Answer:
118.06 mL
Explanation:
The neutralization reaction between HBr (acid) and Ba(OH)₂ (base) is the following:
2HBr + Ba(OH)₂ → BaBr₂ + 2H₂O
According to the equation, 2 moles of HBr react with 1 mol Ba(OH)₂. Thus, at the equivalence point the moles of acid and base react completely:
2 moles HBr = 1 mol Ba(OH)₂
We can replace the moles by the product of the molar concentration (M) and volume (V):
2 x (M HBr) x (V HBr) = M Ba(OH)₂ x V Ba(OH)₂
Now, we introduce the data in the equation to calculate the volume in mL of Ba(OH)₂:
V Ba(OH)₂ = (2 x (M HBr) x (V HBr))/M Ba(OH)₂
= (2 x 0.311 M x 57.7 mL)/(0.304 M)
= 118.06 mL
Therefore, 118 mL of Ba(OH)₂ are needed.
