The p Ka of HCN is 9.30. The pH of a solution prepared by mixing 2.5 moles of KCN and 2.5 moles of HCN in water and making total volume up to 500 ml, is:

Question

The p Ka of HCN is 9.30. The pH of a solution prepared by mixing 2.5 moles of KCN and 2.5 moles of HCN in water and making total volume up to 500 ml, is: (A)  9.30  (B)  7.30  (C)  10.30  (D)  8.30

Tardigrade Admin · Accepted Answer

Conc. of KCN = conc. of HCN =(2.5/500 / 1000)=5 M ∴ log ([K C N]/[H C N])= log (5/5)=0 Now, from Hendersons equation, p H= p Ka+ log ([ text salt ]/[ text acid ]) =9.30+0=9.30

Q. The $_{p} K_{a}$ of $H CN$ is $9.30$ . The pH of a solution prepared by mixing $2.5$ moles of $K CN$ and $2.5$ moles of $H CN$ in water and making total volume up to $500 m l$ , is:

$9.30$

$7.30$

$10.30$

$8.30$

Conc. of $K CN =$ conc. of $H CN$
$= \frac{2.5}{500/1000} = 5 M$
$∴ lo g \frac{[ K CN ]}{[ H CN ]} = lo g \frac{5}{5} = 0$
Now, from Hendersons equation,
$p H =_{p} K_{a} + lo g \frac{[ salt ]}{[ acid ]}$
$= 9.30 + 0 = 9.30$

Q. The p​Ka​ of HCN is 9.30. The pH of a solution prepared by mixing 2.5 moles of KCN and 2.5 moles of HCN in water and making total volume up to 500ml, is:

9.30

7.30

10.30

8.30