## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Section 16.6 - Finding the [H3O+] and pH of Strong and Weak Acid Solutions - For Practice - Page 741: 16.8

#### Answer

The $K_a$ for that weak acid is: $1.8 \times 10^{-6}$

#### Work Step by Step

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HA ]& [ A^- ]& [ H_3O^+ ]\\ Initial& 0.175 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.175 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ A^- ][ H^+ ]}{[ HA ]}$$ $$K_a = \frac{(x)(x)}{[ HA ]_{initial} - x}$$ 3. Using the pH, find the $H_3O^+$ concentration: $$[H_3O^+] = 10^{-pH} = 10^{-3.25} = 5.6 \times 10^{-4} \space M$$ $[H_3O^+] = x = 5.6 \times 10^{-4}$ 4. Substitute the value of x and calculate the $K_a$: $$K_a = \frac{( 5.6 \times 10^{-4} )^2}{ 0.175 - 5.6 \times 10^{-4} }$$ $K_a = 1.8 \times 10^{-6}$

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