# Does Calgon Affect The Composition Of Hard Water Biology Essay

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Hard water is water containing dissolved minerals. Calcium and magnesium ions dissolve in it as the water seeps through the ground. It increases the possibility of the formation of lime scale and therefore reduces the efficiency of the cleaning processes. This occurs because the heat transfer from the element to the water is reduced by the lime scale and makes the appliance less efficient. Many water softener products are available for consumers such as calgon, which are believed to reduce the effect of the hard water on appliances. This is turn would improve the cleaning process. However the process is ambiguous, because we are unsure whether the calgon removes or masks the effect of the dissolved ions. Water hardness is measured in comparison to CaCO3 ppm of water. This is because CaCO3 is a common precipitate and the table below shows how different level of hardness is categrorized.

My preliminary research showed that calgon affects the composition of the water by softening it. [1] This is done by a process called ion exchange where the ion exchanger, calgon, removes the Ca2+ and Mg2+ from the hard water and replaces them with exchangeable Na+ ions. [2] We discuss about the Ca2+ and Mg2+ ions when we consider hard water because these are the two ions with the highest concentration that are responsible for the water hardness. Every ion that is removed from the water is replaced by an equivalent amount of another ionic specie [3] . Therefore two Na+ ions replace either one Ca2+ or Mg2+ ion. There are also other processes that occur due to the chemicals present in calgon that soften the water. This is why calgon is very effective, because many different chemicals simultaneously soften water.

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The levels 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 complexes with the Eriochrome black T indicator, when the pH 10 buffer is added to it and it become a wine red colour. This happens when [H-ErioT] 2- ions bonds with either Ca2+ or Mg2+ ions.

Ca2+ (aq) + [H-ErioT]2-(aq) + H2O(l) [Ca-ErioT]-(aq) + H3O+(aq)

Mg2+ (aq) + [H-ErioT]2-(aq) + H2O(l) [Mg-ErioT]-(aq) + H3O+(aq)

EDTA has pH 10 buffer added to it because it allows the EDTA to disassociate to form [EDTA] 4. This disassociated form means it has the maximum available amount of tetracarboxylate ion, which has plenty of electrons and six available binding sites. The anion, [EDTA] 4-, wraps itself around the calcium and magnesium ions so that the six pairs of electrons are shared with the metal ion.

The colour change from the red wine to a permanent blue, has a colour change of transitory purple in the middle. The permanent blue would be the end point of the titration and this would mean that all the Ca2+ and Mg2+ ions have either been removed or dissolved so that they have no effect on the water. In order for this to occur, the two reactions below would have to occur first where the ions in the hard water bond with the EDTA solution.

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

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

The end point is when the complex [Mg-ErioT]- is completely broken up by the anion [EDTA] 4- releasing blue [H-ErioT]2- ion. The final reaction is shown with the equation below:

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

Aim

To find to what extent:

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Calgon affects the composition of hard water

To work out if it affects it by removing or masking dissolving the dissolved ions.

Hypothesis

The more the calgon that is added, the softer the water would become.

Calgon contains synthetic compounds called zeolites which are aluminosilicate minerals. [6] They have a large surface area as they contain large numbers of holes, which accommodate for cations such as Na+ loosely. 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.

Also the rate at which calgon softens the water will decrease as more calgon is added.

This is because if the volume of the water is kept constant, then there will be similar amounts of Ca2+ and Mg2+ ions in the water 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 investigation, the independent variable was the mass of calgon used to soften the hard water. Initially, a preliminary experiment was carried out to help decide which masses would produce an efficient set of results. The following masses of calgon were chosen for several reasons: 0.0g, 0.1g, 0.2g, 0.3g, 0.5g, 0.8g, 1.0g, 1.2g, and 1.3g. Nine different masses was the maximum possible to do because there was a restriction on time and therefore it would have been very time consuming to investigate a larger range of calgon mass. Another reason for this was because I believed that these masses would show how the rate of ion exchange changed. Furthermore, doing more masses of calgon would have just given similar results as the general trend is clearly visible on the graphs.

