Answer
$pH = 5.70$
Work Step by Step
- Find the numbers of moles:
$C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.05 = 5 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.1* 0.045 = 4.5 \times 10^{-3}$ moles
Write the acid-base reaction:
$CH_3COOH(aq) + NaOH(aq) -- \gt NaCH_3COO(aq) + H_2O(l)$
- Total volume: 0.05 + 0.045 = 0.095L
Since the base is the limiting reactant, only $ 0.0045$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[CH_3COOH] = 0.005 - 0.0045 = 5 \times 10^{-4}$ moles.
Concentration: $\frac{5 \times 10^{-4}}{ 0.095} = 5.263 \times 10^{-3}M$
$[NaOH] = 0.0045 - 0.0045 = 0$
$[NaCH_3COO] = 0 + 0.0045 = 0.0045$ moles.
Concentration: $\frac{ 0.0045}{ 0.095} = 0.04737M$
- Calculate the $pK_a$ for the acid
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.745$
- Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.745 + log(\frac{0.04737}{5.263 \times 10^{-3}})$
$pH = 4.745 + log(9)$
$pH = 4.745 + 0.9542$
$pH = 5.70$
