Question

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

Answer 1

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is
`1.1* 10^6` M.

`HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)`

This value indicates that

A.
`CN^-` is a stronger base than
`ONO^-`

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is
`ONO^-`

D. The conjugate acid of CN- is HCN

Answer: A.
`CN^-` is a stronger base than
`ONO^-`

Step-by-step explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When
`K_(p)>1`; the reaction is product favoured.

When
`K_(p);<1` ; the reaction is reactant favored.

`When K_(p)=1`; the reaction is in equilibrium.

As,
`K_p>>1`, the reaction will be product favoured and as it is a acid base reaction where
`HONO` acts as acid by donating
`H^+` ions and
`CN^-` acts as base by accepting
`H^+`

Thus
`HONO` is a strong acid thus
`ONO^-` will be a weak conjugate base and
`CN^-` is a strong base which has weak
`HCN` conjugate acid.

Thus the high value of K indicates that
`CN^-` is a stronger base than
`ONO^-`