Question

If the pk, of CH NH, is 3.38, what is the phl of 0.70 M solution of CHANH.CH a. b. c. d. e. 5.39 4.67 7.00 8.61 4.11 N t her to be an entropy decrease in the system in which of the following processes might o a. CaCl (s)-> Cal(aq) + 2Cl(aq) b. HO(1) -> H2O(g) e. Fe(s) at 200 K --Fe(s) at 300 K d. N:0.(g) - 2NO, (g) e. 1xg) - 1:(s) 19. When HCl is added to a HF/NaF buffer system, the Ho from the strong acid will react with the producing more a. H.O. HO b. CI. HCI c. HF. F d. FF e. F, HF 20. If the solubility product of scandium(III) fluoride, ScFs, in pure water is 1.0 x 10"What is the solubility of ScF3 a. 1.0 x 10 M b. 1.4 x 10-M c. 6.2 10 M d. 2.0 x 10M e. 2.0 100M

Answer 1

17)

use:

pKb = -log Kb

3.38= -log Kb

Kb = 4.169*10^-4

use:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/4.169*10^-4

Ka = 2.399*10^-11

CH3NH3+ + H2O -----> CH3NH2 + H+

0.7 0 0

0.7-x x x

Ka = [H+][CH3NH2]/[CH3NH3+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.399*10^-11)*0.7) = 4.098*10^-6

since c is much greater than x, our assumption is correct

so, x = 4.098*10^-6 M

So, [H+] = x = 4.098*10^-6 M

use:

pH = -log [H+]

= -log (4.098*10^-6)

= 5.39

Answer: 5.39

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