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Difficult H2O is H2O incorporating dissolved minerals. Calcium and Mg ions dissolve in it as the H2O seeps through the land. It increases the possibility of the formation of lime graduated table and hence reduces the efficiency of the cleansing processes. This occurs because the heat transportation from the component to the H2O is reduced by the lime graduated table and makes the contraption less efficient. Many H2O softener merchandises are available for consumers such as calgon, which are believed to cut down the consequence of the difficult H2O on contraptions. This is bend would better the cleansing procedure. However the procedure is equivocal, because we are diffident whether the calgon removes or masks the consequence of the dissolved ions. Water hardness is measured in comparing to CaCO3 ppm of H2O. This is because CaCO3 is a common precipitate and the tabular array below shows how different degree of hardness is categrorized.

My preliminary research showed that calgon affects the composing of the H2O by softening it.[ 1 ]This is done by a procedure called ion exchange where the ion money changer, calgon, removes the Ca2+ and Mg2+ from the difficult H2O and replaces them with exchangeable Na+ ions.[ 2 ]We discuss about the Ca2+ and Mg2+ ions when we consider difficult H2O because these are the two ions with the highest concentration that are responsible for the H2O hardness. Every ion that is removed from the H2O is replaced by an tantamount sum of another ionic coinage[ 3 ]. Therefore two Na+ ions replace either one Ca2+ or Mg2+ ion. There are besides other procedures that occur due to the chemicals present in calgon that soften the H2O. This is why calgon is really effectual, because many different chemicals at the same time soften H2O.

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The degrees of Ca2+ and Mg2+ ions can be determined by titration with ethylenediaminetetraacetic acid[ 4 ]( EDTA ) which is a chelating agent and a weak acid.

These are reversible reactions.Calcium ions form composites with the Eriochrome black T index, when the pH 10 buffer is added to it and it become a wine ruddy coloring material. This happens when [ H-ErioT ] 2- ions bonds with either Ca2+ or Mg2+ ions.

Ca2+ ( aq ) + [ H-ErioT ] 2- ( aq ) + H2O ( cubic decimeter ) [ Ca-ErioT ] – ( aq ) + H3O+ ( aq )

Mg2+ ( aq ) + [ H-ErioT ] 2- ( aq ) + H2O ( cubic decimeter ) [ Mg-ErioT ] – ( aq ) + H3O+ ( aq )

EDTA has pH 10 buffer added to it because it allows the EDTA to dissociate to organize [ EDTA ] 4. This disassociated signifier means it has the maximal available sum of tetracarboxylate ion, which has plentifulness of negatrons and six available adhering sites. The anion, [ EDTA ] 4- , wraps itself around the Ca and Mg ions so that the six braces of negatrons are shared with the metal ion.

The coloring material alteration from the ruddy vino to a lasting blue, has a color alteration of ephemeral purple in the center. The lasting blue would be the terminal point of the titration and this would intend that all the Ca2+ and Mg2+ ions have either been removed or dissolved so that they have no consequence on the H2O. In order for this to happen, the two reactions at a lower place would hold to happen foremost where the ions in the difficult H2O bond with the EDTA solution.

Ca2+ ( aq ) + [ EDTA ] 4- ( aq ) [ Ca- EDTA ] 2- ( aq )

Mg2+ ( aq ) + [ EDTA ] 4- ( aq ) [ Mg- EDTA ] 2- ( aq )

The terminal point is when the complex [ Mg-ErioT ] – is wholly broken up by the anion [ EDTA ] 4- let go ofing bluish [ H-ErioT ] 2- ion. The concluding reaction is shown with the equation below:

[ Mg-ErioT ] – ( aq ) + [ EDTA ] 4- ( aq ) + H3O+ ( aq ) [ Mg- EDTA ] 2- ( aq ) + [ H-ErioT ] 2- ( aq ) + H2O ( cubic decimeter )[ 5 ]

Purpose

To happen to what extent:

Calgon affects the composing of difficult H2O

To work out if it affects it by taking or dissembling fade outing the dissolved ions.

Hypothesis

The more the calgon that is added, the softer the H2O would go.

Calgon contains man-made compounds called zeolites which are aluminosilicate minerals.[ 6 ]They have a big surface country as they contain big Numberss of holes, which accommodate for cations such as Na+ slackly. These can be readily exchanged for other cations in contact solution. Therefore Ca2+ and Mg2+ can be readily exchanged for Na+ ions in the zeolites.

Besides the rate at which calgon softens the H2O will diminish as more calgon is added.

This is because if the volume of the H2O is kept changeless, so there will be similar sums of Ca2+ and Mg2+ ions in the H2O whilst there would be more Na+ available for ion exchange as more calgon is added. This means that the more the calgon that is added, the faster the rate of ion exchange would be.

