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2 HC₂H₂O₂ + Ba(OH)₂

Ba(C₂H₂O₂)₂
+ 2 H₂O
A sample of barium hydroxide (Ba(OH)₂) 0.67 M is titrated with acetic acid (HC₂H₂O₂) 1.2 M. If 55.00 mL of acetic
acid were required, what was the volume of the sample of barium hydroxide?
153.54 mL Ba(OH)₂
(magenta)
98.51 mL Ba(OH)2
(red)
49.25 mL Ba(OH)₂
(blue)

User Ilyssis
by
8.3k points

1 Answer

2 votes

The balanced chemical equation for the reaction between acetic acid and barium hydroxide is:

2 HC₂H₂O₂ + Ba(OH)₂ → Ba(C₂H₂O₂)₂ + 2 H₂O

From the balanced equation, we can see that 2 moles of acetic acid react with 1 mole of barium hydroxide to produce 1 mole of barium acetate and 2 moles of water.

The number of moles of acetic acid used in the titration can be calculated as follows:

moles of HC₂H₂O₂ = Molarity × volume in liters

moles of HC₂H₂O₂ = 1.2 M × (55.00 mL / 1000 mL/ L)

moles of HC₂H₂O₂ = 0.066 moles

From the balanced equation, we know that 2 moles of acetic acid react with 1 mole of barium hydroxide. Therefore, the number of moles of barium hydroxide present in the titration can be calculated as:

moles of Ba(OH)₂ = 0.066 moles / 2

moles of Ba(OH)₂ = 0.033 moles

The molarity of the barium hydroxide solution can be calculated as:

Molarity = moles / volume in liters

We rearrange this equation to solve for the volume:

volume in liters = moles / Molarity

volume in liters = 0.033 moles / 0.67 M

volume in liters = 0.04925 L

Finally, we convert the volume to milliliters:

volume in mL = 0.04925 L × 1000 mL/L

volume in mL = 49.25 mL

Therefore, the volume of the sample of barium hydroxide used in the titration is 49.25 mL. The answer is blue.

User Markko Paas
by
8.0k points