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 t
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:
A. `H^(+)` ions released by `HOCN`
B. Sum of `H^(+)` ions released by all monoprotic acids
C. Sum of `H^(+)` ions released only the first dissociation of all the acids
D. Overall `H^(+)` ion concentration in the solution.
8 views
1 Answers
For `HOCN^(-),K_(a)=([H^(+)][OCN^(-)])/([HOCN])`,Here `[H^(+)]` =total `H^(+)` concentration of solution.
8 views
Answered