Answer:
[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.
Explanation:
HF + H₂O ⇄ H₃O⁺ + F⁻.
[H₃O⁺] = [F⁻].
∵ [H₃O⁺] = √Ka.C,
Ka = 6.8 x 10⁻⁴, C = 0.710 M.
∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.
∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.2 x 10⁻²) = 4.55 x 10⁻¹³.
The value of concentration of [H₃O⁺] & [F⁻] is 2.2 x 10⁻² M and [OH⁻] is 4.55 x 10⁻¹³M in 0.710M HF.
How we calculate acid dissociation constant?
Acid dissociation constant for any HA acid will be calculated as:
Ka = [H⁺][A⁻] / [HA].
ICE table for given reaction is written as:
HF + H₂O ⇄ H₃O⁺ + F⁻
Initial: 0.710 0 0
Change: -x +x +x
Equilibrium: 0.710-x +x +x
Given value of Ka = 6.8 × 10⁻⁴
Putting all values on the above value of Ka, we get
6.8 × 10⁻⁴ = x.x / 0.710-x
We take the value of 0.710-x as 0.710, because value of x is negligible as compare to 0.710.
6.8 × 10⁻⁴ = x.x / 0.710
x² = 6.8 × 10⁻⁴ × 0.710
x = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M
So, [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M
We know that pH + pOH = 14, we write this equation in base 10 form as:
[H₃O⁺][OH⁻] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ / 2.2 x 10⁻² = 4.55 x 10⁻¹³
Hence, value of [H₃O⁺] & [F⁻] is 2.2 x 10⁻² M and [OH⁻] is 4.55 x 10⁻¹³M.
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