Answer
The pH is approximately 5.05; therefore, the correct answer is (b).
Work Step by Step
1000ml = 1L
1. Find the numbers of moles:
$C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.045 = 4.5 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.1* 0.03 = 3 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$CH_3COOH(aq) + NaOH(aq) -- \gt NaCH_3COO(aq) + H_2O(l)$
- Total volume: 0.045 + 0.03 = 0.075L
3. Since the base is the limiting reactant, only $ 0.003$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[CH_3COOH] = 0.0045 - 0.003 = 1.5 \times 10^{-3}$ moles.
Concentration: $\frac{1.5 \times 10^{-3}}{ 0.075} = 0.02M$
$[NaOH] = 0.003 - 0.003 = 0$
$[NaCH_3COO] = 0 + 0.003 = 0.003$ moles.
Concentration: $\frac{ 0.003}{ 0.075} = 0.04M$
4. Calculate the $pK_a$ for the acid
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.74$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.74 + log(\frac{0.04}{0.02})$
$pH = 4.74 + log(2)$
$pH = 4.74 + 0.301$
$pH = 5.05$