Independent Variable

In this probe, the independent variable was the mass of calgon used to soften the difficult H2O. Initially, a preliminary experiment was carried out to assist make up one’s mind which multitudes would bring forth an efficient set of consequences. The undermentioned multitudes of calgon were chosen for several grounds: 0.0g, 0.1g, 0.2g, 0.3g, 0.5g, 0.8g, 1.0g, 1.2g, and 1.3g. Nine different multitudes was the maximal possible to make because there was a limitation on clip and therefore it would hold been really clip devouring to look into a larger scope of calgon mass. Another ground for this was because I believed that these multitudes would demo how the rate of ion exchange changed. Furthermore, making more multitudes of calgon would hold merely given similar consequences as the general tendency is clearly seeable on the graphs.

Dependant Variable

The volume of EDTA solution required to turn Eriochrome Black T index to a bluish shadiness is the dependent variable. Using this, I can work out the hardness of the H2O and the single measures of Ca2+ and Mg2+ ions, when changing multitudes of calgon is added to it.

Controlled Variables

Temperature- A higher temperature means that the Ca and Mg ions are more likely to go soluble, and the procedure of ion exchange would be more rapid as the rate of diffusion would be faster. The comparative affinity of ions can be reduced by the higher temperatures, which makes it easier for Na+ ions to displace high sums of Ca2+ ions.[ 7 ]This would intend it would interfere with the calgon, and the consequences would be inaccurate. Therefore the temperature was kept at a changeless room temperature ( 25 A°C ) .

pH- The exchange efficiency would be greater in footings of rosin capacity for the ion to be removed from the solution, if there is a higher penchant a rosin exhibits to a peculiar ion. Both Mg2+ and Ca2+ ions have a penchant to a strong acid rosin[ 8 ], nevertheless, even the degree of penchant here varies. Therefore pH 10 buffers were added to the EDTA and to the erio blackchrome T index, in order to avoid the different rosin penchant that would be exhibited on the Mg2+ and Ca2+ ions.

Water Hardness and Volume- The H2O had to be collected from the same beginning, because if it was n’t from the beginning there is a more likely opportunity for the hardness degree of the H2O to change greatly. This would intend there would be different measures of Ca2+ and Mg2+ in the H2O, which would do the probe inaccurate, therefore the H2O was collected from the same pat. The volume of the H2O was besides kept changeless at 25ml to do certain that there were similar sums of Ca and Mg ions in it. This is because a larger volume would theoretically incorporate a higher concentration of Ca2+ and Mg2+ , although this is non guaranteed.

Titration of 25cm3 of Eriochrome Black T solution with 0.01M Ethylenediaminetetraacetic acids to find the alteration in the composing of difficult H2O

Apparatus

25cm3 pipette

50cm3 burette

300cm3 0.01M Ethylenediaminetetraacetic acid

PH metre

pH 10 ammonium hydroxide buffer

16.2g Calgon

25ml Eriochrome Black T index

200ml distilled H2O

675ml tap H2O

Erlenmeyer flask

250cm3 beaker

25ml mensurating cylinder x 3

Spatula

Electronic Balance

Method

1. Pipette 25 cm3 of tap H2O into a conelike flask.

2. Add 2 cm3 buffer solution followed by 3 beads of Eriochrome Black T index solution.

3. Titrate with 0.01 M EDTA until the solution turns from vino red to flip blue with no intimation of

ruddy.

4. Repeat the titration two farther times.

5. Add the specific sum of calgon to the 25cm3 of tap H2O, and delay boulder clay calgon is to the full

dissolved in the H2O and no solute is seeable in the solution.

6. Repeat processes 2-4, and so finish the same for the other calgon multitudes.

Percentage Mistake

It is really likely that there was inaccuracy in the equipment used hence I have to see the experimental mistake as my consequences are derived from experimental informations. The overall per centum mistake, shows the per centum mistake from which my consequences may change in repetitions, due to experimental mistake.

Percentage uncertainty=

25ml of tap H2O was prepared utilizing a 25ml measurement cylinder.

