Calculating pH when weak base is added to an strong acid

Question

$\ce{NH3}$ solution of $\pu{0.1 mol dm-3}$ is being added to a $\pu{25.0 cm3}$ of $\pu{0.1 mol dm-3}$ $\ce{HCl}$ solution. Calculate the pH of the solution when volume of added $\ce{NH3}$ solution is $\pu{25.0 cm3}$ and $\pu{26.0 cm3}$. The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is ${9.25}$.

So I started with the equation:

$$\ce{HCl + NH3 <=> NH4Cl}\tag{1}$$

EDIT #1:

As the initial concentration of both base and acid are same, I calculated how much of base is left in the solution:

$V_\text{miscon} = \pu{26.0 cm3 - 25.0 cm3 = 1.0 cm3}$

That's how I tried to calculate pH of the solution when $\pu{26.0 cm3}$ of $\ce{NH3}$ is added:

$\ce{H+}$ concentration: $$[\ce{H+}]_\text{miscon} = [\ce{H+}]_\text{initial} \frac{V_\text{initial}}{V_\text{miscon}}\tag{2} $$ $$= \pu{(0.1 mol dm-3} \times \pu{0.025 dm3)/0.001 dm3} = \pu{2.5 mol dm-3}$$

But, in this way, the results I am getting are completely different from marking, and even negative pH values.

But, when I tried to find $\ce{H+}$ concentration (to calculate pH) of the solution when $\pu{25.0 cm3}$ of $\ce{NH3}$ is added, I found out that initial $\ce{HCl}$ volume is completely reacted. Only ammonium chloride is left in the solution.

The marking only gives the answers, $\ce{pH = 5.28}$ and $\ce{pH = 7.85}$.

How can I calculate pH of this solution in above two conditions separately?

I can't figure it out, I am stuck in the above equation. Please give me some hints to start solving. Thanks in advance.

EDIT #2:

I finally figured it out (Adding $\pu{25.0 cm3}$ of $\ce{NH3}$).

$$\ce{[NH4+(aq)]}=n(\ce{NH4+})/V_{\mathrm{\ce{total}}}$$ $$=(\pu{0.1 mol dm-3}\cdot\pu{0.025 dm3})/\pu{0.05 dm3}$$ $$=\pu{0.0025 mol}/\pu{0.05 dm3}$$ $$\ \ \ =5.0\times 10^{-2}\ \pu{mol dm-3}$$

$\ce{NH4+}$ reacts with water to produce $\ce{H3O+}$ ions.

$$\ce{NH4+(aq) + H2O(l) <=> H3O+(aq) + NH3(aq)}$$

Using acid dissociation constant for $\ce{NH4+}$:

$$\ce{K_\mathrm{a} = [H3O+(aq)] [NH3(aq)] / [NH4+(aq)]}$$ $$\ce{[H3O+(aq)] = \pu{( K_\mathrm{a} \cdot [\ce{NH4+(aq)}] )0.5}}$$ $$\ce{= \pu{( 5.62\times 10^{-10}\pu{mol dm-3} \cdot 5.0\times 10^{-2}\pu{mol dm-3} )0.5}}$$ $$\ce{= \pu{( 2.81\times 10^{-11}\pu{mol2 dm-6} )0.5}}$$ $$\ \ \ =5.301\times 10^{-6}\ \pu{mol dm-3}$$

$$ \text{pH} = -(\log 5.301 - 6\log 10) = -(0.7244 - 6) = 5.28 $$

Answer 1

The question

This is the question, corrected according to Poutnik's comment, and adding (i) and (ii) to distinguish the two scenarios:

$\ce{NH3}$ solution of $\pu{0.1 mol dm-3}$ is being added to a $\pu{25.0 cm3}$ of $\pu{0.1 mol dm-3}$ $\ce{HCl}$ solution. Calculate the pH of the solution when volume of added $\ce{NH3}$ solution is

(i) $\pu{25.0 cm3}$ and

(ii) $\pu{26.0 cm3}$.

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH4+}$ is ${9.25}$.

The chemical reactions

The chemical equation (1) in the question is incorrect and missing something. It should reflect that in aqueous solution, hydrochloric acid is completely dissociated (strong acid), and ammonia is partially dissociated (weak base). One way of writing this is:

$$\ce{HCl(aq) + NH3(aq) -> NH4+(aq) + Cl-(aq)}\tag{A1}$$ $$\ce{NH4+(aq) <=> NH3(aq) + H+(aq)}\tag{A2}$$

Equation A1 shows that the reaction of a strong acid with a weak base goes to completion (no HCl will be present, chloride does not get protonated). Equation A2 shows that ammonium is a weak acid.

