Chem Question

Question

Consider an aqueous solution of ammonia, NH3, a base. If the resulting pH of this solution is 9.5, then what is the concentration of this ammonia solution? Note that the Ka for ammonium NH4, is 5.56 x 10^(-10). The professor has given the answer to be [NH3]i= 8.7 x 10^(-5) M. I just don't understand how to get this number. Thanks!

Answer 1

You need to understand the relationship between Ka, Kb and Kw.

Kw = 1x10^-14 = Ka x Kb

Thus, Kb for NH3 = 1x10^-14/5.56x10^-10

Kb = 1.8x10^-5

You also need to understand how NH3 reacts in water:  NH3 + H2O ==> NH4⁺ + OH^-

Kb = [NH4⁺][OH^-]/[NH3] = 1.8x10^-5

From the given pH of 9.5, we can find the pOH since pH + pOH = 14

pOH = 14 - 9.5 = 4.5

From the pOH we can find the [OH-]

[OH^-] = 1x10^-4.5 = 3.16x10^-5 M.  This would also be the [NH4+]

Substituting this value into the Kb expression, we can solve for [NH3]

1.8x10^-5 = (3.16x10-5)(3.16x10-5)/[NH3]

[NH3] = 5.6x10^-5 M

Unless I've made a silly math error, I'd have to disagree with the answer given by your professor.