25ml mensurating cylinder= 25ml 0.5ml

% uncertainness =

Electronic balance

0.1g0.005g = 5 %

0.2g0.005g = 2.5 %

0.3g0.005g = 1.667 %

0.5g0.005g = 1 %

0.8g0.005g = 0.625 %

1.0g0.005g = 0.5 %

1.2g0.005g = 0.416 %

1.3g0.005g = 0.385 %

Burette

0.0g calgon = 2.50.05= 2 %

0.1g calgon = 1.860.05=2.68 %

0.2g calgon = 1.50.05= 3.33 %

0.3g calgon = 1.230.05= 4.065 %

0.5g calgon = 1.030.05= 4.854 %

0.8g calgon = 0.860.05= 5.814 %

1.0g calgon = 0.760.05= 6.579 %

1.2g calgon = 0.70.05=7.143 %

1.3g calgon = 0.560.05=8.923 %

Entire per centum uncertainty= % of mensurating cylinder + % of Electronic balance + % of burette

0.0g calgon = 2 % + 2 % = 4 %

0.1g calgon = 2 % +5 % + 2.68 % = 9.68 %

0.2g calgon = 2 % + 2.5 % + 3.33 % = 7.83 %

0.3g calgon = 2 % + 1.66 % 4.065 % = 7.725 %

0.5g calgon = 2 % + 1 % + 4.854 % = 7.854 %

0.8g calgon = 2 % + 0.625 % + 5.814 % = 8.439 %

1.0g calgon = 2 % + 0.5 % + 6.579 % = 9.079 %

1.2g calgon = 2 % + 0.416 % + 7.143 % = 9.559 %

1.3g calgon = 2 % + 0.385 % +8.923 % = 11.308 %

I was unable to cipher the per centum mistake of this experiment because a literature value could non be obtained for this experiment.

Table of consequences demoing how the mass calgon affects the volume of EDTA required to turn Erio blackchrome T into a bluish solution from a vino ruddy coloring material.

Mass of Calgon/g ( A±0.005cm3 )

Tests of Experiment

Mean Titre Volume/cm3 ( A±0.1cm3 )

1

2

3

Initial Volume/cm3 ( A±0.05cm3 )

Final volume/cm3 ( A±0.05cm3 )

Titre Volume/cm3 ( A±0.1cm3 )

Initial Volume/cm3 ( A±0.05cm3 )

Final volume/cm3 ( A±0.05cm3 )

Titre Volume/cm3 ( A±0.1cm3 )

Initial Volume/cm3 ( A±0.05cm3 )

Final volume/cm3 ( A±0.05cm3 )

Titre Volume/cm3 ( A±0.1cm3 )

0

0

2.4

2.4

2.4

4.9

2.5

4.9

7.5

2.6

2.5

0.1

0

1.9

1.9

1.9

3.8

1.9

3.8

5.6

1.8

1.86

0.2

0

1.4

1.6

1.4

2.7

1.4

2.7

4.6

1.9*

1.5

0.3

0

1.1

1.1

1.1

2.4

1.3

2.4

3.7

1.3

1.23

0.5

0

1.2

1.2

1.2

2.1

0.9

2.1

3.1

1

1.03

0.8

0

0.8

0.8

0.8

1.7

0.9

1.7

2.6

0.9

0.86

1.0

0

0.7

0.7

0.7

1.5

0.8

1.5

2.3

0.8

0.76

1.2

0

0.7

0.7

0.7

1.4

0.7

1.4

2.1

0.7

0.7

1.3

0

0.6

0.6

0.6

1.1

0.5

1.1

1.7

0.6

0.56

*Anomaly

Table demoing how the mass of calgon Graph demoing how calgon affects the hardness of tungsten affects the hardness of the H2O H2O

Mass of Calgon/g

Hardness of water/ppm

0

100

0.1

74.4

0.2

60

0.3

49.2

0.5

41.2

0.8

34.4

1

30.4

1.2

28

1.3

24

Calculations of H2O hardness

( The computations for the other calgon multitudes can be found in appendix 1 )

No Calgon

First, I need to change over the volume of the EDTA solution into dm3 from cm3, by spliting it by 1000 so that I can work out the moles of the EDTA solution that was titrated.

2.5/1000 = 0.0025 dm3

Number of moles = concentration ten volume

= 0.01 mol dm3 x 0.025 dm3

= 0.000025 mol

Then I have to change over the moles of the EDTA solution into millimoles by multiplying it by 1000.

0.000025 mol ten 1000 = 0.025 mmol

This means that 25ml of H2O I used contained a sum of 0.025 mmol of Ca2+ and Mg2+ ions. Now, I am traveling to work out the entire metal concentration in units of millimoles per liter of H2O. In order to make this, I have to split the figure of millimoles of Ca2+ and Mg2+ ions ; by the volume of the H2O I used which was 25cm3. This figure now has to be multiplied by 1000 to change over it back into millimeters from liters.

0.025/25 ten 1000 = 1 millimoles per liter of H2O of entire metal concentration

An premise has to be made next that the entire hardness of 1mmol of Ca2+ and Mg2+ ions per liter of H2O is tantamount to 1 mmol of dissolved CaCO3 per liter of H2O. Using the molar mass of CaCOA­3 as 100 gmol-1, I can change over the millimoles equivalent to CaCOA­3 to milligrams tantamount to CaCOA­3. Then we take the figure of millimoles of tantamount CaCOA­A­A­A­3 per liter, divide it by 1000, to change over millimoles into moles, so multiply it by 1000, in order to change over the gms into mgs and so multiply this figure by.

