# 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.

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) / M_{initial} = 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

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.