What is the pH at the equivalence point when 35.0 mL of 0.20 M ammonia is titrated by 0.12M hydrochloric acid? #K_b# for ammonia is #1.8 xx 10^(-5)#
1 Answer
Explanation:
Ammonia,
#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#
The two reactants react in a
In other words, you need equal numbers of moles of ammonia and of hydrochloric acid to get to the equivalence point.
Use the molarity and volume of the ammonia solution to determine how many moles of ammonia it contains
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#
#n_(NH_3) = "0.20 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(35.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#
#n_(NH_3) = "0.0070 moles NH"_3#
This means that the hydrochloric acid solution must also contain
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies V_"solution" = n_"solute"/c)color(white)(a/a)|)))#
You will have
#V_(HCl) = (0.0070 color(red)(cancel(color(black)("moles"))))/(0.12color(red)(cancel(color(black)("mol")))"L"^(-1)) = "0.0583 L" = "58.3 mL"#
The volume of the resulting solution will thus be
#V_"total" = V_(NH_3) + V_(HCl)#
#V_"total" = "35.0 mL" + "58.3 mL" = "93.3 mL"#
Now, notice that the reaction produces aqueous ammonium chloride in
If the reaction consumes
Ammonium chloride dissociates completely in aqueous solution to form ammonium cations,
#"NH"_ 4 "Cl"_ ((aq)) -> "NH"_ (4(aq))^(+) + "Cl"_ ((aq))^(-)#
The concentration of the ammonium cations in the resulting solution will be
#["NH"_4^(+)] = "0.0070 moles"/(93.3 * 10^(-3)"L") = "0.07503 M"#
The ammonium cations will hydrolyze to form hydronium cations ,
#" " "NH"_ (4(aq))^(+) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (3(aq)) " "+" " "H"_ 3"O"_ ((aq))^(+)#
By definition, the acid dissociation constant,
#K_a = (["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])#
In this case, you will have
#K_a = (x * x)/(0.07503 - x) = x^2/(0.07503 - x)#
Now, you know that for aqueous solutions at room temperature, you have the following relationship between
#color(purple)(|bar(ul(color(white)(a/a)color(black)(K_a xx K_b = K_W)color(white)(a/a)|)))#
Here
#K_W = 10^(-14) -># the ionization constant of water
This means that you have
#K_a = K_W/K_b#
#K_a = 10^(-14)/(1.8 * 10^(-5)) = 5.56 * 10^(-10)#
Since
#0.07503 - x ~~ 0.07503#
This will get you
#K_a = x^2/0.07503 = 5.56 * 10^(-10)#
Solve for
#x = sqrt(0.07503 * 5.56 * 10^(-10)) = 6.46 * 10^(-6)#
Since
#["H"_3"O"^(+)] = 6.46 * 10^(-6)"M"#
As you know, the pH of the solution is given by
#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#
This means that you have
#"pH" = - log(6.46 * 10^(-6)) = color(green)(|bar(ul(color(white)(a/a)color(black)(5.19)color(white)(a/a)|)))#
Finally, the result makes sense because neutralizing a weak base such as ammonia with a strong acid will result in the formation of the conjugate acid of the base, which will in turn cause the pH to be lower than