Answer:
[tex][H_3O]^+=2.18x10^{-3}M[/tex]
Explanation:
Hello.
In this case, for the given ionization chemical reaction of acetic, the equilibrium expression is:
[tex]Ka=\frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}[/tex]
Since aqueous species are considered only. Next, by introducing the reaction extent [tex]x[/tex] based on the ICE table, in which it equals the concentration of both H3O+ and CH3COO-, we can also write:
[tex]1.8x10^{-5}=\frac{x*x}{0.265M-x}[/tex]
As the acid ionization constant is small in comparison to 0.265 M, we can make an approximation to compute [tex]x[/tex] as shown below:
[tex]1.8x10^{-5}=\frac{x*x}{0.265M}\\\\x=\sqrt{1.8x10^{-5}*0.265}= 2.18x10^{-3}M[/tex]
It means that the concentration of H3O+ is:
[tex][H_3O]^+=2.18x10^{-3}M[/tex]
Best regards.
A solution of acetic acid (Ka = 1.8 × 10⁻⁵) with an initial concentration of CH₃COOH of 0.265 mol/L has an equilibrium concentration of H₃O⁺ at 25 °C of 2.18 × 10⁻³ M.
Let's consider the reaction for the acid ionization of acetic acid.
CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq)
The acid ionization constant (Ka) is 1.8 × 10⁻⁵ and the initial concentration of the acid (Ca) is 0.265 mol/L.
Considering that it is a weak monoprotic acid, we can calculate the equilibrium concentration of H₃O⁺ at 25 °C using the following expression.
[tex][H_3O^{+} ]= \sqrt{Ka\times Ca } = \sqrt{1.8\times 10^{-5} \times 0.265 } = 2.18 \times 10^{-3} M[/tex]
A solution of acetic acid (Ka = 1.8 × 10⁻⁵) with an initial concentration of CH₃COOH of 0.265 mol/L has an equilibrium concentration of H₃O⁺ at 25 °C of 2.18 × 10⁻³ M.
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