# Chapter 16 - Self-Assessment Quiz - Page 767: Q6

(c) 2.32

#### Work Step by Step

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HC_7H_5O_2 ]& [ C_7H_5{O_2}^- ]& [ H_3O^+ ]\\ Initial& 0.350 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.350 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_7H_5{O_2}^- ][ H_3O^+ ]}{[ HC_7H_5O_2 ]}$$ $$K_a = \frac{(x)(x)}{[ HC_7H_5O_2 ]_{initial} - x}$$ 3. Assuming $0.350 \gt\gt x:$ $$K_a = \frac{x^2}{[ HC_7H_5O_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HC_7H_5O_2 ]_{initial}} = \sqrt{ 6.5 \times 10^{-5} \times 0.350 }$$ $x = 4.8 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 4.8 \times 10^{-3} }{ 0.350 } \times 100\% = 1.4 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 4.8 x 10^{-3}$ 6. $$[H_3O^+] = x = 4.8 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 4.8 \times 10^{-3} ) = 2.32$$

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