The trimethylammonium ion, [(CH₃)₃NH]⁺, is a weak acid - HSC - SSCE Chemistry - Question 20 - 2021 - Paper 1

Since the dissociation of [(CH₃)₃NH]⁺ in water leads to the production of H₃O⁺ and (CH₃)₃N, we can assume that at equilibrium:

[ (CH₃)₃NH]⁺_{equilibrium} = (C - x)$$ Where C is the concentration of trimethylammonium chloride, assumed to be in excess.

From the equation of dissociation, we can write the expression for the acid dissociation constant:

$K_a = \frac{[H₃O⁺][(CH₃)₃N]}{[(CH₃)₃NH]⁺}$

Using the approximate values from equilibrium conditions, we substitute the known quantities. Since the concentration of [(CH₃)₃N] should equal the change in concentration of [(CH₃)₃NH]⁺, where [H₃O⁺] = 3.47 × 10^{-5}:

Substituting values into the Ka expression gives:

$1.55 × 10^{-10} = \frac{(3.47 × 10^{-5})(3.47 × 10^{-5})}{(C - 3.47 × 10^{-5})}$

For the relationship between Kₐ and Kₚ for the salt in solution, we know:

$K_p = K_a \cdot [H₂O]$

Based on the constants given in the problem:

Assuming the concentration of pure water is approximately 55.5 M at 20°C:

$K_p = (1.55 × 10^{-10}) imes (55.5) = 8.60 × 10^{-9}$

Which reflects the dissociation properties confirming option C aligns closely with these calculations when analyzed further.