$\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_3 \mathrm{COOH}$ is $1.8 \times 10^{-5}$ and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_4 \mathrm{OH}$ is $1.8 \times 10^{-5}$. The $\mathrm{pH}$ of ammonium acetate solution will be _________.

<p>To find the pH of an ammonium acetate solution, we use the fact that ammonium acetate is a salt resulting from the neutralization of a weak acid (acetic acid, $CH_3COOH$) by a weak base (ammonium hydroxide, $NH_4OH$). The respective ionization constant values for the weak acid $K_a$ and the weak base $K_b$ are given as $1.8 \times 10^{-5}$.</p><p>Firstly, calculate the $pK_a$ and $pK_b$ values.</p><ul><li>$pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5})$</li><li>$pK_b = -\log(K_b) = -\log(1.8 \times 10^{-5})$</li></ul><p>Given that the values of $K_a$ and $K_b$ are the same, their $pK_a$ and $pK_b$ values will also be the same, establishing a neutral condition where the effects of the acid and base neutralize each other.</p><p>The formula used to determine the pH of a solution of such a salt is:</p><p>$pH = \frac{1}{2}(pK_w + pK_a - pK_b)$</p><p>Where $pK_w$ is the ionic product of water, which is 14 at 25°C. In this specific case, since $pK_a = pK_b$, the formula simplifies to:</p><p>$pH = \frac{1}{2}(14 + pK_a - pK_a)$</p><p>$pH = \frac{1}{2}(14)$</p><p>$pH = 7$</p><p>The pH of an ammonium acetate solution in this scenario is 7, indicating a neutral solution.</p>