Ammonia, NH3 is freely soluble in water, and has the value Kb=1.8x10^-5. If the pH of an ammonia solution is 11.4272, then what original concentration of NH3 (molarity) was used to make up the solution?

Answer 1

NH3 + H2O ==> NH4^+ + OH^-

Kb = (NH4^+)(OH^-)/(NH3)
(x)(x)/(NH3)-x = 1.8 x 10^-5
pH = 11.4272 (where did all these places come from---we only know Kb to 2 places.)
pH = 11.43 pOH = 14-11.43 = 2.57
2.57 = -log(OH^-) and (OH^-) = 2.69 x 10^-3
Substitute into Kb expression and solve for (NH3). You should get something close to 0.4 M but you need to confirm all of the above.

Answer 2

Thank you! I tried all the above and got 0.40 M