Chemistry: The Molecular Science (5th Edition)

$pH = 5.70$
- 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: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.745 + log(\frac{0.04737}{5.263 \times 10^{-3}})$ $pH = 4.745 + log(9)$ $pH = 4.745 + 0.9542$ $pH = 5.70$