Answered: HCN | bartleby

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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HCN (aq) + H₂O (l) ⇄ H₃O⁺ (aq) + CN^- (aq)

K_a = ([H₃O⁺][CN^-])/[HCN] = 6.2 × 10^-10

The equilibrium reaction shown above represents the partial ionization of the weak acid HCN (aq). A 0.200 M HCN (aq) solution has a pH ≈ 4.95. If 0.05 g (0.010 mol) of NaCN (s) is added to 100 mL of 0.200 M HCN (aq), which of the following explains how and why the pH of the solution changes?

The pH will be higher than 4.95 because adding CN^- will disrupt the equilibrium, resulting in an increased production of HCN that decreases the concentration of H₃O⁺.

The pH will be lower than 4.95 because adding CN^- will disrupt the equilibrium, resulting in an increased production of HCN that decreases the concentration of H₃O⁺.

The pH will be higher than 4.95 because CN^- is a strong base that can neutralize HCN.

The pH will remain close to 4.95 because the K_a is so small that hardly any products form.

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