Question

Calculate the pH of a solution containing an amphetamine concentration of 250 mg/

Amphetamine (C9H13N) is a weak base with a pKb of 4.2.

Answer 1

C9H13N + H2O C9H13NH+ + OH- ;

kb= 10^pkb

so kb= 6.3 x 10^-5

[C9H13N]=

=1.85 x 10^-3

Kb = [ C_9H_13NH+] [OH-] / [ C_9H_13N]

6.3 x 10^-5 = (x) (x) / 1.85 x 10^-3- x

6.3 x 10^-5 = x² / 1.85 x 10^-3- x

1.1655 x 10^-7- 6.3 x 10^-5 x = x²

x2 + 6.3 x 10^-5 x - 1.1655 x 10^-7 = 0

If you assume that X << 0.001849 (and the denominator becomes 0.001849), you get X = 0.0003416
So the denominator should have been closer to 0.001849 - 0.000342 = 0.001507 and you will get X = 0.000308
Trying one more time, 0.001849 - 000308 = 0.001541, getting X = 0.000312
pOH = 3.51 so pH = 14.00 - 3.51 = 10.49