## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652c: 55

#### Answer

$Ka (HOCN) = 3.554\times 10^{- 4}$

#### Work Step by Step

1. Calculate the hydronium concentration. $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 2.67}$ $[H_3O^+] = 2.138 \times 10^{- 3}M$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [OCN^-] = x = 2.138 \times 10^{-3}M$ -$[HOCN] = [HOCN]_{initial} - x$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][OCN^-]}{ [HOCN]}$ $Ka = \frac{x^2}{[InitialHOCN] - x}$ $Ka = \frac{( 2.138\times 10^{- 3})^2}{ 0.015- 2.138\times 10^{- 3}}$ $Ka = \frac{ 4.571\times 10^{- 6}}{ 0.01286}$ $Ka = 3.554\times 10^{- 4}$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.