The dissociation constant of HCN is 5× 10-8 . The pH of the solution prepared by mixing 1.5 moles of HCN and 0.15 moles of KCN in water and making up die total volume to 0.5 dm3 is:

Question

The dissociation constant of HCN is 5× 10-8 . The pH of the solution prepared by mixing 1.5 moles of HCN and 0.15 moles of KCN in water and making up die total volume to 0.5 dm3 is: (A) 7.3020 (B) 5.3020 (C) 6.3010 (D) 6.6990

Tardigrade Admin · Accepted Answer

The solution of HCN and KCN is an acidic Buffer solution. Hence, from Hendarsons equation: pH=- log Ka+ log ([Salt]/[Acid]) =- log (5× 10-8)+ log (0.15 text/0.5/1.5 text/0.5) =- log 5- log 10-8+ log (1/10) =- 0.6990+8+(-1) =6.3010

Q. The dissociation constant of HCN is 5×10−8 . The pH of the solution prepared by mixing 1.5 moles of HCN and 0.15 moles of KCN in water and making up die total volume to 0.5dm3 is:

7.3020

5.3020

6.3010

6.6990

The solution of HCN and KCN is an acidic Buffer solution. Hence, from Hendarsons equation: pH=−logKa​+log[Acid][Salt]​ =−log(5×10−8)+log1.5/0.50.15/0.5​ =−log5−log10−8+log101​ =−0.6990+8+(−1) =6.3010

Q. The dissociation constant of HCN is $5 \times 10^{- 8}$ . The pH of the solution prepared by mixing 1.5 moles of HCN and 0.15 moles of KCN in water and making up die total volume to $0.5 d m^{3}$ is: