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Chemistry of Aqueous Solutions – Acids and Bases Part I
  • focuschemistry
  • November 14, 2017
  • Chemistry of Aqueous Solutions

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: HCO3H2O → H3O+ + CO32-
    Base: HCO3 + H2O → CO32- + OH
  3. CO32-: 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:

C6H5CH2NH2+HCI →C6H5CH2NH3++Cl
moles:5 X 10-47.5 X 10-4
LimitingExcess

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:

pH = pKa + lg[conj base]
————–
[acid]
……(1)

Rewritten in our case:

pH = pKa of C6H5CH2NH3+ + lg {[C6H5CH2NH2]/[C6H5CH2NH3+]} ….(2)

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?

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