What is the pH of a solution that results when 0.012 mol HNO₃ is added to 610.0 mL of a solution that is 0.26 M in aqueous ammonia and 0.46M in ammonium nitrate. Assume no volume change. (K_b for NH₃ = 1.8 × 10^-5.) A) 8.96 B) 9.56 C) 5.04 D) 9.01 E) 4.44

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Quizplus · Accepted Answer

The pH of the solution can be calculated using the Henderson-Hasselbalch equation for a buffer solution, which is pH = pKa + log([A-]/[HA]). In this case, NH3 acts as the base (A-) and NH4+ as the acid (HA). The pKa is the negative logarithm of the given Kb for NH3, which is 1.8 × 10^-5. First, we need to convert Kb to Ka for NH4+ using the relation Kw = Ka * Kb, where Kw = 1.0 × 10^-14 at 25°C. Thus, Ka = Kw / Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10, and pKa = -log(Ka) = 9.25. The moles of NH3 (base) initially present are 0.26 mol/L * 0.610 L = 0.1586 mol, and the moles of NH4+ (acid) are 0.46 mol/L * 0.610 L = 0.2806 mol. The addition of 0.012 mol of HNO3 will react completely with an equivalent amount of NH3, forming NH4+ and reducing the NH3 concentration. Thus, the new concentrations are:- [NH3] = (0.1586 mol - 0.012 mol) / 0.610 L = 0.2403 M- [NH4+] = (0.2806 mol + 0.012 mol) / 0.610 L = 0.4787 MPlugging these values into the Henderson-Hasselbalch equation gives:pH = 9.25 + log(0.2403 / 0.4787) = 9.25 + log(0.502) ≈ 9.25 - 0.299 = 8.95, which rounds to 8.96, making option A the correct answer.

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