What is the pH of a 0.280 M solution of trimethylammonium chloride ((CH3CH2)NH3Cl)? Kb=5.6×10^-4 the answer is 5.7 but i need the work shown

Answer 1

Let's look at what happens when you put trimethylammonium chloride in water:

(CH₃)₃NHCl = trimethlammonium chloride (you have an incorrect structure, but it won't matter for the math)

(CH₃)₃NH⁺ + H₂O ==> (CH₃)₃N + H₃O⁺

acid................base............base..........acid......

Kb = 5.6x10^-4 but since the conjugate acid is being hydrolyzed, we want to use the Ka:

Ka = 1x10^-14 / Kb = 1x10^-14 / 5.6x10^-4

Ka = 1.79x10^-11

Ka = [(CH₃)₃N][H₃O⁺]/[(CH₃NH⁺]

1.79x10^-11 = (x)(x) / 0.280-x and if we assume x is small relative to 0.280 we can ignore it in the denominator

x² = 5.0x10^-12

x = 2.2x10^-6 = [H₃O⁺] (which is indeed small compared to 0.280 so we were ok above)

pH = -log [H₃O⁺] = -log 2.2x10^-6

pH = 5.7