## Chemistry 12th Edition

Acid benzoic: Ka = $6.5 \times 10^{-5}$ $Ka = \frac{[H^+][A^-]}{[HA]}$ $[H^+] = [A^-] = x$ $[HA] = [HA]_{initial} - x$ (a) 1. Calculate $[H^+]$ and $[A^-]$ - Since the concentration is too large, compared to the Ka, we can consider, $[HA]_{initial} = [HA] = 0.20M$ Therefore: $6.5 \times 10^{-5} = \frac{x * x}{0.20}$ $1.3 \times 10^{-5} = x^2$ $x = 3.61 \times 10^{-3}M$ $x = [H^+] = [A^-]$ 2. Now, calculate the percent ionization. $\% ionization = \frac{[H^+]}{[HA]} \times 100 \%$ $\% ionization = \frac{3.61 \times 10^{-3}}{0.20M} \times 100\% = 1.805 \%$ (b) 1. Calculate $[H^+]$ and $[A^-]$ $Ka = 6.5 \times 10^{- 5}= \frac{x * x}{ 2 \times 10^{- 4}- x}$ $Ka = 6.5 \times 10^{- 5}= \frac{x^2}{ 2 \times 10^{- 4}- x}$ $1.3 \times 10^{- 8} - 6.5 \times 10^{- 5}x = x^2$ $1.3 \times 10^{- 8} - 6.5 \times 10^{- 5}x - x^2 = 0$ $\Delta = (- 6.5 \times 10^{- 5})^2 - 4 * (-1) *( 1.3 \times 10^{- 8})$ $\Delta = 4.22 \times 10^{- 9} + 5.2 \times 10^{- 8} = 5.62 \times 10^{- 8}$ $x_1 = \frac{ - (- 6.5 \times 10^{- 5})+ \sqrt { 5.62 \times 10^{- 8}}}{2*(-1)}$ or $x_2 = \frac{ - (- 6.5 \times 10^{- 5})- \sqrt { 5.62 \times 10^{- 8}}}{2*(-1)}$ $x_1 = - 1.51 \times 10^{- 4} (Negative)$ $x_2 = 8.6 \times 10^{- 5}$ The concentration can't be negative, therefore: $x = [H^+] = 8.6 \times 10^{- 5}$ $\%ionization = \frac{8.6 \times 10^{- 5}}{0.0002} \times 100\% = 43 \%$