Ammonia ionises in water to form a basic solution according to the following balanced equation: NH₃(g) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) 7.1.1 Is ammonia a WEAK or a STRONG base? Give a reason for the answer - NSC Physical Sciences - Question 7 - 2017 - Paper 2

The conjugate acid of NH₃(g) is NH₄⁺(aq).

Water (H₂O) can act as an ampholyte in this reaction as it can function both as an acid and a base depending on the context.

The soil sample is ACIDIC as indicated by the pH being below 7 in the graph. This shows a higher concentration of H₃O⁺ ions as compared to OH⁻ ions.

To calculate the concentration of OH⁻ after adding 4 cm³ of NH₃(aq), use the equation:

1. Determine pH from the graph after addition (assume pH = 8).

2. Calculate pOH:

   $pOH = 14 - pH = 14 - 8 = 6$

3. Find [OH⁻]:

   $[OH⁻] = 10^{-pOH} = 10^{-6} mol/dm³ = 1 x 10^{-6} mol/dm³$.

Firstly, calculate the number of moles of Na₂CO₃:

n(Na₂CO₃) = rac{m}{M} = rac{0.29 g}{106 g/mol} = 0.00274 mol

According to the balanced reaction, 1 mole of Na₂CO₃ reacts with 2 moles of HCl. Therefore, 0.00274 moles of Na₂CO₃ will react with:

$n(HCl) = 2 imes n(Na₂CO₃) = 2 imes 0.00274 = 0.00548 mol$

The concentration of diluted HCl can now be calculated as:

c(HCl)_{diluted} = rac{n(HCl)}{V} ext{ with } V = 0.050 L

c(HCl)_{diluted} = rac{0.00548 mol}{0.050 L} = 0.109 mol/dm³

Since the original volume was 5 cm³ (0.005 L), the concentration of the concentrated HCl sample is:

c(HCl)_{concentrated} = rac{c(HCl)_{diluted} imes V_{diluted}}{V_{concentrated}} = rac{0.109 imes 0.500}{0.005} = 10.94 mol/dm³.