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)##