Sulfuric acid is a strong dibasic acid - Leaving Cert Chemistry - Question 7 - 2011

According to the Brønsted-Lowry theory:

• Strong Acids: Strong acids are good proton donors, meaning they readily donate protons (H⁺) to other substances in the solution.

• Weak Acids: Weak acids less readily donate protons compared to strong acids. They do not fully ionize, thus the equilibrium shifts to the left, favoring the undissociated form of the acid.

The conjugate base of sulfuric acid (H₂SO₄) is the hydrogen sulfate ion (HSO₄⁻).

The conjugate base of the weak acid HA is the A⁻ ion.

The hydrogen sulfate ion (HSO₄⁻) is the stronger conjugate base compared to A⁻ because:

• HSO₄⁻ can further dissociate to SO₄²⁻ and release a proton.
• In contrast, the A⁻ ion, being the conjugate base of a weak acid, has a lesser tendency to accept protons compared to HSO₄⁻.

The dissociation of sulfuric acid in water can be represented as follows:

1. First dissociation: $H_2SO_4 (aq) \rightarrow H^+ (aq) + HSO_4^- (aq)$

2. Second dissociation: $HSO_4^- (aq) \rightleftharpoons H^+ (aq) + SO_4^{2-} (aq)$

The conduit relationships show that HSO₄⁻ can also donate a proton, confirming that it behaves as an acid.

pH is defined as the negative logarithm to base 10 of the hydrogen ion concentration in a solution:

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

It represents the acidity or alkalinity of a solution, with lower pH values indicating higher acidity.

To find the pH:

1. Write the expression for K_a: $K_a = \frac{[H^+][A^-]}{[HA]}$ Assuming $x$ is the concentration of dissociated acid, $K_a = \frac{x \cdot x}{0.2 - x} \approx \frac{x^2}{0.2}$ (since $x$ is small)

2. Substitute the known $K_a$ value: $6.3 \times 10^{-4} = \frac{x^2}{0.2}$ $x^2 = 6.3 \times 10^{-4} \times 0.2 = 1.26 \times 10^{-4}$ $x = \sqrt{1.26 \times 10^{-4}} \approx 1.12 \times 10^{-2} \text{ M}$

3. Now, calculate the pH: $pH = -\log_{10}(1.12 \times 10^{-2}) \approx 1.95$

To find the concentration of H⁺ ions in the sulfuric acid solution:

1. Since sulfuric acid (balanced reaction) first dissociates completely: $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$ For 1 mole of H₂SO₄, it produces 2 moles of H⁺.

2. Concentration calculation: If the pH is approximately 1.95, then: $[H^+] = 10^{-1.95} \approx 0.0112 \text{ M}$

3. Therefore, The initial concentration of sulfuric acid: $C_{H_2SO_4} = \frac{[H^+]}{2} = \frac{0.0112}{2} \approx 0.0056 ext{ M}$