a) What is the pH of 0.44 M triethylammonium iodide, (C 2 H 5 ) 3 NHI. The K b o

Question

a) What is the pH of 0.44 M triethylammonium iodide, (C2H5)3NHI.
The Kbof triethylamine, (C2H5)3N, is 5.2 x 10-4

b) Calculate the pH of a 0.10 M solution of sodium formate (NaHCOO) given that the Ka of formic acid (HCOOH) is 1.8 x 10-4.

c)

H3PO3(aq) + HSO3-(aq) <=> H2PO3-(aq) + H2SO3(aq)

General chem 2. please help =(

Could you please explain the trends of acids and how to figure out how it can be strong like in question C.

Consider the reaction shown. Using your knowledge of relative acid-base strengths and equilibrium, determine what you can about the size of Kc for the reaction.

H3PO3(aq) + HSO3-(aq) <=> H2PO3-(aq) + H2SO3(aq)

2. Kc < 1

Explanation / Answer

(CH3CH2)3NH+ + H2O <-----> (CH3CH2)2N + H3O+
K = Kw/Kb =1.7 x 10^-11 = x^2/ 0.44-x
x = [H3O+]= 3.1 x 10^-6 M
pH =5.5

HCOOH <----> HCOO- + H+
1.8 x 10^-4 = x^2 / 0.10-x
x = [H+] = 0.0073 M
pH = 2.1

Kc < 1 2

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