1/1000 = 0.001 ten 1000= 1 ten 100 = 100 mgs of tantamount CaCOA­3

I am able to show this is as the figure of parts of tantamount CaCOA­3 per million parts of H2O because the denseness of H2O is 1.00g mL-1

Therefore the hardness of the H2O with no calgon can be expressed as 100 ppm of tantamount CaCOA­3.

Discussion of Consequences

My probe shows that the calgon softens the H2O and that the more the calgon that is added, the softer the H2O becomes. Calgon softens the H2O in several different ways, which is why it is rather effectual as it is illustrated in the graph.

One of the methods is the ion exchange which was mentioned in my hypothesis, where the ions in the difficult H2O such as Ca2+ and Mg2+ are collected by the zeolites in the calgon, and release Na+ in return. The zeolites can be represented as NaA­2Ze[ 9 ], and they retain Ca2+ and Mg2+ from difficult H2O as CaZe and MgZe in their constructions and let go of Na salts as by merchandises. Therefore assorted reactions can happen in the H2O when calgon is added to it by ion exchange.

NaA­A­A­2Ze + Ca ( HCO3 ) 2 CaZe + 2NaHCO3

Na2Ze + Mg ( HCO3 ) 2 MgZe + 2NaHCO3

Na2Ze + MgCl2 MgZe + 2NaCl

Na2Ze + CaCl2 CaZe + 2NaCl

Na2Ze + MgSO4 MgZe + 2Na2SO4

Na2Ze + CaSO4 CaZe + 2Na2SO4

Therefore as more calgon was added, more of these reactions occurred and this in bend meant that more Ca2+ and Mg2+ ions were removed from the difficult H2O.

Another procedure that could happen to soften the H2O is neutralization. This is where the Ca2+ ions in the difficult H2O can be removed by complexing agents. These are anions that form dative covalent bonds by wrapping themselves around the Ca2+ ions itself. This is in bend agencies that they are masked from the difficult H2O efficaciously. Sodium citrate ( sodium 2-hydroxypropane-1,2,3-tricarboxylate )[ 10 ]is one of complexing agents found in calgon. By complexing the Ca2+ ions, it means that it can be kept dissolved in the H2O, alternatively of it going deposited and organizing lime graduated table.

Figure 1.1: The construction of Na 2-hydroxypropane-1,2,3-tricarboxylate

Sodium 2-hydroxypropane-1,2,3-tricarboxylate is known as Cit-3, and it has 3 O ‘s that can bond around the Ca2. They are attracted electro statically to the ions and a dative bond is formed utilizing its lone brace of negatrons. This means that alternatively of being deposited in the H2O and causation calcium hydroxide graduated table, it is dissolved in it. This is because when the Ca ions are removed from the difficult H2O by calgon, no CaCO3 can be deposited because the equilibrium below can non travel to the left, which it has to make in order to lodge the Ca carbonate.

CaCO3 ( s ) + H2O ( cubic decimeter ) + CO2 ( aq ) A Ca2+ ( aq ) + 2HCO3- ( aq )

Another complexing agent nowadays in calgon is sodium tripolyphosphate, nevertheless with this one ; attention has to be taken with the concentration of it. This is because indissoluble Ca and Mg salts are formed when there is a low degree of Na tripolyphosphates.

2Na5P3O10 ( aq ) + 5Ca2+ ( aq ) Ca5 ( P3O10 ) 2 ( s ) + 10Na+ ( aq )

On the other manus, when a higher concentration of Na tripolyphosphate is present, a more soluble composite is formed.[ 11 ]

Na5P3O10 ( aq ) + Ca2+ ( aq ) A [ CaP3O10 ] 3- ( aq ) + 5Na+ ( aq )

A A A A A A A A A A A A A A A A A A A A A

Decision

My hypothesis has been proved right which is apparent in my graph because the hardness of the H2O decreases when higher multitudes of calgon are added to it. Calgon has an immediate consequence on the hardness of the H2O where 0.1g of calgon softens the H2O by 25.6 ppm. The alteration is non relative and the line of best tantrum is a curve. Besides my consequences back up my other hypothesis where the rate at which calgon softens the H2O decreases. This is because the difference in ppm of H2O with no calgon and 0.1g calgon is 25.6 ppm whilst the difference in ppm between 1.2g and 1.3g of calgon is 4 ppm. Although both have a alteration of 0.1g, the consequence calgon has on the composing of difficult H2O is decreased as more calgon is added.

