To determine the Ksp of barium oxalate at 12°C, a 100.0 mL saturated solution was prepared by adding the solid in distilled water maintained at 12°C until it no longer dissolves. The solution was filtered and from this, a 10.0 mL aliquot of the filtrate was drawn. It was found out that 10.70 mL of 5.60 x 104 M HCI was used to reach the endpoint. Given that two moles of HCl react for every mole of dissolved oxalate during the titration, determine the amount of dissolved oxalate in the titration flask (in mol/L). Do not include the unit in your answer below. For those who will be using scientific notation, please follow the format below: Example: 0.00487 -> 4.87E-3 58000 -> 5.8E+4

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To determine the Kgp of barium oxalate at 12°C, a 100.0 mL saturated solution was prepared by adding the
solid in distilled water maintained at 12°C until it no longer dissolves. The solution was filtered and from this, a
10.0 mL aliquot of the filtrate was drawn. It was found out that 10.70 mL of 5.60 x 104 M HCI was used to
reach the endpoint.
Given that two moles of HCl react for every mole of dissolved oxalate during the titration, determine the
amount of dissolved oxalate in the titration flask (in mol/L). Do not include the unit in your answer below.
For those who will be using scientific notation, please follow the format below:
Example:
0.00487 -> 4.87E-3
58000 -> 5.8E+4
Transcribed Image Text:To determine the Kgp of barium oxalate at 12°C, a 100.0 mL saturated solution was prepared by adding the solid in distilled water maintained at 12°C until it no longer dissolves. The solution was filtered and from this, a 10.0 mL aliquot of the filtrate was drawn. It was found out that 10.70 mL of 5.60 x 104 M HCI was used to reach the endpoint. Given that two moles of HCl react for every mole of dissolved oxalate during the titration, determine the amount of dissolved oxalate in the titration flask (in mol/L). Do not include the unit in your answer below. For those who will be using scientific notation, please follow the format below: Example: 0.00487 -> 4.87E-3 58000 -> 5.8E+4
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