What is the pH of a solution that is made of 600 mL of 0.5M acetic acid and 200

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Explanation / Answer

First you have to know the molarity of the weak acid, CH3COOH, and the conjugate base, CH3COO^-. You do this by calculating the moles of each species and then molarity. Initial moles CH3COOH = (0.600 L)(0.5 mol/L) = 0.30 mole CH3COOH(aq) + NaOH(aq) --> CH3COONa(aq) + H2O(l) Moles CH3COO^- produced = moles CH3COOH reacted with NaOH Moles CH3COO^- = (0.200 L)(0.5 mol/L) = 0.10 mole Final moles CH3COOH = 0.30 mole - 0.10 mole = 0.20 mole CH3COOH Final moles CH3COO^- = 0.10 mole Final molarity CH3COOH = (0.20 mol)/(0.80 L) = 0.250 M Final molarity CH3COO^- = (0.10 mol)/(0.80 L) = 0.125 M Now, use the Henderson-Hasselbalch equation to solve the problem. pH = pKa + log[CH3COO^-]/[CH3COOH] pKa = -logKa = -log(1.8 x 10^-5) = 4.74 pH = 4.74 + log(0.125)/(0.250) = 4.74 + (- 0.30) = 4.44

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