Question

Calculate the pH of 0.10 F ethylammonium chloride (CH3CH2NH3Cl) (100% soluble salt). The Kb for ethylamine (CH3CH2NH2) is 4.7 x 10–4. Use the systematic approach to equilibrium to derive in detail the expression you used to calculate either [OH–] or [H+].

Answer 1

pH of 0.10 F of CH3CH2NH3Cl

kb for CH3CH2NH2 = 4.7x10^-4

need to find Ka

Ka x Kb = Kw

Ka x 4.7x10^-4 = 1.0x10^-14

Ka = 2.12x10^-11

now, we need to find H⁺ or OH^- of solution by using ICE table

.......CH₃CH₂NH₃Cl ----------> CH₃CH₂NH₂Cl^- + H⁺
Ini. ...... 0.10 F ................................... 0 .............................. 0
C. ....... -x ........................................... +x ........................... +x
Eq. ...... .(0.10 -x ) ..................................... x .............................. x

Ka = [x] [x] / [0.10-x]

2.12x10^-11 = x² / (0.10-x) ........(assume x << 0.10 so, negligible)

x²= 2.12x10^-12

x = 1.45x10^-6

[H+] = x = 1.45x10^-6

pH = -log[H⁺]

pH = -log [1.45x10^-6]

pH = 5.83