Answer to Question #230247 in General Chemistry for Hasen

Question

If a 0.2M NH4Cl solution has a ph of 5, what is the Ka of NH4?

Answer 1

NH₄Cl dissociates as:

$NH4Cl <--> NH4+ + Cl-$

From here:

$Ka =\frac{ ([NH4^+][Cl^-])}{[NH4Cl]} = \frac{[NH4^+]2}{[NH4Cl]}$

As pH = -log[H⁺]:

[H⁺] = 10^-pH

From here, the concentration of [NH4⁺] in the solution equals:

[H⁺] = 10^-5 M

As [H⁺] = [NH4⁺]= 10^-5 M, [NH₄Cl] = 0.200 M,

Ka of NH4⁺ equals:

$Ka =\frac{ [NH4+]2}{[NH4Cl] }=\frac{ (10-5)2 }{ 0.2} = 5 × 10^{-10}$