## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Section 16.9 - Polyprotic Acids - For Practice - Page 759: 16.17

#### Answer

The pH of this solution is equal to 3.83

#### Work Step by Step

The second proton from $H_2CO_3$ has a negligible contribution to the pH. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ H_2CO_3 ]& [ HC{O_3}^- ]& [ H_3O^+ ]\\ Initial& 0.050 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.050 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ HC{O_3}^- ][ H_3O^+ ]}{[ H_2CO_3 ]}$$ $$K_a = \frac{(x)(x)}{[ H_2CO_3 ]_{initial} - x}$$ 3. Assuming $0.050 \gt\gt x:$ $$K_a = \frac{x^2}{[ H_2CO_3 ]_{initial}}$$ $$x = \sqrt{K_a \times [ H_2CO_3 ]_{initial}} = \sqrt{ 4.3 \times 10^{-7} \times 0.050 }$$ $x = 1.466 \times 10^{-4}$ 4. Test if the assumption was correct: $$\frac{ 1.466 \times 10^{-4} }{ 0.050 } \times 100\% = 0.3 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 1.466 x 10^{-4}$ 6. $$[H_3O^+] = x = 1.466 \times 10^{-4}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 1.466 \times 10^{-4} ) = 3.83$$

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