Unanswered inquiries

My probe has answered my chief inquiry – to what extent does calgon impact the composing of difficult H2O. Despite this merely a general tendency is able to be studied from my consequences, and I would prefer a more precise analysis of the composing of difficult H2O, and of how different chemicals and compounds in calgon affect different component of difficult H2O.

It has been assumed that both Ca2+ and Mg2+ ions are removed from the H2O by calgon ; nevertheless this is non a known fact. The ions could be removed at different rates, which would demo more exactly how calgon affects the composing of difficult H2O. Mg2+ ions can non blend with a complexing agent like Ca because they have one less shell of full negatrons, which means they have no vitamin D orbital energy which is required to organize a bond with the complexing agent.

The continuity of the graph is unknown. The tendency that is seeable on the graph is that the more the calgon that is added to the difficult H2O, the softer the H2O becomes. This line seems to me as if it will plateau off because the rate at which the ion exchange occurs has decreased. The inquiry that could be asked is, could calgon be added to H2O, until it has no Ca or Mg ions left in it. This premise would be supported with the consequences I have gained, nevertheless, the ions are non ever removed, sometimes they are merely dissolved, therefore their presence will ever be. After a piece, the zeolite would be wholly converted into Ca and Mg ions, and hence the softening of the H2O will halt. If this peculiar point is found, so replies could be developed to reply the inquiry of whether uninterrupted add-on of calgon, will to the full deionise the difficult H2O.

Calgon affects the H2O in several ways, but the procedure which has the biggest impact on the composing on the difficult H2O is unknown. If the procedure was known, so the concentration of the chemical that causes this could be alerted so that it can diminish the hardness of the H2O more by the add-on of calgon, whilst being safe at the same clip.

Evaluation

I believe that my probe is valid because most of my repetition reading are similar. The exclusion to this is the anomalous consequence, which was the 3rd titration for when 0.2g of calgon was added to the H2O. 1.9cm3 of EDTA solution was needed on that test to do the solution of Eriochrome black T, difficult H2O and calgon turn bluish. As consequences of the first two tests were rather similar hence the 3rd test was disregarded because I believed that it would hold affected the norm and hence the overall graph.

There is a really likely opportunity of there being inaccuracies in my experiment. Despite holding my largest per centum mistake as 11.3 % for when 1.3g of calgon is added to the H2O, I believe that my probe could be improved. This is because harmonizing to Thames Water, the part where I gained the H2O from has difficult H2O with 308 ppm. As my control, I completed the titration with no calgon, which showed me that the hardness of the H2O was 100 ppm. A possible account for this big difference could be because the Thames Water research took topographic point in 2009. Therefore the hardness of the H2O may hold altered in that clip. This can merely be assumed, nevertheless in my sentiment, I believe that the Thames Water information is more accurate because the H2O to this part travels through chalky land where it is likely to hold Ca and Mg ions dissolve in it, doing the H2O really hard.

Appendix 1

0.1g Calgon

First, I need to change over the volume of the EDTA solution into dm3 from cm3, by spliting it by 1000 so that I can work out the moles of the EDTA solution that was titrated.

1.86/1000 = 0.00186 dm3

Number of moles = concentration ten volume

= 0.01 mol dm3 x 0.00186 dm3

= 0.0000186 mol

Then I have to change over the moles of the EDTA solution into millimoles by multiplying it by 1000.

0.0000186 mol ten 1000 = 0.0186 mmol

This means that 25ml of H2O plus 0.1g calgon contains a sum of 0.0186 mmol of Ca2+ and Mg2+ ions. Now, I am traveling to work out the entire metal concentration in units of millimoles per liter of H2O. In order to make this, I have to split the figure of millimoles of Ca2+ and Mg2+ ions ; by the volume of the H2O I used which was 25cm3. This figure now has to be multiplied by 1000 to change over it back into millimeters from liters.

0.0186/25 ten 1000 = 0.744 millimoles per liter of H2O of entire metal concentration

An premise has to be made next that the entire hardness of mmol of Ca2+ and Mg2+ ions per liter of H2O is tantamount to 1 mmol of dissolved CaCO3 per liter of H2O. Using the molar mass of CaCOA­3 as 100 gmol-1, I can change over the millimoles equivalent to CaCOA­3 to milligrams tantamount to CaCOA­3. Then we take the figure of millimoles of tantamount CaCOA­A­A­A­3 per liter, divide it by 1000, to change over millimoles into moles, so multiply it by 1000, in order to change over the gms into mgs and so multiply this figure by.

