Ionic Equilibrium Practice Problem solutions

Question.1 Write conjugate acid and base of the following :

a) NH₃
b) HCO₃⁻
c) NH₂⁻
d) HPO₄²⁻
e) H₂O
f) HSO₄⁻
g) OH⁻
f) H₂PO₄⁻

Solution

a. NH₃ (Ammonia):

• Conjugate acid: NH₄⁺ (Ammonium ion)
• Conjugate base: NH₂⁻ (Amide ion)

b. HCO₃⁻ (Bicarbonate or Hydrogen carbonate ion):

• Conjugate acid: H₂CO₃ (Carbonic acid)
• Conjugate base: CO₃²⁻ (Carbonate ion)

c. NH₂⁻ (Amide ion):

• Conjugate acid: NH₃ (Ammonia)
• Conjugate base: NH₃ (Ammonia)

d. HPO₄²⁻ (Hydrogen phosphate ion):

• Conjugate acid: H₂PO₄⁻ (Dihydrogen phosphate ion)
• Conjugate base: PO₄³⁻ (Phosphate ion)

e. H₂O (Water):
- Water can act as both an acid and a base in different contexts.

• Conjugate acid: H₃O⁺ (Hydronium ion) in acidic conditions.
• Conjugate base: OH⁻ (Hydroxide ion) in basic conditions.

f. HSO₄⁻ (Hydrogen sulfate ion):

• Conjugate acid: H₂SO₄ (Sulfuric acid)
• Conjugate base: SO₄²⁻ (Sulfate ion)

g. OH⁻ (Hydroxide ion):

• Conjugate acid: H₂O (Water)
• Conjugate base: H₂O (Water)

h. H₂PO₄⁻ (Dihydrogen phosphate ion):

• Conjugate acid: H₃PO₄ (Phosphoric acid)
• Conjugate base: HPO₄²⁻ (Hydrogen phosphate ion)

Remember, the conjugate acid of a base is formed by adding a proton (H⁺), and the conjugate base of an acid is formed by removing a proton.

Question.2 Identify conjugate Acid- Base pair - 

a) H₂SO₄ + H₂O → HSO₄^- + H₃O⁺

b) H₂O + H₂O → H₃O + OH^- 

Solution

a) \( \text{H}_2\text{SO}_4 + \text{H}_2\text{O} \rightarrow \text{HSO}_4^- + \text{H}_3\text{O}^+ \)

In this reaction:

• \( \text{H}_2\text{SO}_4 \) is a strong acid, which donates a proton (\( \text{H}^+ \)) to water (\( \text{H}_2\text{O} \)).
• The resulting species \( \text{HSO}_4^- \) is the conjugate base of \( \text{H}_2\text{SO}_4 \).
• The \( \text{H}_3\text{O}^+ \) (hydronium ion) is formed when water gains a proton and is the conjugate acid of \( \text{H}_2\text{O} \).

So, the conjugate acid-base pair is \( \text{H}_2\text{SO}_4 / \text{HSO}_4^- \) and \( \text{H}_2\text{O} / \text{H}_3\text{O}^+ \).

b) \( \text{H}_2\text{O} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{OH}^- \)

In this reaction:

• One water molecule donates a proton (\( \text{H}^+ \)) to another water molecule.
• The resulting \( \text{H}_3\text{O}^+ \) is the conjugate acid of water (\( \text{H}_2\text{O} \)).
• The \( \text{OH}^- \) (hydroxide ion) is formed when water loses a proton and is the conjugate base of water.

So, the conjugate acid-base pair is \( \text{H}_2\text{O} / \text{H}_3\text{O}^+ \) and \( \text{H}_2\text{O} / \text{OH}^- \).

Question.3 Write example of Lewis Acids & Lewis Bases.

Answer :

Lewis Acids:
1. AlCl₃ (Aluminum chloride)
2. Fe³⁺ (Ferric ion)
3. H⁺ (Hydronium ion)
4. BF₃ (Boron trifluoride)
5. Zn²⁺ (Zinc ion)

Lewis Bases:
1. NH₃ (Ammonia)
2. H₂O (Water)
3. OH⁻ (Hydroxide ion)
4. F⁻ (Fluoride ion)
5. :NH₃ (Ammonia with a lone pair, denoted as a Lewis base)

Vibrant Learning ⁺

^{The Chemical species which acts as donor of lone pair of electron is called Lewis Bases meanwhile, the chemical species which act as electron or lone pair electron acceptor are called Lewis Acids.}

Question.4  State and Derive Ostwald's Dilution Law.

