Respuesta :
Answer:
[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
[tex]K = \dfrac{[\text{H$_{3}$O$^{+}$}][\text{CNO$^{-}$}] }{[\text{HCNO}]} = \dfrac{0.0059^{2}}{0.094} = \mathbf{3.7 \times 10^{-4}}[/tex]
The concentration of [tex]\rm H_3O^+[/tex] is 0.0059 mol/L, [tex]\rm CNO^-[/tex] 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:
[tex]\rm HCNO\;+\;H_2O\;\rightarrow\;H_3O^+\;+\;CNO^-[/tex]
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 [tex]\rm H_3O^+[/tex] 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 = [tex]\rm 0.1\;\times\;\dfrac{5.9}{100}[/tex]
x = 0.0059 mol/L
The concentration of [tex]\rm H_3O^+[/tex] and [tex]\rm CNO^-[/tex] has been x.
[tex]\rm H_3O^+[/tex] = 0.0059 mol/L
[tex]\rm CNO^-[/tex] = 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:
[tex]\rm k\;=\;\dfrac{[H_3O^+]\;[CNO^-]}{[HCHO]}[/tex]
[tex]\rm k\;=\;\dfrac{[0.0059]\;[0.0059]}{[0.094]}[/tex]
k = 0.00037
The concentration of [tex]\rm H_3O^+[/tex] is 0.0059 mol/L, [tex]\rm CNO^-[/tex] 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:
https://brainly.com/question/10217868
