Question #13d4e

It's important to recognize that you're titrating a weak base with a strong acid, so the pH at equivalence point will not be equal to `7`.

That is the case because the complete neutralization of a weak base with a strong acid will result in the formation of the base's conjugate acid, which will then react with water to produce an acidic solution.

So right from the start you can expect the pH at equivalence point to be lower than `7`.

So, ammonia, `"NH"_3`, will react with hydrochloric acid, `"HCl"`, to produce ammonium cations, `"NH"_4^(+)`, chloride anions, `"Cl"^(-)`, and water.

The chloride anions are spectator ions, so they are of no interest to you in this reaction.

The balanced chemical equation for this neutralization reaction looks like this

`"NH"_text(3(aq]) + "H"_3"O"_text((aq])^(+) -> "NH"_text(4(aq])^(+) + "H"_2"O"_text((l])`

The hydronium cation, `"H"_3"O"^(+)`, is produced by the dissociation of the strong acid in aqueous solution.

Notice that you one mole of ammonia will react with one mole of hydronium cations and produce one mole of ammonium cations.

This tells you that for every one mole of reactants that is consumed by the reaction, one mole of ammonium cations will be produced.

Use the molarity and volume of the two solutions to find how many moles of each you're mixing

`color(blue)(c = n/V implies n = c * V)`

`n_(NH_3) = "0.050 mol" color(red)(cancel(color(black)("L"^(-1)))) * 30.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.0015 moles NH"_3`

`n_(H_3O^(+)) = "0.025 mol" color(red)(cancel(color(black)("L"^(-1)))) * 60.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.0015 moles H"_3"O"^(+)`

So, equal numbers of moles of each reactant will result in a complete neutralization of the weak base. Now, if `0.0015` moles of ammonia reacts with `0.0015` moles of hydronium cations, then you can that the reaction will produce `0.0015` moles of ammonium cations.

At equivalence point, your solution will contain

`n_(NH_3) = "0 moles NH"_3 ->` completely consumed

`n_(H_3O^(+)) = "0 moles H"_3"O"^(+) ->` completely consumed

`n_(NH_4^(+)) = "0.0015 moles NH"_4^(+) ->` produced by the reaction

Now, in order to find the pH of the solution, you need to know the acid dissociation constant, `K_a`, for the ammonium cations, which can be found listed here

`K_a = 5.6 * 10^(-10)`

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Use the total volume of the solution to find the molarity of the ammonium cations

`V_"total" = V_(NH_3) + V_(H_3O^(+))`

`V_"total" = "30.0 mL" + "60.0 mL" = "90.0 mL"`

The concentration of ammonium cations will be

`["NH"_4^(+)] = "0.0015 moles"/(90.0 * 10^(-3)"L") = "0.01667 M"`

Use an ICE table to help you find the equilibrium concentration of hydronium ions

`" " "NH"_text(4(aq])^(+) + "H"_2"O"_text((l]) " "rightleftharpoons" " "NH"_text(3(aq]) " "+" " "H"_3"O"_text((aq])^(+)`

`color(purple)("I")" " " "0.01667" " " " " " " " " " " " " " " "0" " " " " " " " " "0`
`color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " " " " "(+x)`
`color(purple)("E")" "0.01667-x" " " " " " " " " " " " " "x" " " " " " " " " "x`

The acid dissociation constant will be equal to

`K_a = (["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])`

This will be equivalent to

`K_a = (x * x)/(0.01667 - x) = x^2/(0.01667 -x) = 5.6 * 10^(-10)`

Because `K_a` has such a small value, you can use the approximation

`0.01667 - x ~~ 0.01667`

This will give you

`x^2/0.01667 = 5.6 * 10^(-10)`

`x = sqrt(5.6 * 0.01667 * 10^(-10)) = 3.06* 10^(-6)`

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

`color(blue)("pH" = - log(["H"_3"O"^(+)]))`

`"pH" = - log(3.06 * 10^(-6)) = color(green)(5.51)`

As predicted, the pH at equivalence point is lower than `7`.