Dependant Variable

The volume of EDTA solution required to turn Eriochrome Black T indicator to a blue shade is the dependent variable. Using this, I can work out the hardness of the water and the individual quantities of Ca2+ and Mg2+ ions, when varying masses of calgon is added to it.

Controlled Variables

Temperature- A higher temperature means that the calcium and magnesium ions are more likely to become soluble, and the process of ion exchange would be more rapid as the rate of diffusion would be faster. The relative affinity of ions can be reduced by the higher temperatures, which makes it easier for Na+ ions to displace high amounts of Ca2+ ions. [7] This would mean it would interfere with the calgon, and the results would be inaccurate. Therefore the temperature was kept at a constant room temperature (25 Â°C).

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

Water Hardness and Volume- The water had to be collected from the same source, because if it wasn't from the source there is a more likely chance for the hardness level of the water to vary greatly. This would mean there would be different quantities of Ca2+ and Mg2+ in the water, which would make the investigation inaccurate, therefore the water was collected from the same tap. The volume of the water was also kept constant at 25ml to make sure that there were similar amounts of calcium and magnesium ions in it. This is because a larger volume would theoretically contain a higher concentration of Ca2+ and Mg2+ , although this is not guaranteed.

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Examples of our workTitration of 25cm3 of Eriochrome Black T solution with 0.01M EDTA to determine the change in the composition of hard water

Apparatus

25cm3 pipette

50cm3 burette

300cm3 0.01M EDTA

PH meter

pH 10 ammonia buffer

16.2g Calgon

25ml Eriochrome Black T indicator

200ml distilled water

675ml tap water

Erlenmeyer flask

250cm3 beaker

25ml measuring cylinder x 3

Spatula

Electronic Balance

Method

1. Pipette 25 cm3 of tap water into a conical flask.

2. Add 2 cm3 buffer solution followed by 3 drops of Eriochrome Black T indicator solution.

3. Titrate with 0.01 M EDTA until the solution turns from wine red to sky blue with no hint of

red.

4. Repeat the titration two further times.

5. Add the specific amount of calgon to the 25cm3 of tap water, and wait till calgon is fully

dissolved in the water and no solute is visible in the solution.

6. Repeat processes 2-4, and then complete the same for the other calgon masses.

Percentage Error

It is very likely that there was inaccuracy in the equipment used therefore I have to consider the experimental error as my results are derived from experimental data. The overall percentage error, shows the percentage error from which my results may vary in repeats, due to experimental error.

Percentage uncertainty=

25ml of tap water was prepared using a 25ml measuring cylinder.

25ml measuring cylinder= 25ml 0.5ml

% uncertainty =

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%

Total percentage uncertainty= % of measuring 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 calculate the percentage error of this experiment because a literature value could not be obtained for this experiment.

## Table of results showing how the mass calgon affects the volume of EDTA required to turn Erio blackchrome T into a blue solution from a wine red colour.

## Mass of Calgon/g (Â±0.005cm3)

## Trials of Experiment

## Mean Titre Volume/cm3 (Â±0.1cm3)

1

2

3

Initial Volume/cm3 (Â±0.05cm3)

Final volume/cm3 (Â±0.05cm3)

Titre Volume/cm3 (Â±0.1cm3)

Initial Volume/cm3 (Â±0.05cm3)

Final volume/cm3 (Â±0.05cm3)

Titre Volume/cm3 (Â±0.1cm3)

Initial Volume/cm3 (Â±0.05cm3)

Final volume/cm3 (Â±0.05cm3)

Titre Volume/cm3 (Â±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 showing how the mass of calgon Graph showing how calgon affects the hardness of w affects the hardness of the water water

## 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 water hardness

(The calculations for the other calgon masses can be found in appendix 1)

No Calgon

First, I need to convert the volume of the EDTA solution into dm3 from cm3, by dividing 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 x volume

= 0.01 mol dm3 x 0.025 dm3

= 0.000025 mol

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

0.000025 mol x 1000 = 0.025 mmol

This means that 25ml of water I used contained a total of 0.025 mmol of Ca2+ and Mg2+ ions. Now, I am going to work out the total metal concentration in units of millimoles per litre of water. In order to do this, I have to divide the number of millimoles of Ca2+ and Mg2+ ions; by the volume of the water I used which was 25cm3. This figure now has to be multiplied by 1000 to convert it back into millimetres from litres.

