I am having issues getting this calculations for my titrations: My solution was

Question

I am having issues getting this calculations for my titrations:

My solution was calcium chloride from solid calcium chloride dihydrate and distilled water

we titrate 0.02 M calcium chloride solution

I used 0.2938 g solid calcium chloride dihydrate

It was taken 10.0 mL of the solution.

and 70 mL of the titrate was used

my titrate was EDTA4-

and we use a buffer with ethanolamine,magnesium,sulfatem and EDTAH4

I need to find from each titration:

1)         Moles of Ca2+ in 10.00 mL

2)         Moles of EDTA4- in 10.00 mL

3)         [EDTA4-]

3a) average [EDTA4-]

4)         %RSD for [EDTA4-]

Even just a sample one will be awasome thank you.

Please be specific with the equations! I want to learn how you got to the answer

Starting Ending Trial 1 10.0 mL 32.4mL Trial 2 0 mL 24.0mL Trial 3 0 mL 23.5mL

Explanation / Answer

1) 1M solution = 1 mol sample in 1 L of solvent

the molecular weight of CaCl2 -2H2O is = 40 + (2 * 35.5) + (2 * (( 2* 1) + 16)) = 40 + 71 + (2 * 18) = 40 + 71 + 36 = 147 g/mol

1000 ml 1M calcium chloride solution contains 147 g of calcium cchloride

1 ml 1M calcium chloride solution contains 147/1000 g of calcium cchloride

10.0 ml 1M calcium chloride solution contains (147/1000) * 10.0 g of calcium cchloride

10.0 ml 0.02 M calcium chloride solution contains (147 / 1000) * 10.0 * 0.02 g = 0.0294 g of calcium cchloride

147 g of calcium chloride contains 1 mol of Ca2+ ion

Therefore, 0.0294 g of calcium chloride contains 0.0294/147 mol = 0.0002 mol of Ca2+ ion

Therefore, Moles of Ca2+ ion in 10.0 ml is = 0.0002 mol

3) From the titration table, the average volume of EDTA4- solution is = (trial 1 volume + trial 2 volume + trial 3 volume)/ 3 = ((32.4 - 10.0) + (24.0 - 0 ) + (23.5 - 0))/3 ml = (22.4 + 24.0 + 23.5)/3 ml = 69.9/3 ml = 23.3 ml

For the titration, V1S1 = V2S2

=> S2 = (V1S1 ) / V2 = (10.0 ml * 0.02 M) / 23.3 ml = 0.0086 M

Therefore, average [EDTA4-] is 0.0086 M

2) molar mass of EDTA is 292 g/mol

1000 ml 1M EDTA solution contains 1mol of EDTA4-

1 ml 1M EDTA solution contains 1/ 1000 mol of EDTA4-

10.0 ml 1M EDTA solution contains (1/1000) * 10.0 mol of EDTA4-

10.0 ml 0.0086 M EDTA solution contains (1 / 1000) * 10.0 * 0.0086 mol = 0.000086 mol of EDTA4-

Therefore, Moles of EDTA4- in 10.00 ml is = 0.000086 mol

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