Calculate the pH of a 0.44 M aqueous solution
of NH4Cl. Kb = 1.8 × 10−5
.
I've tried used the Ka=Kw/Kb equation then used the resulting 5.5e-10 in Kw=([H+][Cl-])/[NH4Cl] , using that to get the [H+], but it isn't correct and I don't know what else to do.
2 answers
Correction. I meant Ka=([H+][Cl-])/[NH4Cl] not Kw
You're right that Ka = Kw/Kb = 5.55E-10
But your equation is wrong.
Here is the hydrolysis equation which you need.
.........NH4^+ + H2O ==> NH3 + H3O^+
I........0.44.............0......0
C.........-x..............x......x
E.......0.44-x............x.......x
Then Ka = (x)(x)/(0.44-x)
Solve for x = (H3O^+) and convert to pH. Most NH4Cl solution are somewhere around pH 5 so you should get close to that (but probably not 5.0).
But your equation is wrong.
Here is the hydrolysis equation which you need.
.........NH4^+ + H2O ==> NH3 + H3O^+
I........0.44.............0......0
C.........-x..............x......x
E.......0.44-x............x.......x
Then Ka = (x)(x)/(0.44-x)
Solve for x = (H3O^+) and convert to pH. Most NH4Cl solution are somewhere around pH 5 so you should get close to that (but probably not 5.0).
7 answers