Calculate the K_a of a monoprotic weak acid if a 0.50 M solution has 1.14% ioniz

Question

Calculate the K_a of a monoprotic 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)_2 NH, if the ph of a 0.975 M (C_2H_5)_2 NH solution is 12.55. a) 4.84times 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

Q26.

Kb for: ethylamine

pH = 12.55, pOH =14-12.55 = 1.45

[OH] = 10^-pOH = 10^-1.45 = 0.035481

now

B + H2O <--> BH+ + OH-

Kb = [BH+][OH-]/[B]

substitute values

Kb = (0.035481)(0.035481)/(0.975-0.035481) = 0.001339942

Kb = 1.339*¨10^-3

Q27

similarly...

KA1 = [H+][A-]/[HA]

1.4*10^-4 = x*x/(0.15-x)

x = [H+] = 0.0045

pH = -log(0.0045) = 2.35

[HA] = M-x = 0.15 - 0.0045 = 0.1455 M

%ion = [H+]/M*100 = 0.0045/0.15*100 = 3 %

note that all samples are true

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