Cal is titrating 57.5 mL of 0.363 M HBr with 0.315 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalence point?

33.1mL of 0.315M Ba(OH) are required to reach the equivalence point.

Explanation:

The reaction of HBr with Ba(OH)₂ is:

2HBr + Ba(OH)₂ →  BaBr₂ + 2H₂O

That means to find the theoretical volume Cal need to add we need to find moles of HBr in solution, convert these to moles of Ba(OH)₂ and with its molar concentration find the volume, thus:

Moles HBr:

57.5mL = 0.0575L * (0.363mol/L) = 0.0209 moles HBr

Moles Ba(OH)₂:

0.0209 moles HBr * (1mol Ba(OH)₂ / 2 mol HBr) = 0.0104 moles Ba(OH)₂

Volume:

0.0104 moles Ba(OH)₂ * (1L / 0.315mol) = 0.0331L =

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