In: Chemistry
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+].
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