Question #62d6f

The idea here is to use the ammonium ion's acid dissociation constant, #K_a#, to find the value of ammonia's base dissociation constant, #K_b#.

The relationship between the base dissocaition constant, the acid dissociation constant, and water's self-ionization constant, #K_W#, is given by the equation

#K_W = K_a xx K_b#

In your case, the base dissociation constant of ammonia will be

#K_b = K_W/K_a = 10^(-14)/(5.62 * 10^(-10)) = 1.78 * 10^(-5)#

To determine the pH of the solution, you first need to know the #p"OH"# of the solution, which in turn requires the cocnentration of hydroxide ions, #"OH"^(-)#.

Ammonia will react with water to form ammonium and hydroxide ions. Use an ICE table to determine the equilibrium concentration of hydroxide ions

#"NH"_text(3(aq]) " "+" " "H"_2"O"_text((l]) " "rightleftharpoons" " "NH"_text(4(aq])^(+) " "+" " "OH"_text((aq])^(-)#

#color(purple)("I")" " " " 0.1" " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " "0#
#color(purple)("C")" " color(white)(x)-x" " " " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)#
#color(purple)("E")" " color(white)(x)0.1-x" " " " " " " " " " " " " " " "color(white)(xx)x" " " " " " " " " "color(white)(x)x#

By definition, the base dissociation constant will be

#K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3]) = (x * x)/(0.1 - x)#

Because the value of #K_b# is so small, you can say that

#(0.1 - x) ~~ 0.1#

This means that you have

#K_b = x^2/0.1 = 1.78 * 10^(-5)#

#x = sqrt(0.1 * 1.78 * 10^(-5)) = 0.001334#

This means that you have

#["OH"^(-)] = x = "0.001334 M"#

The #p"OH"# of the solution will be

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

#p"OH" = -log(0.001334) = 2.87#

The pH of the solution will thus be

#p"H" = 14 - p"OH" = 14 - 2.87 = color(green)(11.13)#