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Calculate final pH of 25 ml buffer 0.1 M CH3COOH/CH3COO, pH=3.50, after the addition of 1.00 ml of 0.1 M HCl.
3.50=4.75+log(base/3.16x10-4)
Base = 1.78x10-5 M
Acid = 3.16x10-4 M
Mole base = 0.000712 - 0.0001 = 0.000612
Mole Acid = 0.01264 +0.0001 = 0.01274
Mole Acid added = 0.0001 Mole
ph=4.75+log(0.000612/0.01274)
ph = 2.43
Is this correct?
3.50=4.75+log(base/3.16x10-4)
Base = 1.78x10-5 M
Acid = 3.16x10-4 M
Mole base = 0.000712 - 0.0001 = 0.000612
Mole Acid = 0.01264 +0.0001 = 0.01274
Mole Acid added = 0.0001 Mole
ph=4.75+log(0.000612/0.01274)
ph = 2.43
Is this correct?