5moldm³ 30cm³ CH₃COOH solution is neutralized by Ba(OH)₂ solution. 0.75dm³ of the solution is used. If 10cm³ of Ba(OH)₂ solution is neutralized by 0.1moldm⁻³ NaOH solution, calculate W.

Initial amount of CH_3COOH: n = 0.50 \, mol \, dm^-3 × 0.025 \, dm^3 = 0.0125 \, mol Equivalence volume of NaOH: V_eq = 0.0125 \, mol0.10 \, mol \, dm^-3 = 0.125 \, dm^3 = 125 \, cm^3 K_a = 1.8 × 10^-5 \, mol \, dm^-3 pK_a = -(1.8 × 10^-5) = 4.74 (i) Before adding NaOH Weak acid: [H^+] = sqrt(K_a C) Substitute: [H^+] = sqrt(1.8 × 10^-5) × 0.50 = sqrt(9.0 × 10^-6) = 3.0 × 10^-3 \, mol \, dm^-3 pH = -(3.0 × 10^-3) = 2.52 2.52 (ii) After adding 10 \, cm^3 NaOH Moles NaOH added: n = 0.10 \, mol \, dm^-3 × 0.010 \, dm^3 = 0.0010 \, mol Moles CH_3COOH left: 0.0125 - 0.0010 = 0.0115 \, mol Moles CH_3COO^-: 0.0010 \, mol Total volume: 25 + 10 = 35 \, cm^3 = 0.035 \, dm^3 Buffer: pH = pK_a + ( n_CH_3COO^-n_CH_3COOH ) Ratio: (0.0010)/(0.0115) = 0.087 (0.087) = -1.06 pH = 4.74 + (-1.06) = 3.68 3.68 (iii) At equivalence point (125 \, cm^3 NaOH) Moles CH_3COO^- = 0.0125 \, mol Total volume: 25 + 125 = 150 \, cm^3 = 0.150 \, dm^3 C_b = (0.0125)/(0.150) = 0.0833 \, mol \, dm^-3 K_b = 1.0 × 10^-141.8 × 10^-5 = 5.56 × 10^-10 \, mol \, dm^-3 Weak base: [OH^-] = sqrt(K_b C_b) Substitute: [OH^-] = sqrt(5.56 × 10^-10) × 0.0833 = sqrt(4.63 × 10^-11) = 6.8 × 10^-6 \, mol \, dm^-3 pOH = -(6.8 × 10^-6) = 5.17 pH = 14.00 - 5.17 = 8.83 8.83 (iv) After adding 30 \, cm^3 NaOH Moles NaOH added: 0.10 × 0.030 = 0.0030 \, mol Moles CH_3COOH left: 0.0125 - 0.0030 = 0.0095 \, mol Moles CH_3COO^-: 0.0030 \, mol Total volume: 25 + 30 = 55 \, cm^3 = 0.055 \, dm^3 Buffer: n_CH_3COO^-n_CH_3COOH = (0.0030)/(0.0095) = 0.316 (0.316) = -0.50 pH = 4.74 + (-0.50) = 4.24 4.24