## Chemistry and Chemical Reactivity (9th Edition)

$pH = 3.25$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [F^-] = x$ -$[HF] = [HF]_{initial} - x = 1 \times 10^{- 3} - x$ For approximation, we consider: $[HF] = 1 \times 10^{- 3}M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][F^-]}{ [HF]}$ $Ka = 7.2 \times 10^{- 4}= \frac{x * x}{ 1\times 10^{- 3}}$ $Ka = 7.2 \times 10^{- 4}= \frac{x^2}{ 1\times 10^{- 3}}$ $7.2 \times 10^{- 7} = x^2$ $x = 8.485 \times 10^{- 4}$ Percent ionization: $\frac{ 8.485 \times 10^{- 4}}{ 1\times 10^{- 3}} \times 100\% = 84.85\%$ %ionization < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Kb = 7.2 \times 10^{- 4}= \frac{x^2}{ 1 \times 10^{- 3}- x}$ $7.2 \times 10^{- 7} - 7.2 \times 10^{- 4}x = x^2$ $7.2 \times 10^{- 7} - 7.2 \times 10^{- 4}x - x^2 = 0$ Bhaskara: $\Delta = (- 7.2 \times 10^{- 4})^2 - 4 * (-1) *( 7.2 \times 10^{- 7})$ $\Delta = 5.184 \times 10^{- 7} + 2.88 \times 10^{- 6} = 3.398 \times 10^{- 6}$ $x_1 = \frac{ - (- 7.2 \times 10^{- 4})+ \sqrt { 3.398 \times 10^{- 6}}}{2*(-1)}$ or $x_2 = \frac{ - (- 7.2 \times 10^{- 4})- \sqrt { 3.398 \times 10^{- 6}}}{2*(-1)}$ $x_1 = - 1.282 \times 10^{- 3} (Negative)$ $x_2 = 5.617 \times 10^{- 4}$ - The concentration can't be negative. $[H_3O^+] = x = 5.617 \times 10^{- 4}$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 5.617 \times 10^{- 4})$ $pH = 3.25$