Question

Consider an aqueous solution, 0.1 M each in HOCN, HCOOH, (COOH)2 and H3PO4, for HOCN, we can write Ka(HOCN) = $\frac { \left[ \mathrm { H } ^ { + } \right] \left[ \mathrm { OCN } ^ { - } \right] } { [ \mathrm { HOCN } ] }$. [H+] in this 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.

Answer 1

Answer: (D)

Overall H+ ion concentration in the solution.

Answer 2

For HOCN– , Ka = $\frac { \left[ \mathrm { H } ^ { + } \right] \left[ \mathrm { OCN } ^ { - } \right] } { [ \mathrm { HOCN } ] }$ ; Here [H+]

= total H+ concentration of solution.