## Chemistry: The Central Science (13th Edition)

The pH of the $HF$ pure solution is equal to $1.58$
1. We get these concentrations at equilibrium: -$[H_3O^+] = [F^-] = x$ -$[HF] = [HF]_{initial} - x = 1 - x$ For approximation, we consider: $[HF] = 1M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][F^-]}{ [HF]}$ $Ka = 6.8 \times 10^{- 4}= \frac{x * x}{ 1}$ $Ka = 6.8 \times 10^{- 4}= \frac{x^2}{ 1}$ $6.8 \times 10^{- 4} = x^2$ $x = 2.608 \times 10^{- 2}$ Percent dissociation: $\frac{ 2.608 \times 10^{- 2}}{ 1} \times 100\% = 2.608\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [F^-] = x = 2.608 \times 10^{- 2}M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.02608)$ $pH = 1.584$