NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
K a for HCN is 5 × 1 0 - 10 at 25 ° C . For maintaining a constant pH of 9, the volume in ml of 5 M KCN solution required to be added to 10 ml of 2 M HCN solution is
Correct answer
2.00
Step-by-step solution
Suppose, volume of KCN added = x ml Total volume of solution after mixing = (10 + x) ml Molarity of HCN in the final solution = M 1 V 1 V 2 = 2 × 10 ( 10 + x ) = 20 ( 10 + x ) M Molarity of HCN in the final solution = M 1 V 1 V 2 = 5 × x ( 10 + x ) = 5 x ( 10 + x ) M pH = - log k a + log [ K C N ] [ H C N ] 9   =   − log   (5×10 − 10 )   +   log   5x 10   +   x 20 10   +   x On solving, x   =   2   mL