Answer:
Step-by-step explanation:
[H+] = 0.1x0.03x2 + 0.2x0.03 = 0.012 mol
[OH-] = 0.001x0.05x2 = 0.0001 mol
=> [H+] dư = 0.012 - 0.0001 =0.0119 mol
pH = -log[H+] = 1.92
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