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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

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