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Acetic acid has a dissociation constant...

Acetic acid has a dissociation constant of `1.8xx10^(-5)`. Calculate the pH value of the decinormal solution of acetic acid.

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To calculate the pH of a decinormal (0.1 M) solution of acetic acid (CH₃COOH) with a dissociation constant (Kₐ) of 1.8 x 10^(-5), we can follow these steps: ### Step 1: Write the dissociation equation of acetic acid Acetic acid dissociates in water as follows: \[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \] ### Step 2: Set up the expression for the dissociation constant (Kₐ) The expression for the dissociation constant (Kₐ) is given by: \[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \] ### Step 3: Define initial concentrations and changes Let the initial concentration of acetic acid be 0.1 M. At equilibrium, if \( x \) is the concentration of \( \text{H}^+ \) ions produced, we can write: - Initial concentration of CH₃COOH = 0.1 M - Change in concentration of CH₃COOH = -x - Change in concentration of CH₃COO⁻ = +x - Change in concentration of H⁺ = +x At equilibrium: - [CH₃COOH] = 0.1 - x - [CH₃COO⁻] = x - [H⁺] = x ### Step 4: Substitute into the Kₐ expression Substituting these values into the Kₐ expression gives: \[ K_a = \frac{x \cdot x}{0.1 - x} = \frac{x^2}{0.1 - x} \] ### Step 5: Assume x is small compared to 0.1 M Since Kₐ is small, we can assume that \( x \) is negligible compared to 0.1 M, thus: \[ K_a \approx \frac{x^2}{0.1} \] ### Step 6: Solve for x Substituting the value of Kₐ: \[ 1.8 \times 10^{-5} = \frac{x^2}{0.1} \] \[ x^2 = 1.8 \times 10^{-5} \times 0.1 \] \[ x^2 = 1.8 \times 10^{-6} \] \[ x = \sqrt{1.8 \times 10^{-6}} \] \[ x \approx 1.34 \times 10^{-3} \, \text{M} \] ### Step 7: Calculate the pH The concentration of H⁺ ions is approximately \( 1.34 \times 10^{-3} \, \text{M} \). The pH is calculated using the formula: \[ \text{pH} = -\log[\text{H}^+] \] \[ \text{pH} = -\log(1.34 \times 10^{-3}) \] \[ \text{pH} \approx 2.87 \] ### Final Answer The pH of the decinormal solution of acetic acid is approximately **2.87**. ---
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