Question

1. The pH of an aqueous solution of 0.285 M ammonium iodide, NH4I (aq), is _____. This solution is (acidic, basic, neutral)

Answer 1

Kb of NH3 = 1.8*10^-5

Lets find Ka of NH4+ which is conjugate acid of NH3

use:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

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

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

Ka = 5.556*10^-10

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

0.285 0 0

0.285-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)*0.285) = 1.258*10^-5

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

so, x = 1.258*10^-5 M

So, [H+] = x = 1.258*10^-5 M

use:

pH = -log [H+]

= -log (1.258*10^-5)

= 4.90

Since pH is less than 7, this is acidic

Answer: 4.90, acidic