## Chemistry (4th Edition)

1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [A^-] = x$ -$[HA] = [HA]_{initial} - x = 0.1M - x$ For approximation, we consider: $[HA] = [HA]_{initial} = 0.1M$ ** HA = Acid, A- = Conjugate base. 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][A^-]}{ [HA]}$ $Ka = 6.5 \times 10^{- 5}= \frac{x * x}{0.1}$ $Ka = 6.5 \times 10^{- 5}= \frac{x^2}{ 0.1}$ $6.5 \times 10^{- 6} = x^2$ $x = 2.54 \times 10^{- 3}$ 5% test: $\frac{ 2.54 \times 10^{- 3}}{0.1} \times 100\% = 2.54\%$ %ionization < 5% : Right approximation. $[H_3O^+] = x = 2.54 \times 10^{-3}$ 3. Find the pH: $pH = -log[H_3O^+]$ $pH = -log( 2.54 \times 10^{- 3})$ $pH = 2.59$