Author Question: In a solution prepared by dissolving 0.100 mole of propanoic acid in enough water to make 1.00 L of ... (Read 96 times)

tingc95 · **on:** Mar 21, 2021

Question 1

If you know K_b for ammonia, NH₃, you can calculate the equilibrium constant, K_a, for the following reaction: NH₄⁺

NH₃ + H⁺ by the equation:

◦ K_a = K_w× K_b.
◦ K_a = K_w / K_b.
◦ K_a = 1 / K_b.
◦ K_a = K_b / K_w.
◦ None of these are correct.

Question 2

In a solution prepared by dissolving 0.100 mole of propanoic acid in enough water to make 1.00 L of solution, the pH is observed to be 2.793. The K_a for propanoic acid (HC₃H₅O₂) is:

◦ 1.61 × 10^–3.
◦ 2.60 × 10^–5.
◦ 1.64 × 10^–2.
◦ 3.79 × 10^–10.
◦ None of these are correct.

johnharpe · **Reply #1 on:** Mar 21, 2021

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