Respuesta :
Answer : The equilibrium concentration of [tex]H_3O^+[/tex] at [tex]25^oC[/tex] is, [tex]2.154\times 10^{-3}m[/tex].
Solution : Given,
Equilibrium constant, [tex]K_c=1.8\times 10^{-5}[/tex]
Initial concentration of [tex]HC_2H_3O_2[/tex] = 0.260 m
Let, the 'x' mol/L of [tex]H_3O^+[/tex] are formed and at same time 'x' mol/L of [tex]HC_2H_3O_2[/tex] are also formed.
The equilibrium reaction is,
[tex]HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)[/tex]
Initially 0.260 m 0 0
At equilibrium (0.260 - x) x x
The expression for equilibrium constant for a given reaction is,
[tex]K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}[/tex]
Now put all the given values in this expression, we get
[tex]1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}[/tex]
By rearranging the terms, we get the value of 'x'.
[tex]x=2.154\times 10^{-3}m[/tex]
Therefore, the equilibrium concentration of [tex]H_3O^+[/tex] at [tex]25^oC[/tex] is, [tex]2.154\times 10^{-3}m[/tex].
A solution of CH₃COOH (Ka = 1.8 × 10⁻⁵) with an initial concentration of CH₃COOH of 0.260 M has an equilibrium concentration of H₃O⁺ at 25 °C of 2.16 × 10⁻³ M.
Let's consider the reaction for the ionization of CH₃COOH.
CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq)
The acid ionization constant (Ka) is 1.8 × 10⁻⁵ and the initial concentration of acetic acid (Ca) is 0.260 M.
For a weak monoprotic acid, we can calculate the equilibrium concentration of H₃O⁺ at 25 °C using the following expression.
[tex][H_3O^{+} ] = \sqrt{Ca \times Ka } = \sqrt{(0.260) \times (1.8 \times 10^{-5} ) } = 2.16 \times 10^{-3} M[/tex]
A solution of CH₃COOH (Ka = 1.8 × 10⁻⁵) with an initial concentration of CH₃COOH of 0.260 M has an equilibrium concentration of H₃O⁺ at 25 °C of 2.16 × 10⁻³ M.
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