[kc1985]

 Hi Everyone,

I wonder if anybody could offer some help on a basic calculation which is causing us some difficulty. I'm sure it is a simple calculation.

We need to work out what theoretical volume of 0.015M EDTA corresponds to 1mg/L Calcium Carbonate in 100mL of sample. We have come up with the following, could somebody verify this or correct the calculation?

For 100ml of 1mg/L CaCO3 (100.0869)  = 0.00000999mol/L
Mole = (volume x concentration) / 1000
            = 100 x 0.00000999 / 1000
            = 0.000000999 mol
Equivalent EDTA
0.000000999 = v x 0.015 / 1000
                      V = 0.0666ml
                          = 66.6uL

Similarly we need to calculate what volume of 0.035M EDTA corresponds to 20mg/L of calcium carbonate in 5.5ml of sample:
 
For 100ml of 20mg/L CaCO3 = 0.0001998 mol/L
Mole = (volume x concentration)/ 1000
            = 100 x 0.0001998 / 1000
            = 0.00001998
Equivalent EDTA
0.00001998 = v x 0.035 / 1000
                    v = 0.5709ml
                       = 570.9uL
For 5.5ml (tube is approx. 5.5 ml)=31.3995µL
Thanks for any help

[Hunter2]

Easier calculation:

1 mg/l  = 0,1 mg/0,1 l = 0,1 mg/100 ml  => 0,001 mg/ml

1 CaCO₃ correspond to 1 EDTA

So it means for 0,015 M = 0,015 mol/l = 0,015 mmol/ml => 0,001mg = 0,0666 ml = 66,6 µl

correct.

With the same method you would also get the 32 µl  for the second question.

[kc1985]

Thanks for your *delete me*!

[kc1985]

I wonder if anybody could provide some more help on this.
Would 0.015M equate to 0.015N?

[kc1985]

Hi,

Could anybody provide the molarity of 0.015N EDTA and an explanation as to why this is so?

Thanks in advance,

KC

[Dan]

It depends on the application. The equivalence factor is not the same for all reactions, which is why I think normality as a measure of concentration should be banned.

http://en.wikipedia.org/wiki/Equivalent_concentration

[Hunter2]

In the case of EDTA yes.

[kc1985]

I agree, molarity would be so much more straight forward for what we are doing:

We are titrating the 0.015N solution against 100ml of 20mg/L CaCO3 solution and want to know what molarity the EDTA corresponds to.

[Dan]

Start with a balanced chemical equation.

[kc1985]

Our problem is that we are titrating against an solution of 300mg/L Hardness as CaCO3. The manufacturer of the test kit recommends using a plastic dropper to add 0.035N EDTA and 1 drop corresponds to 20mg/L so at the end point we multiply the number of drops by 20 to determine the concentration in CaCO3 of the solution.

We find if we follow the manufacturers method then the end point of the titration is after approximately 15 drops so 15x20=300mg/L (pretty accurate this way). We have weighed a drop to be approx 63μL so the total volume to reach the end point for the solution is 945μL.

However if we use the theoretical volume of EDTA needed for 1mg/L of CaCO3 (31.4μL from the above) we need twice the number of drops to reach the end point. So we need 30 drops at 31.4μL per drop which is still 942μL (so the total volume is pretty close) but since  1 drop=20mg/L then the hardness of this solution is 30x20=600mg/L. Twice the actual hardness.

We thought that this could be explained if 0.035N halves when converting to M (ie 0.035N=0.0175M) such as the case with other acids such as H2SO4. This would explain our error in the above calculation as the theoretical value above would become 62.8μL agreeing with the volume of 1 drop from the test kit manufacturer of the test kit.

But if 0.035N=0.035M then we are at a loss. Any ideas would be much appreciated.

[Hunter2]

The normality and molarity of EDTA is the same

http://www.google.fr/imgres?imgurl=http://www.lookchem.com/300w/2010/0617/12264-18-5.jpg&imgrefurl=http://www.lookchem.com/Calcium-edta-complex/&h=363&w=341&tbnid=PZfcKvBwT55fUM:&zoom=1&tbnh=93&tbnw=87&usg=__eSIccIFoSVW5zdyKUzdHsD0Qk2g=&docid=IM_FEcK6DOUqgM&client=firefox-a

One molecule of EDTA catch one Calcium. So you have to do more trouble shooting now.

Something wrong in your calculation as well, if 63 µl = 20 mg/l  then 1 mg/l = 3,14 µl not 31,4 µl.

So to get 300 mg/l determined you need 300 x 3,14 = 942 µl  This devided by 63 µl gives 14,9 drops. Everything is fine.

[kc1985]

Thanks Hunter,

I have tracked down a link to the manufacturers website with http://www.hach.com/kb-parameterfaq
In the hardness section there is mention that 0.02N EDTA= 0.01M EDTA. So if they can explain how they have come to this conclusion it would help.
Like you said the overall volume are both the same but what is throwing us is how they mention that 1 drop of EDTA is equal to 20mg/L of CaCO3 which means if we count drops rather than overall volume as recommended we get twice the concentration.

[Hunter2]

You mean this one:
How are the multipliers derived for the hardness buret titration?