0.025/25 x 1000 = 1 millimoles per litre of water of total metal concentration

An assumption has to be made next that the total hardness of 1mmol of Ca2+ and Mg2+ ions per litre of water is equivalent to 1 mmol of dissolved CaCO3 per litre of water. Using the molar mass of CaCOÂ3 as 100 gmol-1, I can convert the millimoles equivalent to CaCOÂ3 to milligrams equivalent to CaCOÂ3. Then we take the number of millimoles of equivalent CaCOÂÂÂÂ3 per litre, divide it by 1000, to convert millimoles into moles, then multiply it by 1000, in order to convert the grams into milligrams and then multiply this figure by .

1/1000 = 0.001 x 1000= 1 x 100 = 100 milligrams of equivalent CaCOÂ3

I am able to express this is as the number of parts of equivalent CaCOÂ3 per million parts of water because the density of water is 1.00g mL-1

Therefore the hardness of the water with no calgon can be expressed as 100 ppm of equivalent CaCOÂ3.

Discussion of Results

My investigation shows that the calgon softens the water and that the more the calgon that is added, the softer the water becomes. Calgon softens the water in several different ways, which is why it is quite effective 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 hard water such as Ca2+ and Mg2+ are collected by the zeolites in the calgon, and release Na+ in return. The zeolites can be represented as NaÂ2Ze [9] , and they retain Ca2+ and Mg2+ from hard water as CaZe and MgZe in their structures and release sodium salts as by products. Therefore various reactions can occur in the water when calgon is added to it by ion exchange.

NaÂÂÂ2Ze + 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 turn meant that more Ca2+ and Mg2+ ions were removed from the hard water.

Another process that could occur to soften the water is neutralisation. This is where the Ca2+ ions in the hard water 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 turn means that they are masked from the hard water effectively. 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 water, instead of it becoming deposited and forming lime scale.

Figure 1.1: The structure of sodium 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 using its lone pairs of electrons. This means that instead of being deposited in the water and causing lime scale, it is dissolved in it. This is because when the calcium ions are removed from the hard water by calgon, no CaCO3 can be deposited because the equilibrium below cannot move to the left, which it has to do in order to deposit the calcium carbonate.

CaCO3(s) + H2O(l) + CO2(aq) Â Ca2+(aq) + 2HCO3-(aq)

Another complexing agent present in calgon is sodium tripolyphosphate, however with this one; care has to be taken with the concentration of it. This is because insoluble calcium and magnesium salts are formed when there is a low level of sodium tripolyphosphates.

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

On the other hand, when a higher concentration of sodium tripolyphosphate is present, a more soluble complex is formed. [11]

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

## Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Conclusion

My hypothesis has been proved correct which is evident in my graph because the hardness of the water decreases when higher masses of calgon are added to it. Calgon has an immediate effect on the hardness of the water where 0.1g of calgon softens the water by 25.6 ppm. The change is not proportional and the line of best fit is a curve. Also my results support my other hypothesis where the rate at which calgon softens the water decreases. This is because the difference in ppm of water 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 change of 0.1g, the effect calgon has on the composition of hard water is decreased as more calgon is added.

Unanswered questions

My investigation has answered my main question - to what extent does calgon affect the composition of hard water. Despite this only a general trend is able to be studied from my results, and I would prefer a more precise analysis of the composition of hard water, and of how different chemicals and compounds in calgon affect different constituent of hard water.

It has been assumed that both Ca2+ and Mg2+ ions are removed from the water by calgon; however this is not a known fact. The ions could be removed at different rates, which would show more precisely how calgon affects the composition of hard water. Mg2+ ions cannot fuse with a complexing agent like calcium because they have one less shell of full electrons, which means they have no d orbital energy which is required to form a bond with the complexing agent.

