Answer
$pH = 5.39$
Work Step by Step
1. Calculate the molar mass $(NaCH3COOH)$:
22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 16* 1 + 16* 1 + 1.01* 1 = 83.05g/mol
2. Calculate the number of moles $(NaCH3COOH)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 83}{ 83.05}$
$n(moles) = 1$
3. Find the concentration in mol/L $(NaCH3COOH)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 1}{ 1.5} $
$C(mol/L) = 0.67$
4. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.74$
5. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.67}{0.15}$
- 4.4: It is.
6. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.67}{1.8 \times 10^{-5}} = 3.7\times 10^{4}$
- $ \frac{0.15}{1.8 \times 10^{-5}} = 8333$
7. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.74 + log(\frac{0.67}{0.15})$
$pH = 4.74 + log(4.4)$
$pH = 4.74 + 0.65$
$pH = 5.39$
