Question

20 mL of a solution containing equal moles of Na₂CO₃ and NaHCO3 required 16 mL of 0.16 M HCI solution to reach the phenolphthalein end point. What volume (in mL) of 0.1 MH2SO4 solution would have been required if methyl orange been used as an indicator from the beginning?

Answer 1

38.4 mL

Answer 2

Here's how to solve this problem:

1. Determine moles of CO₃²⁻ from phenolphthalein titration: 0.00256 moles.
   At the phenolphthalein end‐point, only the CO₃²⁻ in Na₂CO₃ is neutralized:

   $$\text{CO}_3^{2-} + \text{H}^+ \to \text{HCO}_3^-$$

   Given that 16 mL of 0.16 M HCl was used,

   $$x = 0.016\,L \times 0.16\,M = 0.00256\text{ moles}$$

2. Equal moles of NaHCO₃ are present.

3. Total H⁺ moles needed = 2×0.00256 (for CO₃²⁻) + 0.00256 (for HCO₃⁻) = 0.00768 moles. When titrating with methyl orange, complete neutralization is required. The reactions are:

   $$\text{Na}_2\text{CO}_3 + 2\text{H}^+ \to 2\text{Na}^+ + \text{H}_2\text{O} + \text{CO}_2$$ $$\text{NaHCO}_3 + \text{H}^+ \to \text{Na}^+ + \text{H}_2\text{O} + \text{CO}_2$$

   Total moles of H⁺ required:

   $$\underbrace{0.00256\times 2}_{\text{from Na}_2\text{CO}_3} + \underbrace{0.00256\times 1}_{\text{from NaHCO}_3} = 0.00512 + 0.00256 = 0.00768\text{ moles}$$

4. Since 0.1 M H₂SO₄ gives 2 equivalents, moles H₂SO₄ = 0.00768/2 = 0.00384. In 0.1 M H₂SO₄, each mole supplies 2 moles of H⁺. Therefore, moles of H₂SO₄ needed:

   $$\frac{0.00768}{2} = 0.00384\text{ moles}$$

5. Required volume = 0.00384/0.1 = 38.4 mL. Volume of 0.1 M H₂SO₄ required:

   $$\text{Volume} = \frac{0.00384\text{ moles}}{0.1\,M} = 0.0384\,L = 38.4\,mL$$