## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 64

#### Answer

$[H_3O^+] = 6.0 \times 10^{-3}$ $pH = 2.22$

#### Work Step by Step

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HCHO_2 ]& [ CHO{_2}^- ]& [ H_3O^+ ]\\ Initial& 0.200 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.200 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CHO{_2}^- ][ H^+ ]}{[ HCHO_2 ]}$$ $$K_a = \frac{(x)(x)}{[ HCHO_2 ]_{initial} - x}$$ 3. Assuming $0.200 \gt\gt x:$ $$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.200 }$$ $x = 6.0 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 6.0 \times 10^{-3} }{ 0.200 } \times 100\% = 3.0 \%$$ 5. Thus, it is correct to say that $x = 6.0 x 10^{-3}$ 6. $$[H^+] = x = 6.0 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( \sqrt{ 1.8 \times 10^{-4} \times 0.200 } ) = 2.22$$

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.