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Sagot :
Answer:
The answer is "2%"
Explanation:
Equation:
[tex]HNO_2\ (aq) \leftrightharpoons H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\ K_a = 4.0\times \ 10^{-4}[/tex]
[tex]H^{+}=?[/tex]
Formula:
[tex]Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}[/tex]
Let
[tex][H^{+}] = [NO_2^{-}] = x[/tex] at equilibrium
[tex]x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \ M\\\\[/tex]
therefore,
[tex][H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M[/tex]
Calculating the % ionization:
[tex]= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\[/tex]
The approximate percent ionization of HNO₂ in a 1.0 M HNO₂ (aq) solution is 2%.
How we calculate the % ionization?
% ionization of any compound will be calculated as follow:
% ionization = ([ion]/[acid or base]) ₓ 100
Given chemical reaction with ICE table will be represented as:
HNO₂(aq) → H⁺(aq) + NO₂⁻(aq)
initial 1 0 0
change -x +x +x
equilibrium 1-x x x
Equilibrium constant will be represented as:
Ka = [H⁺][NO₂⁻] / [HNO₂]
Acid dissociation constant for HNO₂ = 4×10⁻⁴
Putting all values in the above equation, we get
4×10⁻⁴ = x² / 1-x
Value of changeable quantity is very less, so we neglect from the concentration of HNO₂.
4×10⁻⁴ = x²
x = 2 × 10⁻²
So, the concentration of H⁺ ion = 2 × 10⁻²M
Now we put all these values in the % ionization equation, we get
% ionization = (0.02/1) × 100 = 2%
Hence , % ionization is 2%.
To know more about % ionization, visit the below link:
https://brainly.com/question/12198017
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