Question

The pH of an aqueous solution of 0.109 M ammonium iodide, NH4I (aq), is ________.

This solution is ( acidic, basic, or neutral)

Answer 1

Kb for NH3 is 1.8*10^-5

Find Ka of NH4+

use:

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/1.8*10^-5

Ka = 5.556*10^-10

NH4+ + H2O -----> NH3 + H+

0.109 0 0

0.109-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*0.109) = 7.782*10^-6

since c is much greater than x, our assumption is correct

so, x = 7.782*10^-6 M

so.[H+] = x = 7.782*10^-6 M

use:

pH = -log [H+]

= -log (7.782*10^-6)

= 5.11

Answer: 5.11

Since pH is less than 7, this is acidic