Calculate the pH of 0.15 M aqueous solution of ammonia?

The of the solution is 11.72.
To solve this problem you need the value of the base dissociation constant of ammonia, ##K_b##, which is listed as being ##1.8 * 10^(-5)##.
Since ammonia is aweak base, it will increase the concentration of ##OH^(-)## ions in solution, so you would expect the solution to have a greater than 7.
Use the ICE table (more here: of ##OH^(-)## ions – labeled as ##x##
##” “NH_(3(aq)) + H_2O_((l)) rightleftharpoons NH_(4(aq))^(+) + OH_((aq))^(-)##I…….0.15………………………………….0…………….0C……(-x)…………………………………..(+x)………..(+x)E…..0.15-x………………………………..x…………….x
Use the definition of the base dissociation constant
##K_b = ([OH^(-)] * [NH_4^(+)])/([NH_3])##
##1.8 * 10^(-5) = ( x * x)/(0.15 – x) = x^2/(0.15-x)##
Because ##K_b## is so small, you can approximate 0.15 – x with 0.15
##1.8 * 18^(-5) = x^2/0.15 => x = 0.005196##
The solution’s pOH will be
##pOH = – log([OH^(-)]) = -log(0.005196) = 2.28##
Therefore, the pH of the solution will be
##pH_”sol” = 14 – pOH = 14 – 2.28 = color(green)(11.72)##