The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)2 is (Assume complete neutralization) (1) 2.5 mL (2) 5.0 mL (1) 10.0 mL (1) 7.5 mL

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Question

Prashant Kotian · Accepted Answer

Correct answer: .. Ba(OH)2 + 2HBr  BaBr2 + 2H2O mmol 0.1 required mmol of HBr = 0.2 = 0.02 × Vml Vml = 10

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