[xo1991]

hey guys, I have come for some assistance with a couple problems I'm confused about. I'm in gen chem. this is for a pre-lab and isn't anything we've reviewed in lecture.


1) A preliminary titration showed that a dilution of river water is required to conserve the total amount of EDTA consumed in your titrations. In a preliminary trial, a 2-mL aliquot of estuarine water required 25.00 mL of EDTA to reach the end point. Calculate how much EDTA a 10-mL aliquot of estuarine water would require.

The answer I got was 125 mL EDTA. Not sure if i did it the right way; I just set up proportions. Is it more complicated than that because of dilutions?

2) Based on the information and answer determined in Question 1, calculate a dilution factor that will allow a titration of a 10-mL aliquot of diluted river water to consume EDTA in the range of 15 to 40 mL. Include how much EDTA would be used if such a dilution were used.

I don't really understand this question or know where to start


3) When a 10.00 mL aliquot of hard water diluted by a factor of 4 is titrated, 38.25 mL of 1.002 E-03 M EDTA (0.001002 M) is required to reach the end point. The balanced equation indicates that one mole of EDTA consumes one mole of calcium. What is the total concentration of calcium carbonate (in mg/L) contained in the hard water sample? The formula weight of calcium carbonate is 100.071 g/mole.


I used M=mol/volume to find the moles of EDTA(3.829E-5 mol). This was part of the unit we just completed. I dont know where to go from there/what to do with the first part of the question (diluted by a factor of 4???)


any advice would help
thanks

[UG]

The answer to the first one looks ok, I don't think there is anything else involved.

2) Based on the information and answer determined in Question 1, calculate a dilution factor that will allow a titration of a 10-mL aliquot of diluted river water to consume EDTA in the range of 15 to 40 mL. Include how much EDTA would be used if such a dilution were used.

I don't really understand this question or know where to start

Taking the numbers from Q1, a 2 mL aliquot of estuarine water required 25.00 mL of EDTA to reach the end point. So you could dilute this solution by a factor of 5 (ie, increase the volume by 5 times) to 10 mL and since the number of moles of ions in the solution doesn't change, you will still need 25.00 mL of EDTA to neutralise this diluted solution. Am I making sense?

3) When a 10.00 mL aliquot of hard water diluted by a factor of 4 is titrated, 38.25 mL of 1.002 E-03 M EDTA (0.001002 M) is required to reach the end point. The balanced equation indicates that one mole of EDTA consumes one mole of calcium. What is the total concentration of calcium carbonate (in mg/L) contained in the hard water sample? The formula weight of calcium carbonate is 100.071 g/mole.


I used M=mol/volume to find the moles of EDTA(3.829E-5 mol). This was part of the unit we just completed. I dont know where to go from there/what to do with the first part of the question (diluted by a factor of 4???)

So the number you calculated is the number of moles of EDTA reacted. One mole of EDTA consumes one mole of calcium ions so you already got number of moles of calcium as well. When the solution is diluted by a factor of 4 (ie, the volume increases 4 times) the concentration of the species in solution decreases by a factor of four. So you can work out original number of moles of calcium in the original solution and then find the mass of calcium carbonate using this.

[xo1991]

Ok. I kind of see what you're saying for number three. With the value I found, I calculated that amount of moles to grams.  I got like .0038.  Then I multiplied by 1000 to convert to milligrams. then divide by 4 for the dilution?

ultimately I got .96mg/L  is that right?

also I still don't really get number two. mind rewording it a little bit?

thanks a lot

[UG]

The wording of the question confuses me a bit, "When a 10.00 mL aliquot of hard water diluted by a factor of 4..." seems to me that the original volume of the solution was 2.500 mL and they diluted it by adding water until it was 10.00 mL. So you have worked out the number of moles of calcium, you can work out the original concentration by (moles of calcium)/(original volume). You can then convert this into mg/L.
But the question might actually mean that they took a 10.00 mL sample of hard water and diluted it to 40.00 mL in which case the answer will turn out differently. My gut feel is that the first way is the right one though.

[UG]

also I still don't really get number two. mind rewording it a little bit?

Ok lets see, using the same numbers as before: a 2-mL aliquot of estuarine water required 25.00 mL of EDTA to reach the end point.
Now you want a 10.00 mL diluted solution to consume EDTA in the range 15 to 40 mL.
Lets say you had a 5-mL aliquot of estuarine water and you diluted by a factor of 2 to give a 10.00 mL solution. But using proportions, the 5-mL aliquot would need 62.5 mL of EDTA so that is outside the range so you can't have dilution factor of 2. Lets say then you had a 4-mL aliquot and you diluted by a factor of 2.5 to give a 10.00 mL solution. Again, using proportions, you would need 50.00 mL of EDTA so thats not allowed. Lets say now you have a 3-mL aliquot and you diluted by a factor of 3.333 to 10.00 mL, using proportions, you would need 37.5 mL of EDTA so this is an 'allowable' dilution factor. And then you had a 2-mL aliquot and dilution factor of 5 to get 10.00 mL, this would consume 25.00 mL EDTA so again, it is an 'allowable' dilution factor.   

[xo1991]

thanks a lot. i finally understand what that whole dilution thing means.

[Borek]

calculate a dilution factor that will allow a titration of a 10-mL aliquot of diluted river water to consume EDTA in the range of 15 to 40 mL.

The idea is that you dilute the original sample, then you take 10 mL of the diluted sample. This is equivalent to taking 10mL/dilution factor of the original sample (yields the same numbers UG posted).