Calculate the K_a of a mono weak acid if a 0.50 M solution has 1.14% ionization.

Question

Calculate the K_a of a mono weak acid if a 0.50 M solution has 1.14% ionization. a) 6.57 times 10^-5 b) 3.84 times 10^-6 c) 2.66 times 10^-4 d) 2.59 times 10^-4 e) 7.49 times 10^-5 How many grams of barium hydroxide would you need to prepare 400 mL of a solution with pH = 12.48? a) 5.17 g b) 4.14 g c) 2.58 g d) 2.07 g e) 1.04 g Calculate the K_b for diethylamine, (C_2H_5)_2NH, if the pH of a 0.975 M (C_2H_5)_2 NH solution is 12.55. a) 4.84 times 10^-4 b) 9.66 times 10^-10 c) 3.91 times 10^-5 d) 7.74 times 10^-26 e) 1.34 times 10^-3 Which of the following is true for a 0.15 M solution of a weak acid, HA, with K_a = 1.4 times 10^-4? a) The acid would be 3.1% dissociated. b) The equilibrium concentration of the acid, HA, would be 0.145 M. c) The solution would have a pH = 2.34 d) The equilibrium concentration of the conjugate base would be 4.58 times 10^-3 M. e) All of the above are true.

Explanation / Answer

24. The equation for the dissociation of HA is:

HA H+ + A¯

The Ka expression is:

Ka = ([H+] [A¯]) / [HA]

[H+] = (0.50 M) (0.0114) = 0.0057 M

1:1 stoichiometry hence:

[A¯] = [H+] = 0.0057 M

Ka = ([H+] [A¯]) / [HA] = [0.0057 M]2 / 0.50 M = 6.5 x 10-5

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