The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous is (Assume complete neutralization) [2023]

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Solution

Q.
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) [2023]

1 2.5 mL

2 7.5 mL

3 5.0 mL

4 10.0 mL

Ans.
(4)

${Ba(OH)}_{2} + 2 HBr \to {BaBr}_{2} + 2 H_{2} O$

$mmol$ $0.1$

$Required mmol of HBr = 0.2 = 0.02 \times V_{mL}$

$V_{mL} = 10 mL = 1.362 L = 1362 mL$

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