Question #96b5e

Ammonia, #"NH"_3#, is a weak base, so right from the start you should expect the pH of the solution to be higher than #7#, the pH of pure water at room temperature.

When dissolved in aqueous solution, ammonia reacts with water to form ammonium cations, #"NH"_4^(+)#, and hydroxide anions, #"OH"^(-)#, according to the equilibrium reaction

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

Since ammonia is a weak base, only a tiny fraction of the ammonia molecules will ionize to form ammonium cations and hydroxide anions. The majority of the molecules will remain unionized, i.e. they will not accept a proton from water.

Now, your goal here is to find the equilibrium concentration of the hydroxide anions, since this will allow you to calculate the solution's pOH, and consequently its pH.

Your tool of choice here will be an ICE table

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

#color(purple)("I")color(white)(aaaaacolor(black)(0.250)aaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0)#
#color(purple)("C")color(white)(aaaacolor(black)((-x))aaaaaaaaaaaaaaaacolor(black)((+x))aaaaaaacolor(black)((+x))#
#color(purple)("E")color(white)(aaacolor(black)(0.250-x)aaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaaaaacolor(black)(x)#

By definition, the base dissociation constant, #K_b#, will be equal to

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

In your case, this will be equal to

#1.85 * 10^(-5) = (x * x)/(0.250 - x) = x^2/(0.250 - x)#

Notice that the value of #K_b# is significantly smaller than the initial concentration of ammonia. This means that you can use the approximation

#0.250 - x ~~ 0.250#

You will thus have

#1.85 * 10^(-5) = x^2/0.250 implies x^2 = 0.250 * 1.85 * 10^(-5)#

This will get you

#x = sqrt(0.250 * 1.85 * 10^(-5)) = 2.15 * 10^(-3)#

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

#["OH"^(-)] = x = 2.15 * 10^(-3)"M"#

The pOH of the solution is defined as

#color(blue)(|bar(ul(color(white)(a/a)"pOH" = - log(["OH"^(-)])color(white)(a/a)|)))#

Plug in your value to find

#"pOH" = - log(2.15 * 10^(-3)) = 2.67#

As you know, an aqueous solution at room temperature has

#color(blue)(|bar(ul(color(white)(a/a)"pH " + " pOH" = 14color(white)(a/a)|)))#

This means that the pH of the solution will be

#"pH" = 14 - 2.67 = color(green)(|bar(ul(color(white)(a/a)color(black)(11.33)color(white)(a/a)|)))#

As predicted, the pH of the solution is indeed higher than #7#.