The dissociation constant of acetic acid is $x\times10^{-5}$. When 25 mL of 0.2 $\mathrm{M~CH_3COONa}$ solution is mixed with 25 mL of 0.02 $\mathrm{M~CH_3COOH}$ solution, the pH of the resultant solution is found to be equal to 5. The value of $x$ is ____________

<p>To find the dissociation constant of acetic acid, we use the Henderson-Hasselbalch equation for the given system and conditions. The equation is as follows:</p> <p>$ \text{pH} = \text{pK}_\text{a} + \log \left(\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}\right) $</p> <p>In the solution mixture:</p> <p><p>We have <strong>25 mL of 0.2 M $\text{CH}_3\text{COONa}$</strong> and <strong>25 mL of 0.02 M $\text{CH}_3\text{COOH}$</strong>.</p></p> <p><p>The concentration ratio $\left(\frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]}\right)$ becomes $\frac{25 \times 0.2}{25 \times 0.02} = 10$.</p></p> <p>Given that the pH of the solution is 5, we substitute into the equation:</p> <p>$ 5 = \text{pK}_\text{a} + \log 10 $</p> <p>Since $\log 10 = 1$, we solve for $\text{pK}_\text{a}$:</p> <p>$ 5 = \text{pK}_\text{a} + 1 \quad \Rightarrow \quad \text{pK}_\text{a} = 4 $</p> <p>Converting from $\text{pK}_\text{a}$ to $K_\text{a}$, we use:</p> <p>$ K_\text{a} = 10^{-\text{pK}_\text{a}} = 10^{-4} $</p> <p>Thus, since the dissociation constant $K_\text{a}$ is given as $x \times 10^{-5}$, compare:</p> <p>$ 10^{-4} = 10 \times 10^{-5} $</p> <p>Therefore, $x = 10$.</p>