NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
One litre of 1 M solution of an acid HA K a = 10 − 4 at 25 o C has pH = 2 . It is diluted by water so the new p H becomes double. The solution was diluted to y × 1 0 z m l . The value of y + z 2 is:
Correct answer
5.5
Step-by-step solution
pH = 2 ⇒ H + 1 = 10 - 2 = c 1 α 1 = 1 × α 1 α 1 = 10 - 2 pH = 4 ⇒ H + 2 = 10 - 4 = 1 × 1000 V × α 2 k a = 10 - 4 = α 2 2 C 2 1 - α 2 = α 2 × α 2 × C 2 1 - α 2 10 - 4 = α 2 × 10 - 4 1 - α 2 ⇒ α 2 = 0 .5 H + 2 = 1 × 1000 V × α 2 = 10 - 4 1 × 1000 V × 0 .5 = 10 - 4 V = 5 × 10 6 ml ⇒ x + y 2 = 5 + 6 2 = 11 2 = 5 .5