Cal is titrating 57.7 mL of 0.311 M HBr with 0.304 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalence point?

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.

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