Sara took 2 mL of dilute NaOH solution in a test tube and added two drops of phenolphthalein solution to it. The solution turned pink in colour. She added dilute H₂SO₄ to the above solution drop by drop until the solution in the test tube became colourless. 40 drops of dilute H₂SO₄ were used for the change in colour from pink to colourless. When Sara added a drop of NaOH to the solution, the colour changed back to pink again.
Sara now tried the activity with different volumes of NaOH and recorded her observation in the table given below :

S. No.	Volume of dil. NaOH taken (mL)	Drops of dil. H2SO4 used
1	2	20
2	3	30
3	4	40

Answer the following questions based on the above information:

(a) If Sara used concentrated H₂SO₄ in place of dilute H₂SO₄, how many drops will be required for the change in colour to be observed?

1. 40
2. < 40
3. > 40

Justify your answer.

(b) Sara measured 20 drops of dil. H₂SO₄ and found its volume to be 1 mL. If Sara observed a change in colour of NaOH solution by using 3 mL of H₂SO₄, how many mL of NaOH did she add to the test tube initially?

OR

Sara takes 10 drops of dilute H₂SO₄ in the test tube and adds two drops of phenolphthalein solution to it. Then she adds NaOH dropwise. Sara observes a change in colour after adding 20 drops of NaOH. What change in colour would she observe and why?

(c) Write a balanced chemical equation for the reaction taking place in the above experiment. Which of the following is true and why? The reaction is a

1. neutralisation and double displacement reaction
2. neutralisation and precipitation reaction
3. precipitation and double displacement reaction
4. neutralisation, double displacement as well as precipitation reaction.

(a) < 40 drops.
Concentrated H₂SO₄ provides more H⁺ ions per drop than dilute acid, so fewer drops are needed to neutralise the same amount of NaOH and discharge the pink colour of phenolphthalein.

(b) From Sara’s note, 20 drops = 1 mL.
If she used 3 mL H₂SO₄, that is 60 drops and her table shows the drops of acid are proportional to the mL of NaOH (20 drops per 2 mL NaOH → 10 drops per 1 mL NaOH).

Thus 60 drops of acid would neutralise 6 mL of NaOH.

OR

If she starts with dilute H₂SO₄ plus phenolphthalein and then adds NaOH dropwise, the solution will change from colourless to pink when the solution becomes basic, because phenolphthalein is colourless in acid and pink in base.

(c) Balanced equation:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

This is a neutralisation and double displacement reaction because the acid and base react to form salt + water (neutralisation). Ions exchange partners (Na⁺ with H⁺/SO₄^2-), so it’s double displacement. It is not a precipitation reaction because sodium sulphate (Na₂SO₄) is soluble in water.