Amphetamine (C9H13N)(C9H13N) is a weak base with a pKbpKb of 4.2. You may want to reference (Pages 710 - 713) Section 16.8 while completing this problem. Part A Calculate the pHpH of a solution containing an amphetamine concentration of 205 mg/Lmg/L. Express your answer to two d

pH = 10.38

Explanation:

∴ molar mass C9H13N = 135.21 g/mol

∴ pKb = - log Kb = 4.2

⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]

∴ C sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M

mass balance:

⇒ C sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N](1)

charge balance:

⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water

⇒ [C9H20O3N+] = [OH-](2)

(2) in (1):

⇒ [C9H13N] = 1.516 E-3 - [OH-]

replacing in Kb:

⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])

⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0

⇒ [OH-] = 2.3985 E-4 M

∴ pOH = - Log [OH-]

⇒ pOH = 3.62

⇒ pH = 14 - pOH = 14 - 3.62 = 10.38

i think its 5049.

i hope this !

gas

explanation:

smoke is the matter of a gas