Home
Class 12
CHEMISTRY
Consider an aqueous solution 0.1 M each ...

Consider an aqueous solution 0.1 M each in HOCN, HCOOH, `(COOH)_2` and `H_3PO_4` for HOCN, we can write `K_a(HOCN)=([H^+][OCN^-])/([HOCN]),[H^+]` in the expression refers to:

Promotional Banner

Similar Questions

Explore conceptually related problems

Write basicity of H_3PO_2,H_3PO_3 and H_3PO_4 .

The degree of hydrolysis of 0.005 M aqueous solution of NaOCN will be ( K_a for HOCN = 4.0xx10^-4 )

The degree of hydrolysis of 0.005 M aqueous solution of NaOCN will be ( K_a for HOCN = 4.0xx10^-4 )

The pH of a solution containing 0.1 M CH_3COONa and 0.1 M (C_2H_5COO)_2 Ba will be K_a(CH_3COOH)=2xx10^(-5). K_a(C_2H_5COOH)=1.5xx10^(-5) :

An aqueous solution of 0.01 M CH_3COOH has Van't Hoff factor 1.01. If pH=-log [H^+] , pH of 0.01 M CH_3COOH solution would be

An aqueous solution of 0.01 M CH_(3) COOH has van't Hoff factor 1.01 . If pH = - "log"[H^(+)] , pH of 0.01 M CH_(3)COOH solution would be :

The pH of the resultant solution of 20 ml of 0.1 M H_3PO_4 and 20 ml 0.1 M Na_3PO_4 is