What is the #"pH"# of a #"1.5-M"# solution of ammonia? The dissociation constant of ammonia at #25.0^@"C"# is #1.80 * 10^-5#.

As you know, ammonia acts as a weak base in aqueous solution, so right from the start, you should expect the #"pH"# of the solution to be #> 7#.

#"NH"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-)#

The ratio that exists between the equilibrium concentrations of the ammonium cations and of the hydroxide anions and the equilibrium concentration of ammonia is given by the base dissociation constant, #K_b#.

#K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3])#

Now, ammonia will only partially ionize to produce ammonium cations and hydroxide anions. If you take #x# #"M"# to be the concentration of ammonia that ionizes, you can say that, at equilibrium, the solution will contain

#["NH"_4^(+)] = ["OH"^(-)] = x quad "M"#

This happens because every mole of ammonia that ionizes produces #1# mole of ammonium cations and #1# mole of hydroxide anions.

So if #x# #"M"# ionizes, you can expect the solution to contain #x# #"M"# of the two ions.

The solution will also contain

#["NH"_3] = (1.5 - x) quad "M"#

When #x# #"M"# ionizes, the initial concentration of ammonia will decrease by #x# #"M"#.

This means that the expression of the base dissociation constant will now take the form

#K_b = (x * x)/(1.5 - x)#

which is equal to

#1.80 * 10^(-5) = x^2/(1.5 - x)#

Notice that the value of the base dissociation constant is significantly smaller than the initial concentration of the base. This tells you that you can use the approximation

#1.5 -x ~~ 1.5#

because the concentration of ammonia that ionizes will be significantly lower than the initial concentration of the base, i.e. the ionization equilibrium will lie to the left.

You now have

#1.80 * 10^(-5) = x^2/1.5#

Rearrange and solve for #x# to get

#x = sqrt(1.5 * 1.80 * 10^(-5)) = 0.005196#

Since #x# #"M"# represents the equilibrium concentration of the hydroxide anions, you can say that

#["OH"^(-)] = "0.005196 M"#

Now, an aqueous solution at #25^@"C"# has

#"pH + pOH = 14"#

Since

#"pOH" = - log(["OH"^(-)])#

you can say that the #"pH"# of the solution is given by

#"pH" = 14 + log(["OH"^(-)])#

Plug in your value to find

#"pH" = 14 + log(0.005196) = color(darkgreen)(ul(color(black)(11.72)))#

The answer is rounded to two decimal places because you have two sig figs for the molarity of the solution.