Using Normality (equivalents/L):

At endpoint, equivalents of EDTA = equivalents of CaCO3
Equivalents/L of CaCO3 = conc (g/L)/equiv. weight
The equivalent weight of CaCO3 = ½ MW CaCO3 = 50 g/equivalent (In a complex-formation reaction, the equiv weight is that weight which reacts with or provides one mole of the reacting cation if it is univalent, or one-half mole if divalent, one-third mole if trivalent, etc.)
Each mL EDTA used = 0.00002 equivalents:

0.001 L x 0.02 equiv. EDTA/L = 0.00002 equiv. EDTA = equiv. CaCO3
Convert to mg CaCO3and divide by sample volume to get sample concentration:
0.00002 equiv. CaCO3 x 50 g/equiv. = 0.001 g or 1 mg CaCO3

Example: Sample volume is 50 mL: 1 mg/0.05 L = 20 mg/L
Therefore 1 mL of 0.02 N EDTA = 20 mg/L CaCO3 if 50 mL sample volume is taken; multiplier is 20 Using Molarity and a 0.02 N EDTA solution:

0.020 N EDTA = 0.01 M EDTA
At endpoint, moles EDTA = moles Ca or CaCO3
Each mL EDTA used = 0.00001 moles CaCO3:

0.001 L x 0.01 mol EDTA/L = 0.00001 moles EDTA = moles Ca or CaCO3
Convert to mg CaCO3 and divide by sample volume to get sample concentration:

0.00001 mol CaCO3 x 100 g CaCO3/mol = 0.001 g or 1 mg CaCO3

Example: Sample volume is 50 mL: 1 mg/0.05 L = 20 mg/L
Therefore 1 mL 0.02 N EDTA = 20 mg/L CaCO3 if 50 mL sample volume is taken; multiplier is 20



The thing gets complicated, because that Disodium-EDTA is bivalent like Calcium it is. So the ratio is still 1:1. So the describtion is not correct.

[kc1985]

Yes that's the one.

We are stumped by the manufacturers claims although there method works when I have performed blind tests on  different analysts.

We are sure that a 300mg/L takes approximately 940μL of 0.035N to reach end point, the next thing we need to decide is what volume corresponds to 20mg/L of CaCO3 in 5.5ml of solution, going by the manufacturers approach then 300/20=15 and 940/15=63μL whereas the calculation at the start of this post would suggest that it should be 32μL which corresponds to 20mg/L CaCO3.

So the question is are the manufacturers correct in determining 1 drop/63uL of 0.035N to equate to 20mg/L of CaCO3 or is the above calculation correct to determine 32uL to correspond to 20mg/L.

Whatever we choose is going to have an effect on our method, ie we use 20 × n= mg/L CaCO3 with n being the number of drops. I'm thinking we have miscalculated at the beginning in some way as 32μL being considered to correspond to 20mg/L does not work whereas applying a 2:1 N:M ratio in the calculation above would give approx. 64uL which matches our manufacturers advice.

I have emailed them and hopefully they can shed some light on why they consider one drop of 0.035N EDTA to be equivalent to 20mg/L CaCO3. I am wondering if what we are using is pure 0.035N EDTA, we are going to prepare some and see if when using this does the 15 drops using a pipette to add 32μL per drop reach the end point.

[kc1985]

I found this post on a previous subject. And it would seem that for water hardness it has been adopted that the normality of EDTA is twice the molarity:

LabChem Catalog # LC13850-1
Product description:  EDTA, 0.01M (0.02N), Titrant for Water Hardness, (1mL = 1mg CaCO3)

I welcome you to check out their fine product line for yourself...

Again, here is the widely accepted definition of normality:

Normality = molarity x n (where n = the number of protons exchanged in a reaction).
 
In short, while there is a relationship between the normality of a solution and the molarity of a solution, the normality can only be determined by examining the reaction, determining the proton exchange and multiplying molarity by that number.

Returning to the question at hand, the hardness test:

Ca+2 + EDTA-4  CaEDTA-2

We can agree that 1 mole of EDTA neutralizes 1 mole of Calcium as CaCO3 (it’s a 1:1 equivalence).  Here is where normality confuses things, since EDTA molarity is not equivalent to normality in this case.  It is very common to see the hardness calculation based on the normality of EDTA, even though in this instance molarity would be much more straightforward since we simply want to convert back to hardness in mg/L.  In the hardness reaction each EDTA molecule exchanges two protons to chelate one divalent calcium cat ion (Ca++).  So, even though there is a 1:1 molecular equivalence using molarity, the EDTA normality is, by definition, twice the molarity for the hardness test.    Again, given that EDTA donates two protons per molecule in the balanced equation for hardness determination, you must multiply EDTA molarity by two to get the correct normality.   For this very reason you don't see standardized EDTA available for purchase as a “normal” solution (unless specified "for hardness" as in the example I gave above).  EDTA can potentially donate up to four protons, depending on the reaction.  The textbook definition of “normality” means that a standardized “normal” solution of EDTA could be very misleading.

The EDTA hardness calculation using a “molar” solution is:

Hardness as CaCO3 mg/L =    Vol (mL) titrant x EDTA (Mol) x 100g/mol CaCO3 x 1,000mg/g   
                                                                                Vol (mL) Sample


This simplifies to:

Hardness as CaCO3 mg/L =   V1 x M1 X 100,000 / V2


To convert the calculation to use a “normal” solution, N/2 must be substituted for M1. Factoring out the 2 in the N/2 term with 100,000, you get:

Hardness as CaCO3 mg/L =   V1 x N X 50,000 / V2