NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
30 ml of 0.2 M NaOH is added with 50 ml 0.2 M C H 3 C O O H solution. The extra volume of 0.2 M N a O H required to make the pH of the solution 5.00 is 10 x .The value of x is. The ionisation constant of C H 3 C O O H = 2 × 1 0 - 5 .
Correct answer
3
Step-by-step solution
Milli.moles of N a O H = 30 × 0.2 = 6 m.m of C H 3 C O O H = 50 × 0.2 = 10 Now: p H = - log ( 2 × 1 0 - 5 ) + log 6 4 = 4.87 Suppose 'v'mL of N a O H is added then m.mol of C H 3 C O O N a = 6 + v × 0.2 m.mol of C H 3 C O O H = 4 - v × 0.2 pH = pK a + log Salt Acid 5 = - log ( 2 × 1 0 - 5 ) + log 6 + 0.2 v 4 - 0.2 v 0.3010 = log 6 + 0.2 v 4 - 0.2 v So 6 + 0.2 v 4 - 0.2 v = 2 ⇒ v = 3.33 m L = 10 3 m L So x = 3