Answer
(a) $pH = 7.21$
(b) $pH = 7.24$
Work Step by Step
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 6.2 \times 10^{- 8})$
$pKa = 7.21$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.132}{0.132}$
- 1: It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.13}{6.2 \times 10^{-8}} = 2.1\times 10^{6}$
- $ \frac{0.13}{6.2 \times 10^{-8}} = 2.1\times 10^{6}$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 7.21 + log(\frac{0.132}{0.132})$
$pH = 7.21 + log(1)$
$pH = 7.21 + 0$
$pH = 7.21$
5. Calculate the molar mass $(NaOH)$:
22.99* 1 + 16* 1 + 1.01* 1 = 40g/mol
6. Calculate the number of moles $(NaOH)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.425}{ 40}$
$n(moles) = 0.0106$
7. Find the concentration in mol/L $(NaOH)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.0106}{ 2} $
$C(mol/L) = 5.3\times 10^{- 3}$
8. The addition of a strong base will reduce the acid concentration, and increase the base concentration by its own value:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 7.21 + log(\frac{[0.132 + 5.3 \times 10^{-3}]}{[0.132 - 5.3 \times 10^{-3}]})$
$pH = 7.21 + log(\frac{0.137}{[0.127]})$
$pH = 7.21 +log(1.08)$
$pH = 7.21 + 0.03$
$pH = 7.24$