The continuity of the graph is unknown. The trend that is visible on the graph is that the more the calgon that is added to the hard water, the softer the water becomes. This line seems to me as if it will plateau off because the rate at which the ion exchange occurs has decreased. The question that could be asked is, could calgon be added to water, until it has no calcium or magnesium ions left in it. This assumption would be supported with the results I have gained, however, the ions are not always removed, sometimes they are just dissolved, therefore their presence will always exist. After a while, the zeolite would be completely converted into calcium and magnesium ions, and therefore the softening of the water will stop. If this particular point is found, then answers could be developed to answer the question of whether continuous addition of calgon, will fully deionise the hard water.

Calgon affects the water in several ways, but the process which has the biggest impact on the composition on the hard water is unknown. If the process was known, then the concentration of the chemical that causes this could be alerted so that it can decrease the hardness of the water more by the addition of calgon, whilst being safe at the same time.

Evaluation

I believe that my investigation is valid because most of my repeat reading are similar. The exception to this is the anomalous result, which was the third titration for when 0.2g of calgon was added to the water. 1.9cm3 of EDTA solution was needed on that trial to make the solution of Eriochrome black t, hard water and calgon turn blue. As results of the first two trials were quite similar therefore the third trial was disregarded because I believed that it would have affected the average and therefore the overall graph.

There is a very likely chance of there being inaccuracies in my experiment. Despite having my largest percentage error as 11.3% for when 1.3g of calgon is added to the water, I believe that my investigation could be improved. This is because according to Thames Water, the region where I gained the water from has hard water with 308 ppm. As my control, I completed the titration with no calgon, which showed me that the hardness of the water was 100 ppm. A possible explanation for this large difference could be because the Thames Water research took place in 2009. Therefore the hardness of the water may have altered in that time. This can only be assumed, however in my opinion, I believe that the Thames Water data is more accurate because the water to this region travels through chalky ground where it is likely to have calcium and magnesium ions dissolve in it, making the water very hard.

Appendix 1

0.1g Calgon

First, I need to convert the volume of the EDTA solution into dm3 from cm3, by dividing 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 x volume

= 0.01 mol dm3 x 0.00186 dm3

= 0.0000186 mol

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

0.0000186 mol x 1000 = 0.0186 mmol

This means that 25ml of water plus 0.1g calgon contains a total of 0.0186 mmol of Ca2+ and Mg2+ ions. Now, I am going to work out the total metal concentration in units of millimoles per litre of water. In order to do this, I have to divide the number of millimoles of Ca2+ and Mg2+ ions; by the volume of the water I used which was 25cm3. This figure now has to be multiplied by 1000 to convert it back into millimetres from litres.

0.0186/25 x 1000 = 0.744 millimoles per litre of water of total metal concentration

An assumption has to be made next that the total hardness of mmol of Ca2+ and Mg2+ ions per litre of water is equivalent to 1 mmol of dissolved CaCO3 per litre of water. Using the molar mass of CaCOÂ3 as 100 gmol-1, I can convert the millimoles equivalent to CaCOÂ3 to milligrams equivalent to CaCOÂ3. Then we take the number of millimoles of equivalent CaCOÂÂÂÂ3 per litre, divide it by 1000, to convert millimoles into moles, then multiply it by 1000, in order to convert the grams into milligrams and then multiply this figure by .

0.744/1000 = 0.000744 x 1000= 0.744 x 100 = 74.4 milligrams of equivalent CaCOÂ3

I am able to express this is as the number of parts of equivalent CaCOÂ3 per million parts of water because the density of water is 1.00g mL-1

Therefore the hardness of the water with no calgon can be added can be expressed as 74.4 ppm of equivalent CaCOÂ3.

