Question

1)The pH of aqueous 0.477 M pyridine (C₅H₅N) ion is 9.43. What is the K_b of this base?

2)Which ionic compound forms a pH-neutral aqueous solution at 25 °C?

	LiF
	LiClO4
	K2S
	NaHCO3
	NH4Cl

Answer 1

2) LiClO4

salt of a strong acid (HClO4) and a strong base (LiOH)

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1)

C5H5N is a weak base

pOH = 14-pH = 14 -9.43 = 4.57

[OH-] = antilog (-pOH) = antilog (-4.57) = 0.0000269 M

C5H5N + H2O <-----> C5H5NH+ + OH-

0.477 0 0

(0.477-x) +x +x

Kb = [C5H5NH+] * [OH-] /[C5H5N]

Kb = x*x /(0.10-x)

[C5H5N] = (0.477-x = 0.477 (appx)

x= [OH-] = 0.0000269 M

Kb= x^2 /0.1

= 0.0000269^2 /0.477

= 1.52*10^-9