Question

Ka for a weak acid, HA, is 1.6x10-6. What are the (a) pH and (b) degree of ionization of the acid in a 10-3 M solution?
(c) Calculate pKa and pKb.

Answer 1

...................HA ==> H^+ + A^-
I.............0.001 M.......0.........0
C................-x.............x..........x
E............0.001-x.........x..........x
Plug the E line into the Ka expression and solve for x = (H^+). Convert H^+ to pH the usual way. You did that in the question just below.
degree of ionization = (H^+)/(HA) in the E line above.
pKa = -log Ka and you have Ka.
You get pKb from pKa + pKb = pKw = 14. You know pKa and pKw, solve for pKb. Post your work if you get stuck.

Answer 2

how did we get the molarity (0.001)
I did not know how to solve the question because there is no molarity can you explain please.

Answer 3

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