Jawabanmu

2014-04-20T21:10:16+07:00
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
(d).

CH3COOH + OH^- → CH3COO^- + H2O

mol CH3COOH = 100 * 0.01 - 0.001 = 0.999 mmol

CH3COO^- = 100 * 0.02 + 0.001 = 2.001 mmol

pH = pKa + log ([CH3COO^-]/[CH3COOH])

pH = 5 + log ([2.001]/[0.999])

pH = 5.3