Question

50 ml, 0.04 solution of sodium sesquicarbonate dihydrate $[Na_2CO_3.NaHCO_3.2H_2O]$ is titrated against 0.05 M HCl solution. X ml of acid is required to reach the phenolphthalein end point while Y ml of same acid was required when methyl orange is used as indication for end point in separate titration.

Which of the following statements are correct?

• A. X + Y = 160 ml
• B. X + Y = 120 ml
• C. Y-X = 80 ml
• D. If the titration is started with phenolphthalein indicator and methhyl orange is added at the end point, 2x ml HCl would be required further to reach end point

Answer 1

Answer: (A), (C), (D)

Options (A), (C), and (D) are correct.

Answer 2

Solution:

Given salt: $Na_2CO_3 \cdot NaHCO_3 \cdot 2H_2O$

• Contains 1 mole of $CO_3^{2-}$ and 1 mole of $HCO_3^-$ per mole of salt.

• Reaction steps:

  • With phenolphthalein: Only $CO_3^{2-}$ is neutralized $CO_3^{2-} + H^+ \rightarrow HCO_3^-$ (1 eq acid per salt molecule)
  • With methyl orange: Complete neutralization of both $HCO_3^-$ groups $HCO_3^- + H^+ \rightarrow H_2CO_3$ (2 eq acid per salt molecule)

Thus, total acid needed per mole salt = 1 + 2 = 3 eq.

For a 50 mL, 0.04 M salt solution:

Moles salt = 0.05 L × 0.04 M = 0.002 moles

Acid needed for phenolphthalein endpoint (X) = 0.002 moles HCl

Acid needed for complete neutralization (Y) = 3 × 0.002 = 0.006 moles HCl

Using 0.05 M HCl:

$X = \frac{0.002 \text{ mol}}{0.05 \text{ mol/L}} = 40 \text{ mL}$

$Y = \frac{0.006 \text{ mol}}{0.05 \text{ mol/L}} = 120 \text{ mL}$

Now check the statements:

(A) X + Y = 40 + 120 = 160 mL → Correct

(B) X + Y = 120 mL → Incorrect

(C) Y – X = 120 – 40 = 80 mL → Correct

(D) If titration starts with phenolphthalein (endpoint at 40 mL), then adding methyl orange gives further acid needed = Y – X = 80 mL, which equals 2X (since 2×40 = 80) → Correct

Summary (Minimal Explanation):

• Mole salt = 0.002 mol
• Phenolphthalein endpoint: 0.002 mol HCl → 40 mL
• Methyl orange endpoint: 0.006 mol HCl → 120 mL
• Hence, X + Y = 160 mL, Y – X = 80 mL and additional acid required after phenolphthalein endpoint = 80 mL = 2×40 mL.