pH of 0.40M triethylammonium chloride is 5.90.
A solution's acidity may be determined by looking at its pH, which is a measurement of hydrogen ion concentration. Pure water slightly separates into ions with roughly equal amounts of hydrogen and hydroxyl (OH) ions. [H+] is 107 for a neutral solution, or pH = 7.
Concentration of triethylammonium chloride = 0.40M
pH = ?
(CH3CH2)3NHCl ------> (CH3CH2)3NH⁺ + Cl⁻
(CH3CH2)3NH⁺ will react with water to give H3O⁺ .
(CH3CH2)3N will have a Kb = 5.2 x 10 ^(-4)
Kw = Kb x Ka
=> Ka = Kw / Kb = 10^(-14) / 5.2 x 10 ^(-4)
=> Ka = 1.92 x 10^(-11)
so by the reaction we have ,
Ka = x²/(0.40 - x)
=> x = 1.2393 x 10 ^(-6)
now, pH = -log( [H3O⁺]) = - log ( 1.2393 x 10 ^(-6)) = 5.906
To learn more about pH :
https://brainly.com/question/15289741
#SPJ4