0.744/1000 = 0.000744 ten 1000= 0.744 ten 100 = 74.4 mgs of tantamount CaCOA­3

I am able to show this is as the figure of parts of tantamount CaCOA­3 per million parts of H2O because the denseness of H2O is 1.00g mL-1

Therefore the hardness of the H2O with no calgon can be added can be expressed as 74.4 ppm of tantamount CaCOA­3.

0.2g Calgon

First, I need to change over the volume of the EDTA solution into dm3 from cm3, by spliting it by 1000 so that I can work out the moles of the EDTA solution that was titrated.

1.5/1000 = 0.0015 dm3

Number of moles = concentration ten volume

= 0.01 mol dm3 x 0.0015 dm3

= 0.000015 mol

Then I have to change over the moles of the EDTA solution into millimoles by multiplying it by 1000.

0.000015 mol ten 1000 = 0.015 mmol

This means that 25ml of H2O plus 0.2g calgon contains a sum of 0.015 mmol of Ca2+ and Mg2+ ions. Now, I am traveling to work out the entire metal concentration in units of millimoles per liter of H2O. In order to make this, I have to split the figure of millimoles of Ca2+ and Mg2+ ions ; by the volume of the H2O I used which was 25cm3. This figure now has to be multiplied by 1000 to change over it back into millimeters from liters.

0.015/25 ten 1000 = 0.6 millimoles per liter of H2O of entire metal concentration

An premise has to be made next that the entire hardness of mmol of Ca2+ and Mg2+ ions per liter of H2O is tantamount to 1 mmol of dissolved CaCO3 per liter of H2O. Using the molar mass of CaCOA­3 as 100 gmol-1, I can change over the millimoles equivalent to CaCOA­3 to milligrams tantamount to CaCOA­3. Then we take the figure of millimoles of tantamount CaCOA­A­A­A­3 per liter, divide it by 1000, to change over millimoles into moles, so multiply it by 1000, in order to change over the gms into mgs and so multiply this figure by.

0.6/1000 = 0.0006 ten 1000= 0.6 ten 100 = 60 mgs of tantamount CaCOA­3

I am able to show this is as the figure of parts of tantamount CaCOA­3 per million parts of H2O because the denseness of H2O is 1.00g mL-1

Therefore the hardness of the H2O with no calgon can be added can be expressed as 60 ppm of tantamount CaCOA­3.

0.3g Calgon

First, I need to change over the volume of the EDTA solution into dm3 from cm3, by spliting it by 1000 so that I can work out the moles of the EDTA solution that was titrated.

1.23/1000 = 0.00123 dm3

Number of moles = concentration ten volume

= 0.01 mol dm3 x 0.00123 dm3

= 0.0000123 mol

Then I have to change over the moles of the EDTA solution into millimoles by multiplying it by 1000.

0.0000123 mol ten 1000 = 0.0123 mmol

This means that 25ml of H2O plus 0.3g calgon contains a sum of 0.0123 mmol of Ca2+ and Mg2+ ions. Now, I am traveling to work out the entire metal concentration in units of millimoles per liter of H2O. In order to make this, I have to split the figure of millimoles of Ca2+ and Mg2+ ions ; by the volume of the H2O I used which was 25cm3. This figure now has to be multiplied by 1000 to change over it back into millimeters from liters.

0.0123/25 ten 1000 = 0.492 millimoles per liter of H2O of entire metal concentration

An premise has to be made next that the entire hardness of mmol of Ca2+ and Mg2+ ions per liter of H2O is tantamount to 1 mmol of dissolved CaCO3 per liter of H2O. Using the molar mass of CaCOA­3 as 100 gmol-1, I can change over the millimoles equivalent to CaCOA­3 to milligrams tantamount to CaCOA­3. Then we take the figure of millimoles of tantamount CaCOA­A­A­A­3 per liter, divide it by 1000, to change over millimoles into moles, so multiply it by 1000, in order to change over the gms into mgs and so multiply this figure by.

0.492/1000 = 0.000492 ten 1000= 0.492 ten 100 = 49.2 mgs of tantamount CaCOA­3

I am able to show this is as the figure of parts of tantamount CaCOA­3 per million parts of H2O because the denseness of H2O is 1.00g mL-1

Therefore the hardness of the H2O with no calgon can be added can be expressed as 49.2 ppm of tantamount CaCOA­3.

0.5g Calgon

First, I need to change over the volume of the EDTA solution into dm3 from cm3, by spliting it by 1000 so that I can work out the moles of the EDTA solution that was titrated.

1.03/1000 = 0.00103 dm3

Number of moles = concentration ten volume

= 0.01 mol dm3 x 0.00103 dm3

= 0.0000103 mol

Then I have to change over the moles of the EDTA solution into millimoles by multiplying it by 1000.

