Reactions of a weak acid

I need a lil help again,

1. Cyanic acid, HCNO, is a weak acid:
HCNO(aq) + H2O(l)  H3O+(aq) + CNO-(aq)
In a 0.20 M aqueous solution of HCNO, the concentration of H3O+ is measured to be 6.5E-3 M. Evaluate Keq.

i know water cancels out, and H3O AND CNO- are equal to 0. After that I don't know what to do.

Given the reaction at equilibrium:
        HCNO(aq) + H2O(l)   H3O+(aq) + CNO-(aq)
The [H3O+] and the [CNO-] must be equal (since for every H3O+ ion formed, a CNO- ion will also be formed.  The [HCNO] at equilibrium will be the initial concentration (.20M) - the concentration of the H3O+ (since for every H3O+ ion formed, one HCNO molecule was broken up).

The Keq = [H3O+] [CNO-] / [HCNO] 
Plug in the concentrations from above and solve for Keq.


I got 0.030, I don't think its right.

i think you missed what he was saying

according to an ICE table, your K equation would look like this

[H3O+] [CNO-] / [HCNO] = (x)(x) / Minitial = Keq

x = [H3O+]

sorry to both you guys again, i am still getting it wrong. I am suppose to sumbit the answer online, I am not getting any hints from e-grade.

One more shot!
Here's the equation:
                      HCNO(aq) + H2O(l)   H3O+(aq) + CNO-(aq)
Initially                .20 M                            0 M                0M
However, some of the acid will ionize (let's say x amount)
Change                -x                                +x                +x
Therefore at equilibrium, the new concentrations will be
Equilibrium            .20 -x                            x                  x

So at equilibrium the [H3O+] and the [CNO-] both equal x.
But, the problem states that at equilibrium the [H3O+] = 6.5 x 10-3, therefore x must equal 6.5 x 10-3.

Therefore at equilibrium:
                [H3O+] = 6.5 x 10E-3 M
                [CNO-] = 6.5 x 10E-3 M
                [HCNO] = .20 - 6.5E-3 = .1935 M

Plug the equilibrium concentrations into the Keq expression
                Keq = [H3O+] [CNO-] / [HCNO] 
and solve for Keq.

got it thanks