Question

calculate pH of c9h13n with a pkb of 4.2 with a concentration of 225 milligrams per liter... but please show me a decent ice table spelled out so I can understand it thank you

Answer 1

C9H13N + H2O <==> C9H13NH+ + OH-

moles of C9H13N per liter = 0.225 g/ 135g/mol

                                      = 1.67*10^-3 mol

Molarity = 0.00167 M

	C9H13N	C9H13NH+	OH-
initial	0.00167 M	0	0
change	-x	+x	+x
equilibrium	0.00167-x	x	x

Kb = 10^-4.2 = 6.31*10^-5

Kb = [C9H13NH+][OH-]/[C9H13N]

or, 6.31*10^-5 = x*x/(0.00167-x)

or, 1.05*10^-7 - 6.31x*10^-5 - x^2 = 0

or, x =2.94*10^-4 M

[OH-] = 2.94*10^-4 M

pOH = -log[2.94*10^-4 M] = 3.53

pH = 14-pOH = 10.47