Question #5a5e8
1 Answer
Explanation:
You're titrating ammonia,
That happens because the neutralization reaction produces ammonium cations,
The neutralization reaction looks like this
#"HCl"_ ((aq)) + "NH"_ (3(aq)) -> "NH"_ 4"Cl"_ ((aq))#
As you can see, ammonia and hydrochloric acid react in a
Now, the problem doesn't provide you with the volumes of the two solutions, but you know for a fact that they must be equal because the solutions have the same molarity.
The thing to remember here si that you don't need the volumes of the two solutions, all you have to do is keep track of the fact that they are equal.
Let's assume that the volume of the solutions is
The number of moles of hydrochloric acid and ammonia will be
#n_"HCl" = "0.20 mol" color(red)(cancel(color(black)("L"^(-1)))) * V color(red)(cancel(color(black)("L"))) = (0.20 * V)" moles HCl"#
#n_("NH"_3) = "0.20 mol" color(red)(cancel(color(black)("L"^(-1)))) * V color(red)(cancel(color(black)("L"))) = (0.20 * V)" moles NH"_3#
So, the reaction consumes
The volume of the resulting solution is now
#V_"final" = V" L" + V" L" = 2V" L"#
The concentration of ammonium cations in the resulting solution will be
#["NH"_4^(+)] = ((0.20 * color(red)(cancel(color(black)(V))))"moles")/(2 * color(red)(cancel(color(black)(V)))" L") = "0.10 mol L"^(-1)# As you can see, the volume of the two solutions is not important here!
Now, the ammonium cation acts as a weak acid in aqueous solution. Use an ICE table to find the equilibrium concentration of hydronium cations,
#" ""NH"_ (4(aq))^(+) + "H"_ 2"O"_ ((l)) " "rightleftharpoons" " "NH"_ (3(aq)) " "+" " "H"_ 3"O"_ ((aq))^(+)#
Now, it's important to realize that you must use the acid dissociation constant,
For aqueous solutions at room temperature, you have
#color(purple)(|bar(ul(color(white)(a/a)color(black)(K_a xx K_b = 10^(-14))color(white)(a/a)|)))#
In your case, you have
#K_a = (10^(-14))/(1.8 * 10^(-5)) = 5.56 * 10^(-10)#
By definition,
#K_a = (["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])#
which is equivalent to
#K_a = (x * x)/(0.10 - x) = 5.56 * 10^(-10)#
Because
#0.10 - x ~~ 0.10#
You thus have
#x^2/0.10 = 5.56 * 10^(-10) implies x = sqrt(0.10 * 5.56 * 10^(-10)) = 7.46 * 10^(-6)#
Since
#["H"_3"O"^(+)] = 7.46 * 10^(-6)"M"#
Finally, the pH of the solution will be
#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#
Plug in your value to find
#"pH" = - log(7.46 * 10^(-6)) = color(green)(|bar(ul(color(white)(a/a)color(black)(5.13)color(white)(a/a)|)))#