1
$\begingroup$

$\ce{NH3}$ 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 $\ce{NH4+}$:

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

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

$\endgroup$
  • $\begingroup$ You should do some basic stoichiometry. Figure out how many moles of each reactant there is, and which species is in excess in either case. $\endgroup$ – Michael Lautman Jun 5 at 17:36
  • $\begingroup$ The task description has an error. pKa=9.25 is for NH4+, not for NH3. For NH3, there is relevant pKb=pKw-pKa=14-9.25=4.75. $\endgroup$ – Poutnik Jun 5 at 18:01
  • $\begingroup$ Why do you think one pH value is acidic, and the other one basic? "Only ammonium chloride is left" - what would be the pH of ammonium chloride, acidic or basic? $\endgroup$ – Karsten Theis Jun 5 at 18:03
  • $\begingroup$ There are given 3 directly related equilibrium constants for reactions $$\begin{align}\\\ce{NH4+ &<=> NH3 + H+}\\ \ce{NH3 + H2O &<=> NH4+ + OH-}\\ \ce{H2O &<=>H+ + OH-}\\ \end{align}$$ $\endgroup$ – Poutnik Jun 5 at 18:10
  • $\begingroup$ Just as a hint: Adding a small amount of weak base to a larger (both in volume AND molarity) amount of strong acid changes basically nothing. $\endgroup$ – Karl Jun 5 at 19:17
1
$\begingroup$

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:

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

$\endgroup$
  • 1
    $\begingroup$ I hope that the answers to all the questions are similar to this clear, explicit answer that takes into account the individual differences and different cultures of the students. Blessed efforts $\endgroup$ – Adnan AL-Amleh Jun 7 at 3:22
  • $\begingroup$ I agree, but there is also fine distinction between giving fish versus teaching fishing. Solving the whole task may be like just giving fish. $\endgroup$ – Poutnik Jun 7 at 6:20
  • 1
    $\begingroup$ The point is, they are not just "a" weak acid and "a" weak base. They are the conjugated pair acid-base by NH3 + H+ <=> NH4+, with equilibrium defined by the given constant. Such pairs are bases of almost all pH buffers - solutions keeping desired pH. If you know the ratio and the constant you know pH. $\endgroup$ – Poutnik Jun 7 at 7:05
  • 1
    $\begingroup$ I see. I am just thinking the participation of the OP could avoid closure. $\endgroup$ – Poutnik Jun 7 at 11:54
  • 1
    $\begingroup$ @AdnanAL-Amleh NH4+ cannot accept protons. As it is an acid, it will only donate protons. (Brønsted-Lowry definition of an acid) Its conjugate base NH3 will accept a proton because of its lone pair. Also keep in mind that water is amphoteric. $\endgroup$ – hpm0615ngvup Jun 8 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.