Question

Including activity coefficients, find the molar concentration of Ba2+ in a 0.0500 M solution of tetramethylammonium iodate (soluble salts which completely dissociates into (CH3)4N + and IO3 - ions in aqueous solution. Hint: the [IO3 - ] from the slightly soluble BaIO3 in negligible.

Answer 1

Concentration of tetramethylammonium iodate = 0.0500 M

Ksp of Ba(IO3)2 = 1.5x10^-9  

Activity Coefficients, (Ba2+) = 0.38 and (IO3-) = 0.775

Since, tetramethylammonium iodate completely dissociates into (CH3)4N + and IO3 - ions in aqueous solution.

So, [IO3-] = 0.0500 M

Now,

Ba(IO3)2      Ba²⁺   + 2IO₃^-
   s 0.0500 + 2s

Ksp = s*(0.0500 + 2s)²

Since, the [IO3 - ] from the slightly soluble BaIO3 in negligible.

2s << 0.050

So, Ksp = s * (0.0500)²  

Mean ionic activity coefficient = [(Ba2+) ][(IO3-)]²  

= (0.38)(0.775)² = 0.228

With the activity coefficients:

Ksp = s * (0.0500)² * Mean ionic activity coefficient

s = Ksp / [(0.0500)² * 0.228]

s = 1.5x10^-9 / [(0.0500)² * 0.228]

s = 2.63x10^-6

Hence,

[Ba2+] = 2.63x10^-6 M