## Chemistry: Molecular Approach (4th Edition)

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HNO_2 ]& [ N{O_2}^- ]& [ H_3O^+ ]\\ Initial& 0.010 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.010 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ N{O_2}^- ][ H_3O^+ ]}{[ HNO_2 ]}$$ $$K_a = \frac{(x)(x)}{[ HNO_2 ]_{initial} - x}$$ 3. Assuming $0.010 \gt\gt x:$ $$K_a = \frac{x^2}{[ HNO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HNO_2 ]_{initial}} = \sqrt{ 4.6 \times 10^{-4} \times 0.010 }$$ $x = 2.1 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 2.1 \times 10^{-3} }{ 0.010 } \times 100\% = 21.0 \%$$ The percent is greater than 5%; therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{x^2}{[ HNO_2 ]_{initial} - x}$$ $$K_a [ HNO_2 ] - K_a x = x^2$$ $$x^2 + K_a x - K_a [ HNO_2 ] = 0$$ $$x_1 = \frac{- 4.6 \times 10^{-4} + \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} ) ( 0.010 )} }{2 (1)}$$ $$x_1 = 1.9 \times 10^{-3}$$ $$x_2 = \frac{- 4.6 \times 10^{-4} - \sqrt{( 4.6 \times 10^{-4} )^2 - 4 (1) (- 4.6 \times 10^{-4} )( 0.010 )} }{2 (1)}$$ $$x_2 = -2.4 \times 10^{-3}$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 1.9 \times 10^{-3}$$ 6. $$[H_3O^+] = x = 1.9 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 1.9 \times 10^{-3} ) = 2.72$$