0.0000103 mol ten 1000 = 0.0103 mmol

This means that 25ml of H2O plus 0.5g calgon contains a sum of 0.0103 mmol of Ca2+ and Mg2+ ions. Now, I am traveling to work out the entire metal concentration in units of millimoles per liter of H2O. In order to make this, I have to split the figure of millimoles of Ca2+ and Mg2+ ions ; by the volume of the H2O I used which was 25cm3. This figure now has to be multiplied by 1000 to change over it back into millimeters from liters.

0.0103/25 ten 1000 = 0.412 millimoles per liter of H2O of entire metal concentration

An premise has to be made next that the entire hardness of mmol of Ca2+ and Mg2+ ions per liter of H2O is tantamount to 1 mmol of dissolved CaCO3 per liter of H2O. Using the molar mass of CaCOA­3 as 100 gmol-1, I can change over the millimoles equivalent to CaCOA­3 to milligrams tantamount to CaCOA­3. Then we take the figure of millimoles of tantamount CaCOA­A­A­A­3 per liter, divide it by 1000, to change over millimoles into moles, so multiply it by 1000, in order to change over the gms into mgs and so multiply this figure by.

0.412/1000 = 0.000412 ten 1000= 0.412 ten 100 = 41.2 mgs of tantamount CaCOA­3

I am able to show this is as the figure of parts of tantamount CaCOA­3 per million parts of H2O because the denseness of H2O is 1.00g mL-1

Therefore the hardness of the H2O with no calgon can be added can be expressed as 41.2 ppm of tantamount CaCOA­3.

0.8g Calgon

First, I need to change over the volume of the EDTA solution into dm3 from cm3, by spliting it by 1000 so that I can work out the moles of the EDTA solution that was titrated.

0.86/1000 = 0.00086 dm3

Number of moles = concentration ten volume

= 0.01 mol dm3 x 0.00086 dm3

= 0.0000086 mol

Then I have to change over the moles of the EDTA solution into millimoles by multiplying it by 1000.

0.0000086 mol ten 1000 = 0.0086 mmol

This means that 25ml of H2O plus 0.8g calgon contains a sum of 0.0086 mmol of Ca2+ and Mg2+ ions. Now, I am traveling to work out the entire metal concentration in units of millimoles per liter of H2O. In order to make this, I have to split the figure of millimoles of Ca2+ and Mg2+ ions ; by the volume of the H2O I used which was 25cm3. This figure now has to be multiplied by 1000 to change over it back into millimeters from liters.

0.0086/25 ten 1000 = 0.344 millimoles per liter of H2O of entire metal concentration

An premise has to be made next that the entire hardness of mmol of Ca2+ and Mg2+ ions per liter of H2O is tantamount to 1 mmol of dissolved CaCO3 per liter of H2O. Using the molar mass of CaCOA­3 as 100 gmol-1, I can change over the millimoles equivalent to CaCOA­3 to milligrams tantamount to CaCOA­3. Then we take the figure of millimoles of tantamount CaCOA­A­A­A­3 per liter, divide it by 1000, to change over millimoles into moles, so multiply it by 1000, in order to change over the gms into mgs and so multiply this figure by.

0.344/1000 = 0.000344 ten 1000= 0.344 ten 100 = 34.4 mgs of tantamount CaCOA­3

I am able to show this is as the figure of parts of tantamount CaCOA­3 per million parts of H2O because the denseness of H2O is 1.00g mL-1

Therefore the hardness of the H2O with no calgon can be added can be expressed as 34.4 ppm of tantamount CaCOA­3.

1g Calgon

First, I need to change over the volume of the EDTA solution into dm3 from cm3, by spliting it by 1000 so that I can work out the moles of the EDTA solution that was titrated.

0.76/1000 = 0.00076 dm3

Number of moles = concentration ten volume

= 0.01 mol dm3 x 0.00076 dm3

= 0.0000076 mol

Then I have to change over the moles of the EDTA solution into millimoles by multiplying it by 1000.

0.0000076 mol ten 1000 = 0.0076 mmol

This means that 25ml of H2O plus 1.0g calgon contains a sum of 0.0076 mmol of Ca2+ and Mg2+ ions. Now, I am traveling to work out the entire metal concentration in units of millimoles per liter of H2O. In order to make this, I have to split the figure of millimoles of Ca2+ and Mg2+ ions ; by the volume of the H2O I used which was 25cm3. This figure now has to be multiplied by 1000 to change over it back into millimeters from liters.

