# Chapter 17 - Section 17.1 - The Common Ion Effect - Checkpoint - Page 782: 17.1.2

(b) $pH = 2.8$

#### Work Step by Step

- 0.05 mole of $NaF$ in 1.0L = 0.05M NaF. $[NaF] = [F^-] = 0.05M$ (Completely dissociated salt) 1. Drawing the ICE table, we get these concentrations at the equilibrium: $HF(aq) + H_2O(l) \lt -- \gt F^-(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x Products = Initial Concentration + x $[HF] = 0.1 M - x$ $[F^-] = 0.05M + x$ $[H_3O^+] = 0 + x$ 2. Calculate 'x' using the $K_a$ expression. $7.1\times 10^{- 4} = \frac{[F^-][H_3O^+]}{[HF]}$ $7.1\times 10^{- 4} = \frac{( 0.05 + x )* x}{ 0.1 - x}$ Considering 'x' has a very small value. $7.1\times 10^{- 4} = \frac{ 0.05 * x}{ 0.1}$ $7.1\times 10^{- 4} = 0.5x$ $\frac{ 7.1\times 10^{- 4}}{ 0.5} = x$ $x = 1.4\times 10^{- 3}$ Percent dissociation: $\frac{ 1.4\times 10^{- 3}}{ 0.1} \times 100\% = 1.4\%$ x = $[H_3O^+]$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.4 \times 10^{- 3})$ $pH = 2.85$

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