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An analyte containing 0.12 L of a 0.025 M solution of a strong acid is titrated with a 0.220 M solution of a strong base. For

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No of millimoles of auid = 0-12X1000 X0:025 - 3 No of millimoles of base = 0.920 X 3 = 0.660 No of moles of and > No of moles

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