Consider the following reactions that take place in aqueous solution and relate

Question

Consider the following reactions that take place in aqueous solution and relate information. H_2CO_3 + HS- H_2S + HCO_3^- HSO_4 + HCO_3^- H_2CO_3 + SO_4^2- H_2CO_3 (K_a1 = 4.2 times 10^-7, K_a2 = 4.8 times 10^-11) H_2S (K_a1 = 1 times 10^-7, K_a2 = 1 times 10^-19) H_2SO_4 (K_a1 = Very Large, K_a2 = 1.2 times 10^-2) The strongest base in these reactions is. H_2S HCO_3^-(C) HSO_4^- H_2CO_3 HS^- In general, for a diprotic acid K_a1 >> K_a2. This is reasonable because: It is more difficult to remove a proton from the - 1 ion than it is from the neutral acid K_a1 times K_a2 = K for the combined equilibrium Diprotic acids nave 2 corresponds to a weak acid K_a1 corresponds to a strong acid while K_a2 corresponds to a weak acid None of the above A ___ M solution of Ba(OH)_2 has a pH of 12.7. 2.5 times 10^-2 1.0 times 10^-13 2.0 times 10^-13 5.0 times 10^-2 1.27 times 10^1

Explanation / Answer

14)The base which is conjugate base of weakest acid is strongest base

The weakest acid is H2S ( Ka = 1×10^-7)

So, the answer is e) HS-

15) The answer is a) it is more difficult to remove a proton from the -1 ion than it is from the neutral acid

c) The answer is d) 5.0× 10^-2 M

Explanation

pKb1= 0.15

Kb1 = 0.71

Kb1 = [B(OH)+][OH-]/[B(OH)2]

0.71 = X × X/(0.05-X)

X^2 +0.71X - 0.0355=0

X = 0.0469

[OH-] = 0.0469M

pOH = 1.33

pH = 14 -pOH = 14-1.33 = 12.7

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