Environmental Chemistry - Water Hardness PDF

**[Environmental Chemistry -- Water]** **[Hardness of Water]** \*\*Definition: Hard water is water that will not easily form a lather with soap. Hardness is water is caused by Ca^2+^ or Mg^2+^ ions dissolved in the water.\*\* \- One of the most common substances in soap is a chemical called Sodiu...

**[Environmental Chemistry -- Water]** **[Hardness of Water]** \*\*Definition: Hard water is water that will not easily form a lather with soap. Hardness is water is caused by Ca^2+^ or Mg^2+^ ions dissolved in the water.\*\* \- One of the most common substances in soap is a chemical called Sodium Stearate C~17~H~35~COONa. When soap is added to Hard Water, the calcium or magnesium ions in the water react with stearate ions to form a grey insoluble compound called calcium stearate which floats on the water\ Ca^2+^ + 2C~17~H~35~COO^-^ (C~17~H~35~COO)~2~Ca↓ Hardness in water may be defined as temporary or permanent \*\*Definition: Temporary hardness is hardness that can be removed by boiling the water.\*\* \- Rainwater is slightly acidic due to the fact that it dissolves Carbon Dioxide as it falls through the air H~2~O + CO~2~ H~2~CO~3~ (carbonic acid) \- The carbonic acid then reacts with limestone forming calcium hydrogen carbonate. CaCO~3~ + H~2~CO~3~ Ca(HCO~3~)~2~ (calcium hydrogen carbonate) \- Temporary hardness is caused by Ca(HCO~3~)~2~ and Mg(HCO~3~)~2~ (Calcium hydrogencarbonate and Magnesium hydrogencarbonate) \- This can be removed by boiling: Ca(HCO~3~)~2~ CaCO~3~ + CO~2~ + H~2~O \- The chemical reaction of boiling removes Ca^2+^ ions which softens the water \*\*Definition: Permanent hardness is hardness which cannot be removed by boiling the water.\*\* Permanent hardness is caused by the presence of calcium sulfate, CaSO~4~, or magnesium sulfate, MgSO~4~, in the water. When water containing these substances is heated, no chemical reaction occurs to precipitate the Ca^2+^ or Mg^2+^ ions. Hardness is not removed by boiling. **[Methods of Removing Hardness from water]** \- Distillation: Water is boiled. All dissolved solids which cause water hardness are left behind in the residue. This is too expensive to do on a large scale.\ - Using Washing Soda: Washing soda (hydrated sodium carbonate, Na~2~CO~3~.10H~2~O) reacts with Ca^2+^ ions in the water to form calcium carbonate, which is insoluble in water\ - Ion-exchange resin: The Ca^2+^ and Mg^2+^ ions in water are exchanged for Na^+^ ions, which do not cause water hardness. Because this resin exchanges positive ions, it is called a cation exchange resin. **[Softening Water Within a House (Using Ion-Exchange Resin)]** \- A cation-exchange resin exchanges the positive ions in the water for Na^+^ ions\ - Each Ca^2+^ ion in the hard water is replaced by two Na^+^ ions from the resin\ - A cation-exchange resin is a fairly complicated molecule byt may be represented as RNa. The exchange is represented as:\ Ca^2+^ + 2RNa R~2~Ca + 2Na^\ ^- The Ca^2+^ ions remain behind in the resin. Eventually, all the resin loses all of its Na^+^ ions, and it needs to be replenished by passing a concentrated sodium chloride through it. The Ca^2+^ and Mg^2+^ ions in the resin are replaced by the Na^+^ ions and the resin is ready for use again. **[Producing Deionised Water for Laboratory Use (Using Ion-Exchange Resin)\ ]**- Deionised water Is produced by passing ordinary water through a water deioniser that contains a mixture of cation exchange resin and anion exchange resin. This is called a mixed-bed resin\ - The cation-exchange resin removes all the positive ions and the anion-exchange resin removes all the negative ions from the water\ - The cation-exchange resin, represented as RH, replaces the Na^+^ ions with H^+^ ions as follows:\ RH + Na+ RNa + H^+^ \- The Cl^-^ ions are replaced by the anion-exchange resin as follows:\ ROH + Cl^-^ RCl + OH^-\ ^- The positive ions in the water are replaced by H^+^ ions and all the negative ions in the water are replaced by OH^-^ ions. The H^+^ and OH^-^ ions then combine together to from water. **[Deionised and Distilled Water]** \- Deionised water has all suspended and dissolved solids removed, but there may still be gases dissolved in the water.