Question

Question #5a5e8

Answer 1

`"pH" = 5.13`

Explanation:

You're titrating ammonia, `"NH"_3`, a weak base, with hydrochloric acid, `"HCl"`, a strong acid, so right from the start you should know that the pH at the equivalence point is lower than `7`.

That happens because the neutralization reaction produces ammonium cations, `"NH"_4^(+)`, the conjugate acid of ammonia.

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 `1:1` mole ratio, which implies that at the equivalence point, the reaction consumed equal numbers of moles of ammonia and of hydrochloric acid.

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 `V` liters. According to the balanced chemical equation, the reaction consumes `1` mole of hydrochloric acid and `1` mole of ammonia to produce `1` mole of aqueous ammonium chloride.

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 `(0.20 * V)` moles of hydrochloric acid and ammonia, which can only mean that it produced `(0.20 * V)` moles of ammonium chloride, i.e. of ammonium cations, `"NH"_4^(+)`.

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, `"H"_3'O"^(+)`

`" ""NH"_ (4(aq))^(+) + "H"_ 2"O"_ ((l)) " "rightleftharpoons" " "NH"_ (3(aq)) " "+" " "H"_ 3"O"_ ((aq))^(+)`

`color(purple)("I")color(white)(aaaaacolor(black)(0.10)aaaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaaacolor(black)(0)`
`color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaaaaaaaaaaaaaacolor(black)((+x))aaaaaaaacolor(black)((+x))`
`color(purple)("E")color(white)(aaacolor(black)(0.10-x)aaaaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaaaaaacolor(black)(x)`

Now, it's important to realize that you must use the acid dissociation constant, `K_a`, not the base dissociation constant, `K_b`.

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` is equal to

`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 `K_a` is so small compared with the initial concentration of the ammonium cations, you can use the approximation

`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 `x` represents the equilibrium concentration of hydronium cations, you will have

`["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)|)))`