0.0076/25 ten 1000 = 0.304 millimoles per liter of H2O of entire metal concentration

An premise has to be made next that the entire hardness of mmol of Ca2+ and Mg2+ ions per liter of H2O is tantamount to 1 mmol of dissolved CaCO3 per liter of H2O. Using the molar mass of CaCOA­3 as 100 gmol-1, I can change over the millimoles equivalent to CaCOA­3 to milligrams tantamount to CaCOA­3. Then we take the figure of millimoles of tantamount CaCOA­A­A­A­3 per liter, divide it by 1000, to change over millimoles into moles, so multiply it by 1000, in order to change over the gms into mgs and so multiply this figure by.

0.304/1000 = 0.000304 ten 1000= 0.304 ten 100 = 30.4 mgs of tantamount CaCOA­3

I am able to show this is as the figure of parts of tantamount CaCOA­3 per million parts of H2O because the denseness of H2O is 1.00g mL-1

Therefore the hardness of the H2O with no calgon can be added can be expressed as 30.4 ppm of tantamount CaCOA­3.

1.2g Calgon

First, I need to change over the volume of the EDTA solution into dm3 from cm3, by spliting it by 1000 so that I can work out the moles of the EDTA solution that was titrated.

0.7/1000 = 0.0007 dm3

Number of moles = concentration ten volume

= 0.01 mol dm3 x 0.0007 dm3

= 0.000007 mol

Then I have to change over the moles of the EDTA solution into millimoles by multiplying it by 1000.

0.000007 mol ten 1000 = 0.007 mmol

This means that 25ml of H2O plus 1.2g calgon contains a sum of 0.007 mmol of Ca2+ and Mg2+ ions. Now, I am traveling to work out the entire metal concentration in units of millimoles per liter of H2O. In order to make this, I have to split the figure of millimoles of Ca2+ and Mg2+ ions ; by the volume of the H2O I used which was 25cm3. This figure now has to be multiplied by 1000 to change over it back into millimeters from liters.

0.007/25 ten 1000 = 0.28 millimoles per liter of H2O of entire metal concentration

An premise has to be made next that the entire hardness of mmol of Ca2+ and Mg2+ ions per liter of H2O is tantamount to 1 mmol of dissolved CaCO3 per liter of H2O. Using the molar mass of CaCOA­3 as 100 gmol-1, I can change over the millimoles equivalent to CaCOA­3 to milligrams tantamount to CaCOA­3. Then we take the figure of millimoles of tantamount CaCOA­A­A­A­3 per liter, divide it by 1000, to change over millimoles into moles, so multiply it by 1000, in order to change over the gms into mgs and so multiply this figure by.

0.28/1000 = 0.00028 ten 1000= 0.28 ten 100 = 28 mgs of tantamount CaCOA­3

I am able to show this is as the figure of parts of tantamount CaCOA­3 per million parts of H2O because the denseness of H2O is 1.00g mL-1

Therefore the hardness of the H2O with no calgon can be added can be expressed as 28 ppm of tantamount CaCOA­3.

1.3g Calgon

First, I need to change over the volume of the EDTA solution into dm3 from cm3, by spliting it by 1000 so that I can work out the moles of the EDTA solution that was titrated.

0.6/1000 = 0.0006 dm3

Number of moles = concentration ten volume

= 0.01 mol dm3 x 0.0006 dm3

= 0.000006 mol

Then I have to change over the moles of the EDTA solution into millimoles by multiplying it by 1000.

0.000006 mol ten 1000 = 0.006 mmol

This means that 25ml of H2O plus 1.3g calgon contains a sum of 0.006 mmol of Ca2+ and Mg2+ ions. Now, I am traveling to work out the entire metal concentration in units of millimoles per liter of H2O. In order to make this, I have to split the figure of millimoles of Ca2+ and Mg2+ ions ; by the volume of the H2O I used which was 25cm3. This figure now has to be multiplied by 1000 to change over it back into millimeters from liters.

0.006/25 ten 1000 = 0.24 millimoles per liter of H2O of entire metal concentration

An premise has to be made next that the entire hardness of mmol of Ca2+ and Mg2+ ions per liter of H2O is tantamount to 1 mmol of dissolved CaCO3 per liter of H2O. Using the molar mass of CaCOA­3 as 100 gmol-1, I can change over the millimoles equivalent to CaCOA­3 to milligrams tantamount to CaCOA­3. Then we take the figure of millimoles of tantamount CaCOA­A­A­A­3 per liter, divide it by 1000, to change over millimoles into moles, so multiply it by 1000, in order to change over the gms into mgs and so multiply this figure by.

0.24/1000 = 0.00024 ten 1000= 0.24 ten 100 = 24 mgs of tantamount CaCOA­3

I am able to show this is as the figure of parts of tantamount CaCOA­3 per million parts of H2O because the denseness of H2O is 1.00g mL-1

Therefore the hardness of the H2O with no calgon can be added can be expressed as 24 ppm of tantamount CaCOA­3.

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