\ - Distilled water is the purest form of water as all suspended and dissolved solids, along with any dissolved gases are removed from the water. [Advantages and Disadvantages of Hard Water\ ] Advantages Disadvantages --- -------------------------------------- --------------------------------------- 1 Provides Calcium for teeth and bones Blocks pipes, leaves scale on kettles 2 Nicer taste Wastes soap 3 Good for brewing and tanning leather Produces scum **[Mandatory Experiment: To determine the total hardness in a water sample using ethylenediaminetetraacetic acid (edta)]** 1. Wash the pipette, burette and conical flask with deionised water. Rinse the burette with the edta solution and the pipette with the hard water. 2. Using the funnel, fill the burette with the edta solution. Open the tap briefly to fill the part below the tap. Remove the funnel. Adjust the level of the solution to the zero mark. Make sure that the burette is vertical. 3. Use the pipette to transfer 50 cm^3^ of the hard water sample to the conical flask. Add 2-3 cm^3^ of the buffer (pH 10) solution (measured out using the graduated cylinder). 4. Add 0.03 g of the solid indicator to the contents of the flask in the following manner: Add gradually to the flask, swirling after each addition. A deep wine red colour is obtained. 5. Carry out one \'rough\' titration to find the approximate end point, followed by a number of accurate titrations until two titres agree to within 0.1 cm^3^. At the end point, the colour should be dark blue, with no tinge of wine-red colour. 6. From the data, calculate the total hardness of the water sample A sample of water was analysed in order to determine the total amount of hardness in the water. 50 cm^3^ of the water were placed in a conical flask and some buffer solution and eriochrome black t were added. The solution was titrated against 0.01 M edta solution. The average titration figure was 15.2 cm^3^. Calculate the total amount of hardness in the water. In calculations involving edta, the mole ratio is always 1:1 V~ca~ = 50 [Vca x Mca] = [V~ed~ x M~ed~]\ M~ca~ = ? n~ca~ n~ed~\ n~ca~ = 1 [50 x M~ca~ ] = [15.2 x 0.01]\ V~ed~ = 15.2 1 1\ M~ed~ = 0.01 M~ca~ = [15.2 x 0.01]\ n~ed~ = 1 50 M~ca~ = 0.003 moles/L CaCO~3~\ = 0.004 x 100 g/L (rel. Molecular mass of CaCO~3~ = 100)\ = 0.3 g/L CaCO~3~\ = 0.3 x 1000 mg/L CaCO~3~\ = 300 mg/L CaCO~3~\ = 300 ppm CaCO~3~ Total hardness of water =300 ppm **[Water Treatment]** [1. Screening\ ]- The water is passed through a wire mesh to remove any floating debris e.g., [ ] branches and plastic bags [2. Flocculation\ ]- Small suspended particles in water are coagulated i.e., forming larger particles for better sedimentation.\ - Flocculating agents \[Al~2~(SO~4~)~3~\] are responsible for this. [3. Sedimentation\ ]- The water is passed into a settlement tank.\ - Water goes into the bottom of the tank and rises to allow for maximum settlement.\ - 90% of particles are removed this way. [4. Filtration\ ]- Water now passes through beds of sand to remove the remaining particles. [5. Chlorination\ ]- Chlorine is added to remove to the water to sterilise it i.e., kill any harmful microorganisms. [6. Fluoridation\ ]- The addition of fluorine helps prevent tooth decay.\ - It is required by law to be added in Ireland [7. pH adjustment\ ]- Optimum pH is 7.2, acids or bases may need to be added to raise or lower the pH Name of Chemical Purpose Problems caused by adding excess ------------------- ------------------------------------------- --------------------------------------------- Aluminium Sulfate Coagulation of small, suspended particles Taste of water affected, corrosion of pipes Chlorine Sterilise water Taste and smell of water affected Fluorine compound Reduce tooth decay Staining of teeth Calcium hydroxide Raise pH Hardness of water Sodium carbonate Soften water Taste of water affected Sulfuric acid Lower pH Corrosion of pipes **[Mandatory Experiment: To determine (a) the total suspended solids in (in ppm) of a sample of water by filtration, (b) the total dissolved solids (in ppm) of a sample of water by evaporation. In part (c) the pH of a sample of water will be measured.]