Answer
(a) $pH = 3.27$
(b) $pH = 4.84$
Work Step by Step
(a)
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 7.2 \times 10^{- 4})$
$pKa = 3.143$
2. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 3.143 + log(\frac{0.2}{0.15})$
$pH = 3.143 + log(1.333)$
$pH = 3.143 + 0.1249$
$pH = 3.268$
(b)
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.745$
2. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
** $0.025M Ba(CH_3COO)_2 = 0.050M CH_3COO^-$
$pH = 4.745 + log(\frac{0.05}{0.04})$
$pH = 4.745 + log(1.25)$
$pH = 4.745 + 0.09691$
$pH = 4.842$