7.1 Sulphuric acid, H₂SO₄, ionises into two steps as follows: I H₂SO₄(aq) + H₂O(l) ⇌ HSO₄⁻(aq) + H₃O⁺(aq) K₁ = 1 × 10³ II HSO₄⁻(aq) + H₂O(l) ⇌ SO₄²⁻(aq) + H₃O⁺(aq) K₂ = 1 × 10² 7.1.1 Define an acid in terms of the Lowry-Bronsted theory - NSC Physical Sciences - Question 7 - 2021 - Paper 2

The substance that acts as an ampholite in the equations is HSO₄⁻ (hydrogen sulfate ion). An ampholite can act either as an acid or a base depending on the reaction conditions.

The solution of HSO₄⁻(aq) will have a LOWER conductivity compared to the H₂SO₄(aq) solution. This is because H₂SO₄ is a strong acid and fully ionizes to produce more ions in solution, contributing to higher conductivity. In contrast, HSO₄⁻ is a weaker acid that does not completely ionize, resulting in fewer ions and, consequently, lower conductivity.

To find the concentration of HCl, use the formula:

$pH = - ext{log}[H^+]$

From the pH, calculate [H⁺]:

$1.02 = - ext{log}[H^+] \implies [H^+] = 10^{-1.02} \approx 0.0955 \, ext{mol·dm}^{-3}$

Thus, the concentration of HCl is approximately 0.0955 mol·dm⁻³.

First, calculate the moles of HCl and Na₂CO₃:

n(HCl) = c × V = 0.0955 \, 	ext{mol·dm}^{-3} × 0.050 \, 	ext{dm}³ = 0.004775 \, 	ext{mol}

n(Na₂CO₃) = c × V = 0.075 \, 	ext{mol·dm}^{-3} × 0.025 \, 	ext{dm}³ = 0.001875 \, 	ext{mol}

From the balanced equation, 2 moles of HCl react with 1 mole of Na₂CO₃:

n(HCl)_{used} = 2 × n(Na₂CO₃) = 2 × 0.001875 \, 	ext{mol} = 0.00375 \, 	ext{mol}

Thus, moles of excess HCl = n(HCl){initial} - n(HCl){used}:

excess HCl = 0.004775 \, 	ext{mol} - 0.00375 \, 	ext{mol} = 0.001025 \, 	ext{mol}

Now, to find the concentration in the total volume (50 cm³ + 25 cm³ = 75 cm³ = 0.075 dm³):

$c_{excess HCl} = \frac{n_{excess HCl}}{V_{total}} = \frac{0.001025 \, ext{mol}}{0.075 \, ext{dm}^{3}} \approx 0.0137 \, ext{mol·dm}^{-3}$