If you find it confusing that ammonia first picks up a proton, and then partially loses it again, you could also write it like this:

$$\ce{HCl(aq) -> H+(aq) + Cl-(aq)}\tag{B1}$$ $$\ce{NH3 + H+ <=> NH4+(aq)}\tag{B2}$$

If you learned not to write $\ce{H+}$ but instead $\ce{H3O+}$, you could write the B-equations slightly differently:

$$\ce{HCl(aq) + H2O(l)-> H3O+(aq) + Cl-(aq)}\tag{B1'}$$ $$\ce{NH3 + H3O+(aq) <=> NH4+(aq) + H2O(l)}\tag{B2'}$$

In the end, each pair of equations says the same thing: Hydrochloric acid is as a strong acid, ammonia is a weak base. If ammonia is a weak base, we also know that the ammonium cation is a weak acid (with ammonia its conjugate base).

The calculation

There are two steps. You missed one (the one that involves the equilibrium). Let's work with equations A1 and A2 to work through the calculations.

You had the correct intuition about what happens with the hydrochloric acid: When adding 25 mL ammonia solution to the hydrochloric acid solution, the reactants are present at stoichiometric ratio, and the only species that is left to influence the pH is ammonium cation. When adding 26 mL ammonia solution, however, the ammonia is in excess, so some of it turns into the ammonium cation, and some remains. Scenario (i) is called a weak acid in solution, scenario (ii) is called a buffer solution. We can ignore the chloride ion, it is a spectator ion.

The amount of hydrochloric acid present initially can be determined from initial concentration and volume:

$$n_{\mathrm{\ce{HCl},initial}} = c_{\mathrm{\ce{HCl},initial}} \cdot V_{\mathrm{\ce{initial}}}$$

$$\ \ \ =0.10\ \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.025\ \mathrm{L}$$

$$\ \ \ =2.5\times 10^{-3}\ \mathrm{mol}$$

• Scenario (i): weak acid

The amount of ammonia present initially can be determined from initial concentration and volume:

$$n_{\mathrm{\ce{NH3},initial}} = c_{\mathrm{\ce{NH3},initial}} \cdot V_{\mathrm{\ce{initial}}}$$

$$\ \ \ =0.10\ \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.025\ \mathrm{L}$$

$$\ \ \ =2.5\times 10^{-3}\ \mathrm{mol}$$

You calculate the concentration of species by first figuring out the amount, and then dividing by the volume of the solution, which is 50 mL. In scenario (i), all of the ammonia was turned into ammonium, so you know:

$$n_{\mathrm{\ce{NH4+}}}=2.5\times 10^{-3}\ \mathrm{mol}$$

$$c_{\mathrm{\ce{NH4+}}}= \frac{n_{\mathrm{\ce{NH4+}}}}{V_\text{solution}} = \frac{2.5\times 10^{-3}\ \mathrm{mol}}{\pu{50 mL}} = \pu{0.05 mol/L}$$

This concentration is lower by a factor two compared to the initial concentration of ammonia because the volume doubled when we added hydrochloric acid.

Now you can calculate the pH (pH of a weak acid with a pKa of 9.25 and a concentration of 0.05 mol/L:

$$ \text{pH} = \frac{1}{2} 9.25 - \frac{1}{2} \log 0.05 = 5.28 $$

• Scenario (ii): buffer

$$n_{\mathrm{\ce{NH3},initial}} = c_{\mathrm{\ce{NH3},initial}} \cdot V_{\mathrm{\ce{initial}}}$$

$$\ \ \ =0.10\ \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.026\ \mathrm{L}$$

$$\ \ \ =2.6\times 10^{-3}\ \mathrm{mol}$$

HCl is limiting in reaction A1, so that determines the amount of product:

$$ n_{\mathrm{\ce{NH4+}}} = n_{\mathrm{\ce{HCl},initial}} = \pu{2.5e-3 mol}$$

Some of the ammonia is still present: $$ n_{\mathrm{\ce{NH3}}} = n_{\mathrm{\ce{NH3},initial}} - n_{\mathrm{\ce{NH4+}}} = \pu{2.6e-3 mol} - \pu{2.5e-3 mol} = \pu{0.1e-3 mol} $$

You can plug these concentrations into the Henderson-Hasselbalch equation (buffer equation) to get the pH:

$$ \text{pH} = 9.25 + \log \frac{0.1}{2.5} = 7.85$$