# Calculate the pH of the solution below. 30.00mL of 0.100 M "HCl" is mixed with 25.00 mL of 0.100 M "NH"_3 ?

Sep 27, 2016

$\text{pH} = 2.041$

#### Explanation:

The first thing to mention here is that you're not dealing with a buffer solution. Here's why.

Ammonia and hydrochloric acid react in a $1 : 1$ mole ratio to produce aqueous ammonium chloride, $\text{NH"_4"Cl}$.

${\text{NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl}}_{\left(a q\right)}$

The chloride anions coming from the acid are spectator ions, meaning that you can rewrite the chemical equation as

${\text{NH"_ (3(aq)) + "H"_ ((aq))^(+) -> "NH}}_{4 \left(a q\right)}^{+}$

So, for every mole of ammonia present in solution, you need $1$ mole of hydrochloric acid to neutralize it. Use the molarities and volumes of the two solutions to see how many moles of each you're mixing

30.00 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.100 moles H"^(+)/(1color(red)(cancel(color(black)("L")))) = "0.00300 moles HCl"

25.00 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.100 moles NH"_3/(1color(red)(cancel(color(black)("L")))) = "0.00250 moles NH"_3

Notice that you have fewer moles of ammonia than of hydrochloric acid. This implies that ammonia will act as a limiting reagent, i.e. it will be completely consumed by the reaction.

Once the neutralization reaction is complete, the resulting solution will contain

n_ ("NH"_ 3) = "0.00250 moles" - "0.00250 moles" = "0 moles"

n_("H"^(+)) = "0.00300 moles" - "0.00250 moles" = "0.00050 moles H"^(+)

Now, the reaction produces ammonium cations, ${\text{NH}}_{4}^{+}$, the conjugate acid of ammonia, in $1 : 1$ mole ratios with the two reactants.

This means that the resulting solution will also contain

n_ ("NH"_ 4^(+)) = "0.00250 moles NH"_4^(+)

The total volume of the solution will be

${V}_{\text{total" = "30.00 mL" + "25.00 mL" = "55.00 mL}}$

The concentration of ammonium cations and of hydrogen ions in the final solution will be

["NH"_4^(+)] = "0.00250 moles"/(55.00 * 10^(-3)"L") = "0.04545 M"

["H"^(+)] = "0.00050 moles"/(55.00 * 10^(-3)"L") = "0.009091 M"

Now, the ammonium cation can act as an acid in aqueous solution to release hydrogen ions and reform ammonia

${\text{NH"_ (4(aq))^(+) rightleftharpoons "H"_ ((aq))^(+) + "NH}}_{3 \left(a q\right)}$

However, the acid dissociation constant, ${K}_{a}$, for the ammonium cation is equal to

${K}_{a} = 5.6 \cdot {10}^{- 10}$

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

This means that a very, very small number of ammonium cations actually release hydrogen ions in solution.

Moreover, because the solution already contains hydrogen ions, the above equilibrium will shift even more to the left $\to$ think Le Chatelier's Principle here.

You can thus say that the pH of the solution will be determined exclusively by the unreacted hydrogen ions that were delivered by the hydrochloric acid, meaning that the above equilibrium will produce a very small* amount of hydrogen ions compared with what's already present in the solution.

Therefore, you will have

"pH" = - log(["H"^(+)])

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH} = - \log \left(0.009091\right) = 2.041} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, why is this not a buffer solution?

As you know, buffers contain a weak acid and its conjugate base or a weak base and its conjugate acid in comparable amounts.

In this case, the neutralization reaction consumes all the ammonia and produces its conjugate acid. So one important element of a buffer, i.e. the weak base, is missing here.

Moreover, you have an excess of hydrochloric acid, a strong acid, so right from the start you should be able to tell that the pH of the solution will be quite acidic.