Question

10. Ka for HCN is 5 x 10^-10 at 25°C.
a constant pH of 9, the volume of 5M KCN solution
required to be added to 10mL of 2M HCN solution
is-
(1) 4 mL
(2) 7.95 mL
(3) 2 mL
(4) 9.3 mL
​

Answer 1

$\huge\bigstar\underline\orange{Answer}$

★ HCN and KCN will form buffer.

↪️ given Pka = 5×10^-10

Ph = Pka + log [SALT]/[ACID]

★ ph = -log (5×10^-10) + log [(5V/V+10)/(2×10/V+10)]

=> 9 = -log (5×10^-10) + log [V/4]

=> 9 = log 1/5 × 10^10 × (V/4)

taking antilog on LHS

10^9 = V/20 × 10^10

V/20 = 1/10

V = 2

$\boxed{option\:c\:is\:correct}$

Happy To Help (: