Calculate the pH of a 0.430 M solution of NH4I?

Answer 1

0.430M NH4I has 0.430 M [H+]

pH = -log [H+] = -log(0.43) = 0.36

Added:

Yeah, this can't be right

at .1M soln has a pH of 4.6 ref: http://www.chemicalland21.com/industrialchem/inorganic/AMMONIUM%20IODIDE.htm

Here it is:

NH4I --> NH4+ + I-

NH4+ is an acidic ion and I- is a neutral ion; solution will be acidic.

NH4+(aq) + H2O(l) --> NH3(aq) + H3O+(aq)

Ka = [NH3][H3O+]/[NH4+]

Kw = 10-14 and Kb for ammonia = 1.8 *10-5

Ka = 10-14/1.8*10-5 = 5.6 *10-10

let x = equilibruim concentrations of NH3 and H3O+

then 0.43-x =[NH4+] left at equilibrium

and x2/(0.43-x) = 5.6 *10-10

or x = 2.36*10-5 = [H+]

then pH = -log[H+]= 4.63

Answer 2

NH4I goes to NH4+ + I-

Then,
NH4+ <-> H+ + NH3
Initial: .430 <-> 0 + 0
Change: -x <-> +x + x
_____________________________________
Equilibrium: .430-x<-> x + x

Ka=5.6 *10-10

Ka=concentration of products/reactants
5.6e10^-10=x^2/.430-x
solve for x
x=2.36*10-5

pH=-log(2.36*10^-5)
pH=4.63

Answer 3

pKb of NH3 or NH4OH is required.