What is the pH of a solution prepared by adding 0.254 g of ammonium chloride to 115 mL of water? K_b of NH₃ is 1.8 × 10^-5. A)5.32 B)3.07 C)7.00 D)8.68 E)10.93

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Quizplus · Accepted Answer

The pH of the solution can be calculated using the Kb of NH3 and the mass of NH4Cl added to water. First, calculate the concentration of NH4Cl in the solution. The molar mass of NH4Cl is 53.49 g/mol, so 0.254 g corresponds to 0.254 g / 53.49 g/mol = 0.00475 mol. This amount in 115 mL (0.115 L) of water gives a concentration of 0.00475 mol / 0.115 L = 0.0413 M. Given Kb for NH3 is 1.8 × 10^-5, we can find the Ka for NH4+ using the relation Kw = Ka * Kb, where Kw is the ion product of water (1.0 × 10^-14 at 25°C). Thus, Ka = Kw / Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10.Using the formula for the pH of a weak acid (pH = 1/2 * (pKa - log[C])), where [C] is the concentration of NH4+, we find:pKa = -log(Ka) = -log(5.56 × 10^-10) = 9.25,pH = 1/2 * (9.25 - log[0.0413]) ≈ 5.32.Therefore, the pH of the solution is approximately 5.32.

What Is the PH of a Solution Prepared by Adding