Chemistry of Aqueous Solutions – Acids and Bases Part I

Chemistry of Aqueous Solutions is usually one of the BIGGEST FEAR many A-level students have, reason being that schools introduce too many jargons and concepts, such that students cannot filter out the essence from all the noise.

Chemistry of Aqueous Solutions is primarily divided into 2 sections

1. Acids and Bases.
2. Sparingly soluble salts.

In this section, we will touch briefly on acids and bases.

We have always been taught, since secondary school days, that there are acids, bases and salts. But to simplify matters, it’s easier to see that there are only acids and bases. Salts can behave as acids or bases, and hence there is NO need to create another category to confuse ourselves.

The World of Acids and Bases, according to Bronsted and Lowry:

Acid: H⁺(or proton) donor (those without H⁺ cannot be acids; acids MUST have H⁺)
Base : H⁺ acceptor (those with or without H+ can be bases)

Eg:

1. HCN: Acid (H⁺ present)
2. HCO₃^–: can be acid or base
   ie Acid: HCO₃^–H₂O → H₃O⁺ + CO₃^2-
   Base: HCO₃^– + H₂O → CO₃^2- + OH^–
3. CO₃^2-: base (no H⁺ present)

Conjugate Acid – Base Pairs

Pairs of compounds with ALMOST SIMILAR structures with the exception of ONE H⁺DIFFERENCE.

Eg. CH₃COOH (1 H⁺more : acid) and CH₃COO^– (1 H⁺ less : base)
CH₃NH₃⁺ (1 H⁺ more : acid) and CH₃NH₂ (1 H⁺less : base)

Only 3 types of Scenarios can be assessed in exams:

1. Either of the conjugate acid-base present
2. Both the conjugate acid-base present
3. Combination of strong and weak acids / bases present

Tackling Exam questions

Steps

1. Recognize which scenario (in previous blog) the question belong to
2. Employ the corresponding technique
3. Solve mathematically

Example: CJC 2012 P3

Calculate the pH of the resulting solution when 25 cm³ of a 0.0200 moldm^-3HCl is added to 25 cm³ of 0.0300 moldm^-3 
1-phenylmethanamine (C₆H₅CH₂NH₂)
(The Kb value of 1-phenylmethanamine is 2.19 x 10^-5 moldm^-3)

Method:
Step 1 : Determine which scenario this question belongs to.
To do that, the amounts of the 2 compounds have to be determined.

No. of moles of HCl (acid) = 5 x 10^-4
No. of moles of 1-phenylmethanamine (base) = 7.5 x 10^-4

Since the acid and base are both monobasic, reaction mole ratio is 1:1
Thus base is in excess:

After the reaction, only excess C₆H₅CH₂NH₂ and its salt (or conj acid), C₆H₅CH₂NH₃⁺ remain.

The scenario that this question belongs to is one of BOTH conjugate acid-base pair are both present. The pair is also known as a buffer system. The way to solve the pH is to use the Henderson-Hassalbalch equation.

Solution:
Henderson Hassalbalch equation:

Rewritten in our case:

pKa = 14 – pKb = 14 – (-lg (2.19 x 10^-5)) = 14 – 4.6596 = 9.340
[C₆H₅CH₂NH₂] = (7.5-5) x10^-4 / ((25+25)/1000) = 5 x 10^-3
[C₆H₅CH₂NH₃⁺] = 5 x 10^-4/ ((25+25)/1000) = 1 x 10^-2

Substituting into Eq. (2),
pH = 9.340 + lg (5×10^-3/1×10^-2) = 9.04 (3 SF)

Additional Question :
How will the scenario change if the volume of HCl is changed to

1. 0 cm³
2. 37.5 cm³
3. 50 cm³

The scenario of the question would have changed to

1. Only the base (C₆H₅CH₂NH₂)is found
2. Only the conjugate acid (C₆H₅CH₂NH₃⁺) is found
3. Strong acid (HCl) and weak conjugate acid (C₆H₅CH₂NH₃⁺) found

How do we solve these scenarios then?

Sign up for Focus Chemistry’s JC level chemistry lessons to find out!