# Chapter 14 - Acids and Bases - Exercises - Page 704: 65

$[H_3O^+] = [F^-] = 3.5 \times 10^{-3}M$ $[HF] = 0.017M$ $[OH^-] = 2.9 \times 10^{-12}M$ $pH = 2.46$

#### Work Step by Step

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $HF(aq) + H_2O(l) \lt -- \gt F^-(aq) + H_3O^+(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $F^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M -$[H_3O^+] = [F^-] = 0 + x = x$ -$[HF] = [HF]_{initial} - x$ For approximation, we are going to consider $[HF]_{initial} = [HF]$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][F^-]}{ [HF]}$ $Ka = 7.2 \times 10^{- 4}= \frac{x * x}{ 0.02}$ $Ka = 7.2 \times 10^{- 4}= \frac{x^2}{ 0.02}$ $x^2 = 0.02 \times 7.2 \times 10^{-4}$ $x = \sqrt { 0.02 \times 7.2 \times 10^{-4}} = 3.8 \times 10^{-3}$ Percent dissociation: $\frac{ 3.8 \times 10^{- 3}}{ 0.02} \times 100\% = 19\%$ %dissociation > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Ka = 7.2 \times 10^{- 4}= \frac{x^2}{ 0.02- x}$ $1.4 \times 10^{- 5} - 7.2 \times 10^{- 4}x = x^2$ $1.4 \times 10^{- 5} - 7.2 \times 10^{- 4}x - x^2 = 0$ Bhaskara: $\Delta = (- 7.2 \times 10^{- 4})^2 - 4 * (-1) *( 1.4 \times 10^{- 5})$ $\Delta = 5.2 \times 10^{- 7} + 5.8 \times 10^{- 5} = 5.8 \times 10^{- 5}$ $x_1 = \frac{ - (- 7.2 \times 10^{- 4})+ \sqrt { 5.8 \times 10^{- 5}}}{2*(-1)}$ or $x_2 = \frac{ - (- 7.2 \times 10^{- 4})- \sqrt { 5.8 \times 10^{- 5}}}{2*(-1)}$ $x_1 = - 4.2 \times 10^{- 3} (Negative)$ $x_2 = 3.5 \times 10^{- 3}$ - The concentration can't be negative, so $x$ = $x_2$ $= 3.5 \times 10^{-3}$ $[H_3O^+] = [F^-] = x = 3.5 \times 10^{-3}M$ $[HF] = [HF]_{initial} - x = (0.020) - (3.5 \times 10^{-3}) = 0.017M$ 3. Calculate the hydroxide ion concentration: $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $3.5 \times 10^{-3} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{3.5 \times 10^{-3}}$ $[OH^-] = 2.9 \times 10^{-12}M$ 4. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 3.5 \times 10^{- 3})$ $pH = 2.46$

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