## Chemistry: Molecular Approach (4th Edition)

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HClO2 ]& [ Cl{O_2}^- ]& [ H_3O^+ ]\\ Initial& 0.155 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.155 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ Cl{O_2}^- ][ H_3O^+ ]}{[ HClO2 ]}$$ $$K_a = \frac{(x)(x)}{[ HClO2 ]_{initial} - x}$$ 3. Assuming $0.155 \gt\gt x:$ $$K_a = \frac{x^2}{[ HClO2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HClO2 ]_{initial}} = \sqrt{ 0.011 \times 0.155 }$$ $x = 0.041$ 4. Test if the assumption was correct: $$\frac{ 0.041 }{ 0.155 } \times 100\% = 26.0 \%$$ The percent is greater than 5%; therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HClO2 ]_{initial} - x}$$ $$K_a [ HClO2 ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HClO2 ] = 0$$ $$x_1 = \frac{- 0.011 + \sqrt{( 0.011 )^2 - 4 (1) (- 0.011 ) ( 0.155 )} }{2 (1)}$$ $$x_1 = 0.036$$ $$x_2 = \frac{- 0.011 - \sqrt{( 0.011 )^2 - 4 (1) (- 0.011 )( 0.155 )} }{2 (1)}$$ $$x_2 = -0.047$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 0.036$$ 6. $$[H_3O^+] = x = 0.036$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.036 ) = 1.44$$