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sergiy2304 [10]
3 years ago
9

15 points

Chemistry
1 answer:
finlep [7]3 years ago
5 0

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

K = \dfrac{[\text{H$_{3}$O$^{+}$}][\text{CNO$^{-}$}] }{[\text{HCNO}]} = \dfrac{0.0059^{2}}{0.094} = \mathbf{3.7 \times 10^{-4}}

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