Question

In: Chemistry

A 25.00 mL solution containing 0.3000 M Na2C2O4 (sodium oxalate) was titrated with 0.1500 M Ca(NO3)2...

A 25.00 mL solution containing 0.3000 M Na2C2O4 (sodium oxalate) was titrated with 0.1500 M Ca(NO3)2 to precipitate calcium oxalate: Use Excel to draw a graph of pCa vs. VCa from 0 to 75 mL, using a suitable resolution to show a sharp endpoint. Label the axes and show where the end point lies.

Solutions

Expert Solution

Values:

About Na2C2O4:

Volume solution: 25 mL (0.025L)

Molarity: 0.3000 M

Fw: 134 g/mol

About Ca(NO3)2:

Volume solution: x mL

Molarity: 0.1500 M

Fw: 164 g/mol

IF:

THEN:

What volume of solution of 0.15M calcium nitrate is needed to fully holder 25 mL of a solution of sodium oxalate 0.30M ?

First: the number of grams is known in 25mL of a solution of sodium oxalate 0.3000M

Second: by reaction stoichiometry the amount of calcium nitrate consumed for reacting with the mass of sodium oxalate defined holder.

Third: the amount defined by volume of a solution of 0.1500M calcium nitrate solution used as titrant .

Now we get the title , which is nothing more than the equivalent of the amount of titrating solution and the amount of product generated .

Now, use Excel to draw a graph of pCa vs. VCa from 0 to 75 mL, using a suitable resolution to show a sharp endpoint. Label the axes and show where the end point lines.


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