0.2g Calgon

First, I need to convert the volume of the EDTA solution into dm3 from cm3, by dividing 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 x volume

= 0.01 mol dm3 x 0.0015 dm3

= 0.000015 mol

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

0.000015 mol x 1000 = 0.015 mmol

This means that 25ml of water plus 0.2g calgon contains a total of 0.015 mmol of Ca2+ and Mg2+ ions. Now, I am going to work out the total metal concentration in units of millimoles per litre of water. In order to do this, I have to divide the number of millimoles of Ca2+ and Mg2+ ions; by the volume of the water I used which was 25cm3. This figure now has to be multiplied by 1000 to convert it back into millimetres from litres.

0.015/25 x 1000 = 0.6 millimoles per litre of water of total metal concentration

An assumption has to be made next that the total hardness of mmol of Ca2+ and Mg2+ ions per litre of water is equivalent to 1 mmol of dissolved CaCO3 per litre of water. Using the molar mass of CaCOÂ3 as 100 gmol-1, I can convert the millimoles equivalent to CaCOÂ3 to milligrams equivalent to CaCOÂ3. Then we take the number of millimoles of equivalent CaCOÂÂÂÂ3 per litre, divide it by 1000, to convert millimoles into moles, then multiply it by 1000, in order to convert the grams into milligrams and then multiply this figure by .

0.6/1000 = 0.0006 x 1000= 0.6 x 100 = 60 milligrams of equivalent CaCOÂ3

I am able to express this is as the number of parts of equivalent CaCOÂ3 per million parts of water because the density of water is 1.00g mL-1

Therefore the hardness of the water with no calgon can be added can be expressed as 60 ppm of equivalent CaCOÂ3.

0.3g Calgon

1.23/1000 = 0.00123 dm3

Number of moles = concentration x volume

= 0.01 mol dm3 x 0.00123 dm3

= 0.0000123 mol

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

0.0000123 mol x 1000 = 0.0123 mmol

This means that 25ml of water plus 0.3g calgon contains a total of 0.0123 mmol of Ca2+ and Mg2+ ions. Now, I am going to work out the total metal concentration in units of millimoles per litre of water. In order to do this, I have to divide the number of millimoles of Ca2+ and Mg2+ ions; by the volume of the water I used which was 25cm3. This figure now has to be multiplied by 1000 to convert it back into millimetres from litres.

0.0123/25 x 1000 = 0.492 millimoles per litre of water of total metal concentration

An assumption has to be made next that the total hardness of mmol of Ca2+ and Mg2+ ions per litre of water is equivalent to 1 mmol of dissolved CaCO3 per litre of water. Using the molar mass of CaCOÂ3 as 100 gmol-1, I can convert the millimoles equivalent to CaCOÂ3 to milligrams equivalent to CaCOÂ3. Then we take the number of millimoles of equivalent CaCOÂÂÂÂ3 per litre, divide it by 1000, to convert millimoles into moles, then multiply it by 1000, in order to convert the grams into milligrams and then multiply this figure by .

0.492/1000 = 0.000492 x 1000= 0.492 x 100 = 49.2 milligrams of equivalent CaCOÂ3

Therefore the hardness of the water with no calgon can be added can be expressed as 49.2 ppm of equivalent CaCOÂ3.

0.5g Calgon

1.03/1000 = 0.00103 dm3

Number of moles = concentration x volume

= 0.01 mol dm3 x 0.00103 dm3

= 0.0000103 mol

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

0.0000103 mol x 1000 = 0.0103 mmol

This means that 25ml of water plus 0.5g calgon contains a total of 0.0103 mmol of Ca2+ and Mg2+ ions. Now, I am going to work out the total metal concentration in units of millimoles per litre of water. In order to do this, I have to divide the number of millimoles of Ca2+ and Mg2+ ions; by the volume of the water I used which was 25cm3. This figure now has to be multiplied by 1000 to convert it back into millimetres from litres.

0.0103/25 x 1000 = 0.412 millimoles per litre of water of total metal concentration

0.412/1000 = 0.000412 x 1000= 0.412 x 100 = 41.2 milligrams of equivalent CaCOÂ3

Therefore the hardness of the water with no calgon can be added can be expressed as 41.2 ppm of equivalent CaCOÂ3.

