## General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal

# Chapter 16 - Acids and Bases - Example 16-5 - Determining a Value of K(a) from the pH of a Solution of a Weak Acid - Page 712: Practice Example A

#### Answer

$K_a (HOCl) = 2.9 \times 10^{-8}$

#### Work Step by Step

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HOCl ]& [ ClO^- ]& [ H_3O^+ ]\\ Initial& 0.150 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.150 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ ClO^- ][ H^+ ]}{[ HOCl ]}$$ $$K_a = \frac{(x)(x)}{[ HOCl ]_{initial} - x}$$ 3. Using the pH, find the $H_3O^+$ concentration: $$[H_3O^+] = 10^{-pH} = 10^{-4.18} = 6.6 \times 10^{-5} \space M$$ $[H_3O^+] = x = 6.6 \times 10^{-5}$ 4. Substitute the value of x and calculate the $K_a$: $$K_a = \frac{( 6.6 \times 10^{-5} )^2}{ 0.150 - 6.6 \times 10^{-5} }$$ $K_a = 2.9 \times 10^{-8}$

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