Learners prepare a solution of known concentration by dissolving 2 g pure sodium hydroxide crystals, NaOH, in water in a 250 cm³ volumetric flask - NSC Physical Sciences - Question 7 - 2021 - Paper 2

To determine the concentration of the sodium hydroxide solution, we use the formula:

$c = \frac{n}{V}$

1. First, calculate the number of moles of NaOH:

   • Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol.
   • Moles of NaOH = $n = \frac{m}{M} = \frac{2\, \text{g}}{40\, \text{g/mol}} = 0.05\, \text{mol}$

2. Convert the volume from cm³ to dm³:

   • Volume = 250 cm³ = 0.250 dm³.

3. Now calculate the concentration:

   • $c = \frac{0.05\, \text{mol}}{0.250\, \text{dm}^3} = 0.20\, \text{mol/dm}^3$

To find the pH of the sodium hydroxide solution, we first calculate the concentration of hydroxide ions [OH⁻]. Since NaOH is a strong base, it fully dissociates:

$[OH^-] = 0.20\, \text{mol/dm}^3$

Using the relationship between hydroxide concentration and pH, we can find the pOH:

1. Calculate pOH:

   • $\text{pOH} = -\log[OH^-] = -\log(0.20) = 0.699$

2. Now calculate pH using the formula:

   • $\text{pH} + \text{pOH} = 14$
   • $\text{pH} = 14 - 0.699 = 13.301$

To find the initial concentration of the dilute HCl, we follow these steps:

1. Calculate the number of moles of CaCO₃:

   • Molar mass of CaCO₃ = 100 g/mol.
   • Moles of CaCO₃ = $n = \frac{1.5\, \text{g}}{100\, \text{g/mol}} = 0.015\, \text{mol}$

2. From the reaction, the stoichiometry shows 2 moles of HCl react with 1 mole of CaCO₃:

   • Moles of HCl reacted = 2 × (0.015 mol) = 0.03 mol.

3. Calculate moles of HCl excess:

   • Moles of NaOH used = $0.20\, \text{mol/dm}^3 × 0.025\, \text{dm}^3 = 0.005\, \text{mol}$
   • Moles of HCl initially = moles reacted + moles of NaOH used = 0.03 mol + 0.005 mol = 0.035 mol.

4. Finally, calculate the concentration of the initial HCl:

   • Volume of dilute HCl = 50 cm³ = 0.050 dm³.
   • Initial concentration = $c = \frac{0.035\, \text{mol}}{0.050\, \text{dm}^3} = 0.70\, \text{mol/dm}^3$