0.8g Calgon

0.86/1000 = 0.00086 dm3

Number of moles = concentration x volume

= 0.01 mol dm3 x 0.00086 dm3

= 0.0000086 mol

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

0.0000086 mol x 1000 = 0.0086 mmol

This means that 25ml of water plus 0.8g calgon contains a total of 0.0086 mmol of Ca2+ and Mg2+ ions. Now, I am going to work out the total metal concentration in units of millimoles per litre of water. In order to do this, I have to divide the number of millimoles of Ca2+ and Mg2+ ions; by the volume of the water I used which was 25cm3. This figure now has to be multiplied by 1000 to convert it back into millimetres from litres.

0.0086/25 x 1000 = 0.344 millimoles per litre of water of total metal concentration

0.344/1000 = 0.000344 x 1000= 0.344 x 100 = 34.4 milligrams of equivalent CaCOÂ3

Therefore the hardness of the water with no calgon can be added can be expressed as 34.4 ppm of equivalent CaCOÂ3.

1g Calgon

0.76/1000 = 0.00076 dm3

Number of moles = concentration x volume

= 0.01 mol dm3 x 0.00076 dm3

= 0.0000076 mol

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

0.0000076 mol x 1000 = 0.0076 mmol

This means that 25ml of water plus 1.0g calgon contains a total of 0.0076 mmol of Ca2+ and Mg2+ ions. Now, I am going to work out the total metal concentration in units of millimoles per litre of water. In order to do this, I have to divide the number of millimoles of Ca2+ and Mg2+ ions; by the volume of the water I used which was 25cm3. This figure now has to be multiplied by 1000 to convert it back into millimetres from litres.

0.0076/25 x 1000 = 0.304 millimoles per litre of water of total metal concentration

0.304/1000 = 0.000304 x 1000= 0.304 x 100 = 30.4 milligrams of equivalent CaCOÂ3

Therefore the hardness of the water with no calgon can be added can be expressed as 30.4 ppm of equivalent CaCOÂ3.

1.2g Calgon

0.7/1000 = 0.0007 dm3

Number of moles = concentration x volume

= 0.01 mol dm3 x 0.0007 dm3

= 0.000007 mol

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

0.000007 mol x 1000 = 0.007 mmol

This means that 25ml of water plus 1.2g calgon contains a total of 0.007 mmol of Ca2+ and Mg2+ ions. Now, I am going to work out the total metal concentration in units of millimoles per litre of water. In order to do this, I have to divide the number of millimoles of Ca2+ and Mg2+ ions; by the volume of the water I used which was 25cm3. This figure now has to be multiplied by 1000 to convert it back into millimetres from litres.

0.007/25 x 1000 = 0.28 millimoles per litre of water of total metal concentration

0.28/1000 = 0.00028 x 1000= 0.28 x 100 = 28 milligrams of equivalent CaCOÂ3

Therefore the hardness of the water with no calgon can be added can be expressed as 28 ppm of equivalent CaCOÂ3.

1.3g Calgon

0.6/1000 = 0.0006 dm3

Number of moles = concentration x volume

= 0.01 mol dm3 x 0.0006 dm3

= 0.000006 mol

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

0.000006 mol x 1000 = 0.006 mmol

This means that 25ml of water plus 1.3g calgon contains a total of 0.006 mmol of Ca2+ and Mg2+ ions. Now, I am going to work out the total metal concentration in units of millimoles per litre of water. In order to do this, I have to divide the number of millimoles of Ca2+ and Mg2+ ions; by the volume of the water I used which was 25cm3. This figure now has to be multiplied by 1000 to convert it back into millimetres from litres.

0.006/25 x 1000 = 0.24 millimoles per litre of water of total metal concentration

0.24/1000 = 0.00024 x 1000= 0.24 x 100 = 24 milligrams of equivalent CaCOÂ3

Therefore the hardness of the water with no calgon can be added can be expressed as 24 ppm of equivalent CaCOÂ3.