What is the pH of a 0.300 M NH₃ solution that has K_b = 1.8 × 10^-5? The equation for the dissociation of NH₃ is NH₃(aq)+ H₂O(l)⇌ NH₄⁺(aq)+ OH^-(aq). A)2.11 B)2.63 C)11.89 D)11.37

Question

Quizplus · Accepted Answer

The dissociation of NH4+ in water is an acid dissociation reaction, NH4+ ⇌ NH3 + H+. The given $K_a = 1.8 \times 10^{-5}$. To find the pH, first calculate the $pK_a = -\log(1.8 \times 10^{-5})$, then use the formula $pH = pK_a + \log\left(\frac{[NH_3]}{[NH_4^+]}\right)$. Since NH4+ is the acid and it dissociates completely, the concentration of NH3 and NH4+ at equilibrium can be considered equal, simplifying the log term to 0. Thus, $pH = pK_a$. However, since NH4+ is acting as an acid, we actually need to find the pOH first and then convert to pH. The correct calculation involves recognizing that the given $K_a$ value is for the dissociation of NH4+, and we need to find the concentration of OH- produced, then calculate pOH, and finally convert to pH. The pH of a solution is related to pOH by the equation pH + pOH = 14. Given the nature of the dissociation and the provided $K_a$, the calculation should correctly account for the acidic nature of NH4+ and the basic nature of the resulting solution, leading to a pH greater than 7. The correct answer involves recognizing the acidic dissociation and calculating accordingly, which leads to a pH of 11.37, indicating a basic solution as expected from the dissociation of NH4+.

What Is the PH of a 0