Question #a8da2

To solve this problem, you need the value of the base dissociation constant of ammonia, #K_b#, which is listed as being equal to #1.8 * 10^(-5)# #-># see here.

Since ammonia is a weak base, it will increase the concentration of #"OH"^(-)# ions in the solution, so you would expect the solution to have a pH greater than #7#.

Use the ICE table (more here) to solve for the concentration of #"OH"^(-)# ions, which are labeled here as #x#

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

#"I"color(white)(aaaaacolor(black)(0.15)aaaaaaaaaaaaaaaa color(black)(0) aaaaaaa color(black)(0)#
#"C"color(white)(aaaaacolor(black)(-x)aaaaaaaaaaaaaaa color(black)(+x) aaaaa color(black)(+x)#
#"E"color(white)(aaacolor(black)(0.15-x)aaaaaaaaaaaaaa color(black)(x) aaaaaaa color(black)(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 compared with the initial concentration of ammonia, you can use the approximation

#0.15 - x ~~ 0.15#

You will have

#1.8 * 10^(-5) = x^2/0.15#

#x = sqrt(0.15 * 1.8 * 10^(-5))#

#x = 0.001643#

This means that you have

#["OH"^(-)] = "0.001643 M"#

The solution's #"pOH"# will be

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

#"pOH" = -log(0.001643) = 2.78#

Therefore, the #"pH"# of the solution will be

#"pH" = 14 - "pOH"#

#"pH" = 14 - 2.78 = color(green)(11.22)#