# Chapter 17 - Questions and Problems - Page 824: 17.10

(a) $pH = 4.75$ (b) $pH = 4.75$ - The more effective buffer is the first one, because of the greater concentrations.

#### Work Step by Step

(a) 1. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 5})$ $pKa = 4.745$ 2. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{2}{2}$ - 1: It is. 3. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{2}{1.8 \times 10^{-5}} = 1.111\times 10^{5}$ - $\frac{2}{1.8 \times 10^{-5}} = 1.111\times 10^{5}$ 4. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.745 + log(\frac{2}{2})$ $pH = 4.745 + log(1)$ $pH = 4.745 + 0$ $pH = 4.745$ ----- (b) 1. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.2}{1.8 \times 10^{-5}} = 1.111\times 10^{4}$ - $\frac{0.2}{1.8 \times 10^{-5}} = 1.111\times 10^{4}$ 2. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.745 + log(\frac{0.2}{0.2})$ $pH = 4.745 + log(1)$ $pH = 4.745 + 0$ $pH = 4.745$ Since the first buffer has a higher concentratrion of base and acid, the addition of another compound will not effect it as much as it will with the second one.

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