10^-4. answer the following questions: What is the concentration of H_3O^+ ions in this solution? What is the concentration of C_9H_7O_4^-(A^-) in this solution? Assume the auto-ionization of water is negligible. What is the concentration of un-ionized acetylsalicylic acid (HA) still present in the solution? How many moles of acetylsalicylic acid (HC_9H_7O_4) must have been dissolved in 1.00 L of water to prepare this solution? A second solution is prepared by dissolving 2.00 moles of acetylsalicylic acid in 1.00 L of water. What would be the pH of this second solution? Comment on the effect of concentration on the pH of the solution. Are these two quantities proportional?

Explanation / Answer

1) Ka = acid dissociation constant or equilibrium constant for the dissociation of an acid.
HA + H2O ------> A- + H3O+
Ka = [A-][H3O+]/[HA][H2O]
Ka = [A-][H3O+]/[HA]
pKa = -log[Ka]
only acids whic under go weak dissociation have the Ka vales. using these values we can calculate the pH of the acid
pH = 1/2(pKa -logC)
2) Kb = base dissociation constant or equilibrium constant for the dissociation of a base
BOH -------> B+ + OH-
Kb = [B=][OH-]/[BOH]
pKb = -log[Kb]
pOH = 1/2(pKb - logC)
pH = 14 - pOH = 14 - 1/2(pKb - logC)
3)pH = 1/2(pKa - log C)
1.74 = 1/2(-log(3.3*10^-4)-log(C))
C = 1.0034 M
HA -----> H+ + A-
1.0034     0    0
(1.0034-x) x    x
3.3*10^-4 = x*x/(1.0034-x)
x = 0.018
[H3O+] = 0.018 = [C9H7O4-]
un ionised part of [HA] = 1.0034 - 0.018 = 0.9854
1.0034 = n/1L
n = 1.0034 moles of acetylsalcylic acid

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