Question #750c8
1 Answer
Here's what I got.
Explanation:
The problem wants you to use the base dissociation constant,
As you know, ammonia is a weak base, which means that it does not ionize completely in aqueous solution. Simply put, some molecules of ammonia will accept a proton from water, but most will not.
So, the balanced chemical equation for the ionization of ammonia looks like this
#"NH"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "NH"_text(4(aq])^(+) + "OH"_text((aq])^(-)#
Notice that you have a
This means that every molecule of ammonia that ionizes will produce one molecule of ammonium cations and one molecule of hydroxide anions.
In order to determine the percent ionization of the base, you need to know two things
- the initial concentration of the base
- either the equilibrium concentration of hydroxide anions or the concentration of ammonium cations
Use the solution's pH to help you find the equilibrium concentration of hydroxide anions. Determine the solution's pOH first
#color(blue)("pH " + " pOH" = 14)#
#"pOH" = 14 - 11 = 3#
Then use it to find
#color(blue)("pOH" = - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH"))#
In your case, you have
#["OH"^(-)] = 10^(-3)"M"#
So, you know that the initial
Assuming that
If you were to write an ICE table for the ionization of the base, you would have
#" " "NH"_text(3(aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "NH"_text(4(aq])^(+) " "+" " "OH"_text((aq])^(-)#
The base dissociation constant for this equilibrium reaction will be
#color(blue)(K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3])) -># uses equilibrium concentrations
In your case, you will have
#K_b = (x * x)/(["NH"_3]_0 - x) = 1.76 * 10^(-5)#
But you know that
#1.76 * 10^(-5) = (10^(-3))^2/(["NH"_3]_0 - 10^(-3)#
Rearrange to solve for
#["NH"_3]_0 + 10^(-3) = 10^(-6)/(1.76 * 10^(-5))#
#["NH"_3]_0 = "0.05682 M" + 10^(-3)"M" = "0.05782 M"#
So, what this means is that out of an initial concentration of
So, percent ionization will be equal to
#color(blue)("% ionization" = "what ionizes"/"what you start with" xx 100)#
In your case, you have
#"% ionization" = (10^(-3)color(red)(cancel(color(black)("M"))))/(0.05782color(red)(cancel(color(black)("M")))) xx 100 = color(green)(1.73%)#
This tells you that out of
SIDE NOTE I assume that your book doesn't provide the actual percent ionization, since
#10^(-3)/0.05782 = 1.73 * 10^(-2)#