## Chemistry and Chemical Reactivity (9th Edition)

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: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.74 + log(\frac{0.04}{0.02})$ $pH = 4.74 + log(2)$ $pH = 4.74 + 0.301$ $pH = 5.05$