Question #2c896
1 Answer
Explanation:
Ammonium chloride,
#"NH"_4"Cl"_text((aq]) -> "NH"_text(4(aq])^(+) + "Cl"_text((aq])^(-)#
Notice that the salt dissociates in a
So, if
#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,
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])^(+)#
Now, notice that the poroblem provides you with the base dissociation constant,
#color(blue)(K_a * K_b = K_W)" "# , where
to find the acid dissociation constant,
#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
#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
#["H"_3"O"^(+)] = color(green)(1.7 * 10^(-5)"M")#