Definition - At constant temperature, the degree of dissociation (Ionization) of weak electrolyte is directly proportional to square root of its dilution. The law is named after the German chemist Wilhelm Ostwald.

Derivation of Ostwald's Dilution Law

Question.5 Find pH of - 

i) 10^-5 N HCL

ii) 0.001 N H₂SO₄

iii) 4 X 10^-4 N NaOH

iv) 0.01 N Ca(OH)₂

v) 0.001 M Ba(OH)₂

Solutions

i) \(10^{-5}\) N HCl:

\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]

The concentration of \(\text{H}^+\) is \(10^{-5}\) M.

\[ \text{pH} = -\log_{10}(10^{-5}) = 5 \]

ii) \(0.001\) N H2SO4:

\[ \text{pH} = -\log_{10}(2 \times 0.001) \]

\[ \text{pH} = -\log_{10}(0.002) \]

\[ \text{pH} \approx 2.7 \]

iii) \(4 \times 10^{-4}\) N NaOH:

\[ \text{pOH} = -\log_{10}(4 \times 10^{-4}) \]

\[ \text{pOH} \approx 3.4 \]

\[ \text{pH} = 14 - \text{pOH} \]

\[ \text{pH} \approx 10.6 \]

iv) \(0.01\) N Ca(OH)2:

\[ \text{pOH} = -\log_{10}(2 \times 0.01) \]

\[ \text{pOH} = -\log_{10}(0.02) \]

\[ \text{pOH} \approx 1.7 \]

\[ \text{pH} = 14 - \text{pOH} \]

\[ \text{pH} \approx 12.3 \]

v) \(0.001\) M Ba(OH)2:

\[ \text{pOH} = -\log_{10}(2 \times 0.001) \]

\[ \text{pOH} \approx 3.7 \]

\[ \text{pH} = 14 - \text{pOH} \]

\[ \text{pH} \approx 10.3 \]

So, the answers are:

i) \(10^{-5}\) N HCl: \(\text{pH} = 5\)

ii) \(0.001\) N H2SO4: \(\text{pH} \approx 2.7\)

iii) \(4 \times 10^{-4}\) N NaOH: \(\text{pH} \approx 10.6\)

iv) \(0.01\) N Ca(OH)2: \(\text{pH} \approx 12.3\)

v) \(0.001\) M Ba(OH)2: \(\text{pH} \approx 10.3\)

Question.6 0.098 g H2SO4 is dissolved in 2L solution. Find pH.

Solution

Step 1: Calculate moles of \(H_2SO_4\):

\[ \text{Moles of } H_2SO_4 = \frac{0.098 \, \text{g}}{98.09 \, \text{g/mol}} \]

\[ \text{Moles of } H_2SO_4 \approx \frac{0.001 \, \text{mol}}{1 \, \text{mol}} \]

\[ \text{Moles of } H_2SO_4 \approx 0.001 \, \text{mol} \]

Step 2: Calculate molarity (\(M\)):

\[ M = \frac{\text{Moles of } H_2SO_4}{\text{Liters of solution}} \]

\[ M = \frac{0.001 \, \text{mol}}{2 \, \text{L}} \]

\[ M = 0.0005 \, \text{M} \]

Step 3: Calculate concentration of \(H^+\):

\[ [\text{H}^+] = 2 \times M \]

\[ [\text{H}^+] = 2 \times 0.0005 \, \text{M} \]

\[ [\text{H}^+] = 0.001 \, \text{M} \]

Step 4: Calculate pH:

\[ \text{pH} = -\log_{10}[\text{H}^+] \]

\[ \text{pH} = -\log_{10}(0.001) \]

Using a calculator:

\[ \text{pH} \approx -(-3) \]

\[ \text{pH} \approx 3 \]

Therefore, the pH of the solution is approximately 3.

Question.7 pH of H2SO4 solution is 2. How much H2SO4 should be dissolved to make 500 ml solution?