Question #2c896

1 Answer
Dec 1, 2015

Answer:

#["H"_3"O"^(+)] = 1.7 * 10^(-5)"M"#

Explanation:

Ammonium chloride, #"NH"_4"Cl"#, will dissociate completely in aqueous solution to produce ammonium cations, #"NH"_4^(+)#, and chloride anions, #"Cl"^(-)#

#"NH"_4"Cl"_text((aq]) -> "NH"_text(4(aq])^(+) + "Cl"_text((aq])^(-)#

Notice that the salt dissociates in a #1:1# mole ratio with the ammonium ion. This means that you get one mole of the latter for every mole of the former that is dissolved in solution.

So, if #0.20# moles of ammonium chloride are used to make this solution, it follows that the molarity of the ammonium ions will be

#n_(NH_4^(+)) = n_(NH_4Cl) = "0.20 moles"#

#c = "0.20 moles"/(400.0 * 10^(-3)"L") = "0.50 M"#

Once in aqueous solution, the ammonium ions will act as a weak acid and react with the water molecules to form ammonia, #"NH"_3#, a weak base, and hydronium ions, #"H"_3"O"^(+)#.

To determine the equilibrium concentration of the hydronium ions, use an ICE table

#" " "NH"_text(4(aq])^(+) + "H"_2"O"_text((l]) " "rightleftharpoons" " "NH"_text(3(aq]) " "+" " "H"_3"O"_text((aq])^(+)#

#color(purple)("I")" " " "0.50" " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " "0#
#color(purple)("C")" "color(white)(x)(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)#
#color(purple)("E")" "0.50-x" " " " " " " " " " " " " " " "x" " " " " " " " " " " " "x#

Now, notice that the poroblem provides you with the base dissociation constant, #K_b#, of ammonia. This means that you're going to have to use the equation

#color(blue)(K_a * K_b = K_W)" "#, where

#K_W# - the self-ionization constant of water, equal to #10^(-14)# at room temperature

to find the acid dissociation constant, #K_a#, of the ammonium ion.

#K_a * K_b = K_W implies K_a = K_W/K_a#

#K_a = 10^(-14)/(1.8 * 10^(-5)) = 5.6 * 10^(-10)#

By definition, the acid dissociation constant is equal to

#K_a = (["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])#

In your case, this will be equal to

#K_a = (x * x)/(0.50 - x) = x^2/(0.50 - x)#

Since the value of #K_a# is so small, you can use the approximation

#0.50 - x ~~ 0.50#

to get

#K_a = x^2/0.50 = 5.6 * 10^(-10)#

This means that you have

#x = sqrt(0.50 * 5.6 * 10^(-10)) = 1.7 * 10^(-5)#

Since #x# represents the equilibrium concentrations of ammonia and hydronium ions, it follows that you have

#["H"_3"O"^(+)] = color(green)(1.7 * 10^(-5)"M")#