60mL of 0.1M CH3COOH is titrated with 0.2M NaOh, what will the pH equal at these

Question

60mL of 0.1M CH3COOH is titrated with 0.2M NaOh, what will the pH equal at these three points: 1. Initial, before adding NaOH 2. 15mL of NaOH is added 3. 30mL of NaOH is added

Explanation / Answer

find moles: 0.025 L of 0.2 mol / Litre CH3COOH = 0.0050 moles of acid 0.030 L of 0.1 mol / Litre NaOH = 0.0030 moles of base by the equation: 1 CH3COOH & 1 NaOH --> 1 Na CH3CO2 & 1 H2O 0.0030 moles of base reacts with an equal number of moles of acid = 0.0030 moles of acid so 0.0050 moles of acid - 0.0030 moles of acid reacted = 0.0020 moles of acid remain by the equation: 1 CH3COOH & 1 NaOH --> 1 Na CH3CO2 & 1 H2O 0.0030 mol base produces an equal number of mol acetate ion = 0.0030 moles of (CH3CO2)- ion by the dissociation: CH3COOH H+ & (CH3CO2)- Ka = [H+] [(CH3CO2)-] / {CH3COOH ] 1.8 e-5 = [H+] [(0.0030] / [0.0020] [H+] = 1.2 e-5 Molar pH = - log of 1.2 e-5 Molar pH = 4.92

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