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) 5.0 mL
(2) 10.0 mL
(3) 2.5 mL
(4) 7.5 mL

Answer: Option 2
Explanation: To determine the volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)2, we need to consider the balanced chemical equation for the reaction between HBr and Ba(OH)2:

2HBr + Ba(OH)2 → BaBr2 + 2H2O

From the balanced equation, we can see that the stoichiometric ratio between HBr and Ba(OH)2 is 2:1. This means that 2 moles of HBr react with 1 mole of Ba(OH)2.

First, let’s determine the number of moles of Ba(OH)2 in 10.0 mL of 0.01 M solution:
Moles of Ba(OH)2 = concentration (M) × volume (L)
Moles of Ba(OH)2 = 0.01 M × 0.010 L = 0.0001 moles

According to the stoichiometry, 1 mole of Ba(OH)2 requires 2 moles of HBr for complete neutralization. Therefore, 0.0001 moles of Ba(OH)2 will require 2 × 0.0001 = 0.0002 moles of HBr.

Now, let’s calculate the volume of 0.02 M HBr needed to provide 0.0002 moles:
Volume (L) = moles / concentration (M)
Volume (L) = 0.0002 moles / 0.02 M = 0.01 L = 10.0 mL

Therefore, the volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)2 is 10.0 mL.

Hence, the correct option is (2) 10.0 mL.


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