diketahui terdapat 100mL larutan yang mengandung CH3COOH 0,01mol dan CH3COONa 0,02mol dan Ka = 10^-5. tentukan: a. ph larutan b. ph larutan bila ditambah 0,001m

pH = pKa + log ([CH3COO^-]/[CH3COOH]) pH = 5 + log (0.1/0.1) pH = 5 setelah penambahan HCl, CH3COO^- + H^+ → CH3COOH mol CH3COOH = 900 * 0.1 + 100 * 0.1 = 100 mmol CH3COO^- = 900 * 0.1 - 100 * 0.1 = 80 mmol pH = pKa + log ([CH3COO^-]/[CH3COOH]) pH = 5 + log (80/100) pH = 4.9