## Chemistry: Atoms First (2nd Edition)

$[HC_3H_5O_2] \approx 0.1M$ $[H_3O^+] = [C_3H_5{O_2}^-] = x = 1.14 \times 10^{- 3}M$ $pH = 2.943$ Percent dissociation: $= 1.14\%$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [C_3H_5{O_2}^-] = x$ -$[HC_3H_5O_2] = [HC_3H_5O_2]_{initial} - x = 0.1 - x$ For approximation, we consider: $[HC_3H_5O_2] = 0.1M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_3H_5{O_2}^-]}{ [HC_3H_5O_2]}$ $Ka = 1.3 \times 10^{- 5}= \frac{x * x}{ 0.1}$ $Ka = 1.3 \times 10^{- 5}= \frac{x^2}{ 0.1}$ $1.3 \times 10^{- 6} = x^2$ $x = 1.14 \times 10^{- 3}$ Percent dissociation: $\frac{ 1.14 \times 10^{- 3}}{ 0.1} \times 100\% = 1.14\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [C_3H_5{O_2}^-] = x = 1.14 \times 10^{- 3}M$ And, since 'x' has a very small value (compared to the initial concentration): $[HC_3H_5O_2] \approx 0.1M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.14 \times 10^{- 3})$ $pH = 2.943$