Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 16 - Acid-Base Equilibria - Exercises - Page 720: 16.80a

Answer

$[OH^-] = 1.24 \times 10^{- 6}M $ $pH = 8.09$

Work Step by Step

- Identify the significant electrolytes in $NaF$: - $Na^+$ : Insignificant - $F^-$: Weak base. Therefore, we just need to calculate the pH of a $0.105M$ $F^-$ solution. 1. Since $F^-$ is the conjugate base of $HF$ , we can calculate its kb by using this equation: $K_a * K_b = K_w = 10^{-14}$ $ 6.8\times 10^{- 4} * K_b = 10^{-14}$ $K_b = \frac{10^{-14}}{ 6.8\times 10^{- 4}}$ $K_b = 1.471\times 10^{- 11}$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [HF] = x$ -$[F^-] = [F^-]_{initial} - x = 0.105 - x$ For approximation, we consider: $[F^-] = 0.105M$ 3. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][HF]}{ [F^-]}$ $Kb = 1.471 \times 10^{- 11}= \frac{x * x}{ 0.105}$ $Kb = 1.471 \times 10^{- 11}= \frac{x^2}{ 0.105}$ $ 1.544 \times 10^{- 12} = x^2$ $x = 1.243 \times 10^{- 6}$ Percent ionization: $\frac{ 1.243 \times 10^{- 6}}{ 0.105} \times 100\% = 0.001183\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [HF] = x = 1.243 \times 10^{- 6}M $ $[F^-] \approx 0.105M$ 4. Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 1.243 \times 10^{- 6})$ $pOH = 5.906$ $pH + pOH = 14$ $pH + 5.906 = 14$ $pH = 8.094$
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