#### Answer

(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:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.745 + log(\frac{2}{2})$
$pH = 4.745 + log(1)$
$pH = 4.745 + 0$
$pH = 4.745$
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(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.