15 points
a 0.100 m solution of cyanic acid (hcno) is 5.9% ionized. using this information, calculate
[cno], [h3o+], [hcno] and k, for cyanic acid.

The concentration of is 0.0059 mol/L, is 0.0059 mol/L, and HCHO is 0.094 mol/L. The value of k for cyanic acid has been 0.00037.

The reaction of the cyanic acid can be as follows:

The initial concentration of the cyanic acid has been = 0.1 M.

The ionization at equilibrium = 5.9 %

The ICE for the setup has bee:

HCNO          CNO

0.1            0              0            I

-x             +x             +x          C

0.1-x          x              x           E

The x is the concentration at equilibrium. It is 5.9% of the initial concentration.

x =

x = 0.0059 mol/L

The concentration of and has been x.

= 0.0059 mol/L

= 0.0059 mol/L

HCNO = 0.1 - x

HCNO = 0.1 - 0.0059 mol/L

HCNO = 0.094 mol/L.

The value of ionization constant (k) can be calculated as:

k = 0.00037

The concentration of is 0.0059 mol/L, is 0.0059 mol/L, and HCHO is 0.094 mol/L. The value of k for cyanic acid has been 0.00037.

For more information about the concentration and ionization constants, refer to the link:

link

[H₃O⁺] = [CNO⁻] = 0.0059 mol·L⁻¹; [HCNO] = 0.094 mol·L⁻¹; K = 3.7 × 10⁻⁴

Explanation:

1. Set up an ICE table.

                   HCNO + H₂O ⟶ H₃O⁺ + CNO⁻

I/mol·L⁻¹:      0.100                      0           0

C/mol·L⁻¹:        -x                        +x         +x

E/mol·L⁻¹:  0.100-x                     x           x

2. Concentrations

We know that x = 5.9 % of the initial concentration.

x = 0.059 × 0.100 = 0.0059 mol·L⁻¹

[H₃O⁺] = [CNO⁻] = x mol·L⁻¹ = 0.0059 mol·L⁻¹

[HCNO] = 0.100 - 0.0059   = 0.094    mol·L⁻¹

3. Ionization constant

i belive it is water