## Chemistry and Chemical Reactivity (9th Edition)

$pH = 4.38$
1000ml = 1L 50ml = 0.05 L 30ml = 0.03 L 1. Find the numbers of moles: $C(C_6H_5COOH) * V(C_6H_5COOH) = 0.015* 0.05 = 7.5 \times 10^{-4}$ moles $C(KOH) * V(KOH) = 0.015* 0.03 = 4.5 \times 10^{-4}$ moles 2. Write the acid-base reaction: $C_6H_5COOH(aq) + KOH(aq) -- \gt KC_6H_5COO(aq) + H_2O(l)$ - Total volume: 0.05 + 0.03 = 0.08L 3. Since the base is the limiting reactant, only $0.00045$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[C_6H_5COOH] = 0.00075 - 0.00045 = 3 \times 10^{-4}$ moles. Concentration: $\frac{3 \times 10^{-4}}{ 0.08} = 3.8 \times 10^{-3}M$ $[KOH] = 0.00045 - 0.00045 = 0$ $[KC_6H_5COO] = 0 + 0.00045 = 0.00045$ moles. Concentration: $\frac{ 0.00045}{ 0.08} = 5.6 \times 10^{-3}M$ 4. Calculate the $pK_a$ for the acid $pKa = -log(Ka)$ $pKa = -log( 6.3 \times 10^{- 5})$ $pKa = 4.2$ 5. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.2 + log(\frac{5.6 \times 10^{-3}}{3.8 \times 10^{-3}})$ $pH = 4.2 + log(1.5)$ $pH = 4.2 + 0.176$ $pH = 4.38$