## Chemistry 10th Edition

1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [F^-] = x$ -$[HF] = [HF]_{initial} - x = 0.38 - x$ For approximation, we consider: $[HF] = 0.38M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][F^-]}{ [HF]}$ $Ka = 7.2 \times 10^{- 4}= \frac{x * x}{ 3.8 \times 10^{- 1}}$ $Ka = 7.2 \times 10^{- 4}= \frac{x^2}{ 3.8 \times 10^{- 1}}$ $2.736 \times 10^{- 4} = x^2$ $x = 1.654 \times 10^{- 2}$ Percent ionization: $\frac{ 1.654 \times 10^{- 2}}{ 3.8 \times 10^{- 1}} \times 100\% = 4.353\%$ %ionization < 5% : Right approximation. Therefore: $[H_3O^+] = x = 1.654 \times 10^{- 2}$ 3. Find the pH: $pH = -log[H_3O^+]$ $pH = -log( 0.01654)$ $pH = 1.781$