Question

What is the pH of a buffer that is 0.100 M methylamine and 0.175 M methylammonium chloride at 25 °C? pkb for methylamine = 3.36.

Answer 1

pH = 10.396

Step-by-step explanation:

CH3NH2 + H2O ↔ CH3NH3+ + OH-

⇒ pKb = -log ( Kb ) = 3.36

⇒ Kb = 4.365 E-4 = ( [ CH3NH3+ ] * [ OH- ] ) / [ CH3NH2 ]

CH3NH3Cl ↔ CH3NH3+ + Cl-

0.175 M 0.175 M 0.175 M

mass balance:

M CH3NH2 + M CH3NH3Cl = [ CH3NH3+ ] + [ CH3NH2 ] = 0.100 + 0.175 = 0.275

load balance:

[ CH3NH3+ ] = [Cl- ] + [ OH- ] ......... [ OH- ] from water

∴ [ Cl- ] = 0.175 M

⇒ [ CH3NH3+ ] = 0.175

in the mass balance:

⇒ 0.275 = 0.175 + [ CH3NH2 ]

⇒ 0.1 = [ CH3NH2 ]

⇒ Kb = 4.365 E-4 = (( 0.175 ) * [ OH- ] ) / (0.1 )

⇒ 4.365 E-5= 0.175 [ OH-]

⇒ [ OH- ] = 2.494 E-4

⇒ pOH = - log ( 2.494 E-4)

⇒ pOH = 3.6

⇒ pH = 14 - pOH

⇒ pH = 10.396