A 50.0 ml water sample required 5.14ml of 0.0100 M Na2H2Y to reach the endpoint for a titration procedure.?

Question

Calculate the moles of hardening ions in the water sample. Assuming the hardness is due to exclusively to CaCO3, express the hardness concentration in mg CaCO3/L sample.

I do not know how to set this up. My book suggests with an example sort of like this one to use the mv=mv equation. Should I use this or another equation? Please help

thank you

the Na2H2Y i do not know how to put subscripts on this page so I just used caps

I just used normal numbers not subscripts

It is not an acid per say it is disodium ethylenediaminetetraacetate or Na2H2Y the numbers are supposed to be subscripts

here is the procedure:

1. measure about .5 g of Na2H2Y*2H2O mm=372.24g/mol transfer it to a 250 ml volumetric flask containing 100 ml of deionized water to dissolve. dilute to the mark of the volumetric flask with deionized water.
2. prepare a buret for titration. rinse the buret with the Na2H2Y solution and then fill. record the volume of the titrant.
3. obtain about 80 ml of the standard Ca2+ solution from the reagent shelf and record the molar concetration. pipet 25 ml into a 125 ml erlenmeyer flask, add 1 ml of buffer (pH=10) solution, and 2 drops of EBT indicator.
4. titrate the standard Ca2+ solution with the Na2H2Y titrant; swirl continously. near the endpoint slow the rate of addition to drops, teh last few drops should be added at 3-5 intervals. the solution should change from wine-red to purple to blue. the solution is blue at the endpoint.
5. calculate the molar concetration of the Na2H2Y solution

the ratio is 1:1 stoichiometric ratio and Y is Ytturium elemet

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i got for my answer .001028 but it did not seem right

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very much

Answer 1

hey sport, what's your acid there?

can you clarify? Na2H2Y?
i'll be back to check again in a bit.

so Y is an unknown element?

i'm not sure if you're hooking it up with enough info..
how about some dissociation constant? equilibrium constant? anything?

assuming the question is not expecting you to think about this type of stuff..

can we assume it's a diprotic acid that completely dissociates? [which doesn't make sense, but i guess it's just a concept question]

diprotic, 2 H+ ions per acid molecule.

yeah.. since this question is asking you to deal with a carbonate, that's annoying to calculate considering it exists in that solution as carbon dioxide, carbonate, bicarbonate, and carbonic acid.
So i guess we will assume carbonate is, what.. going to react with two H+'s?

ok, so 1:1 ratio.

yes, mv=mv is legit.

0.0100 M * 0.00514L / 0.0500L = X
X = 0.00103M

oops, it wants it in mg/L

lame

(0.00103 mol/L) * (12.01+16.00*3 + 40.08)g/mol * (1000mg/g)
= 103 mg CaCO3 / L


ballin'

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It is not an acid per say it is disodium ethylenediaminetetraacetate or Na2H2Y the numbers are supposed to be subscripts

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oh, s -hit.
i figured this was a lower level question written by a bad prof or something, haha
i've never done a EDTA hard water problem before, sorry man

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ok, making progress.

i think i've almost figured it out. but also dude, come on, no way Y is the element Y. Y represents EDTA for sure. check the molar mass assuming Y=Yttrium element. it's like 1/3 of what it is supposed to be, ~100g/mol range. also, i think you named the coordination compound wrong. i believe it's disodium dihydrogen ethylenediaminetetraacetate* dihydrate. anyways, doesn't really matter.

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ok, assuming 372.24g/mol is right. i don't really feel like confirming it, and i don't really know if it takes the hydration into account [or even whether it's supposed to or not].

hm, so, maybe it's this simple:

ratio of Na2H2Y and CaCO3 is 1:1,

(0.0100 M * 0.00514L) = 5.14E-5 mol Na2H2Y
= 5.14E-5 mol CaCO3

(5.14E-5 mol) CaCO3 = 0.00514g

0.00514g * 1000(mg/g) / 0.0500L=
= 103 mg CaCO3 / L , preserving significant figures.

what do you think?

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cool, looks like the dude below me got the same answer.

Answer 2

Y here is EDTA and not an element.

The reaction is 1:1 so yes you should use the equation

M1V1 = M2V2 =>
0.01*5.14*10^-3 = M2*50 *10^-3 =>
M2= 0.0514/50 = 0.001028 mole/L
The molecular weight of CaCO3 is MW=100 g/mole

mole=mass/MW => mass= mole*MW, so the concentration
0.001028 mole/L => 0.001028*100 g/L= 0.1028g/L =102.8 mg/L

In the additional details you say that you do a titration to find the exact Na2H2Y concentration, which means that for more accurate results you should use that value instead of 0.0100M.
Let's calculate it. The reaction is the same but now the CaCO3 is known and we determine the Na2H2Y. So we use the same equation

M1'V1'=M2'V2'

M1' will be the true concentration of Na2H2Y
V1' the mL of Na2H2Y that you added to the Ca+2 standard solution in order to titrate it
M2' the concentration of the Ca+2 standard solution
V2' =25 mL (the volume of the standard Ca+2 solution you used).

You don't give us the other numbers (V1', M2') so we can't go on.
However you know them. So plug them in the equation and find M1'. Then use M1' instead of 0.0100 in the equation at the very top of my answer, find the more accurate M2 and then convert it to mg/L.

Answer 3

Coupla things first mate......

1. This an EDTA titration. The Y is shorthand for EDTA and it has nothing to do with Yttrium. Your titrant is the sodium salt of ethylene diamine tetra acetic acid (EDTA).

2. The method you outlined is the standardisation of the prepared EDTA solution against a standard (known concentration) solution of calcium ions. Your problem quotes the strength as 0.01M - you would have calculated molarity from your method.

3. Set up the following relationship assuming H is the hardness in mg/l CaCO3

50 / 1000 x H / 100000 = 5.14 x 0.01 / 1000

Noting 50 ml sample MW of CaCO3 is 100000 mg, 5.14 is the titre in mls and 0.01 M is the EDTA strength.

Solving for H = 102.8 mg/L CaCO3

4. 0.01 M EDTA is commonly used in the water industry as 1 ml of solution is equal to 1 mg of CaCO3. This makes the calculation easy - the hardness is titre (mls) x 1000 / sample volume mls. The result is total hardness in mg/l CaCO3.

Answer 4

if it demands 25.0 ml of one million.0 M NaOH then you definately paintings out what share moles of the hydroxide ion you have, hence you need to transform ml to L so take 25/one thousand = .0.5 M of hydroxide ion, which skill you have 0.0.5 M of the hydronium ion. convert the a hundred ml of HCl into L so which you have 0.one million L, and you comprehend which you have .0.5 moles of H+ on your a hundred ml pattern, situations that by skill of ten to get 100ml --> one million.0 L (what molarity is according to) and you get 0.25M answer of Hydrochloric acid.