** (A)1. Find the mass of a dry filter paper. 2\. Filter 1 litre of the water sample 3\. Dry the filter paper at about 105 ^o^C overnight. 4\. Find the mass of the dried filter paper. 5\. Calculate the mass of total suspended solids in mg/l (p.p.m.). (B)6. Find the mass of a clean dry 250cm^3^ beaker. 7\. Add 100cm^3^ of a filtered water sample to the beaker. 8\. Evaporate to dryness in a dust-free oven at 105 ^o^C. 9\. Find the mass of the cool dry beaker. 10\. Calculate the total dissolved solids in the water sample in mg/l (p.p.m.). (C)11. Find the pH of each water sample by taking the respective readings on the pH meter, or by comparing the colours obtained using pH paper or universal indicator solution with the appropriate colour charts. 12\. Record your results A student filtered 500 cm^3^ of water and found that the mass of the filter paper after drying had increased by 0.68 g. 100 cm^3^ of the filtered water were then evaporated to dryness in a beaker and it was found that the mass of the beaker had increased by 0.13 g.\ What is the amount of suspended solids and the amount of dissolved solids in the water expressed in parts per million (ppm)? Total suspended solids = 0.68 g/500 cm^3\ ^ = 0.68 x 2 g/L\ = 0.68 x 2 x 1000 g/L\ = 1360 p.p.m\ Total dissolved solids = 0.13 g/100 cm^3^\ = 0.13 x 10 g/L\ = 0.13 x 10 x 1000 mg/L\ = 1300 ppm Total suspended solids = 1360 ppm\ Total dissolved solids = 1300 ppm **[Water Pollution\ ]**- Water contains dissolved oxygen due to aquatic plants photosynthesising under water. This is necessary for all living things under water.\ - Oxygen levels in water need to be monitored to ensure the safety of a habitat.\ - B.O.D. tests are undertaken to measure the amount of dissolved oxygen present in water. \*\*Definition: Biochemical Oxygen Demand (B.O.D) is defined as the amount of dissolved oxygen, consumed by biological action when a sample of water is kept at 20^o^C in the dark for five days.\*\* [B.O.D Testing] \- Completely fill two bottles with the water to be tested\ - The test is carried out immediately on one of the bottles so potential microorganisms in the water can\'t affect the figure\ - The Winkler method is carried out or use an oxygen sensor\ - The second bottle is kept in complete darkness for 5 days this prevents possible photosynthesis\ - Fixed temperature ensures fair comparisons.\ - After 5 days, the second bottle is tested. This allows time for microorganisms to act\ - The B.O.D. is the difference in the two dissolved oxygen levels as this is the amount of oxygen that has been used up in the test\ - High B.O.D. indicates low oxygen levels, meaning a high level of microorganism's present **[Pollution caused by Eutrophication]** \*\*Definition: Eutrophication is the enrichment of water with nutrients, leading to excessive growth of algae and other plants.\*\* \- Caused by fertilisers and sewage which contain high levels of nitrates and phosphates. \- An increase in nitrates and phosphates to a body of water causes plants and algae to grow rapidly\ - By doing so they cover the surface of the water which in turn prevents sunlight entering the water\ - Plants need sunlight to photosynthesis and produce oxygen **[Pollution caused by 'Heavy Metal' ions]** \- Lead ions (Pb^2+^) are heavy metal ions that could have serious consequences for organisms if they build up in a body of water.\ - Atomic absorption spectrometry (AAS) is used to detect these metals in water.\ - A process called precipitation is used to remove heavy metal ions. These metals are found in most water however, it is only once they go above a certain concentration that they become dangerous. **[Mandatory Experiments: To measure the amount of dissolved oxygen in a sample of water by means of a redox reaction]** 1. Rinse a 250 cm^3^ reagent bottle with deionised water, shaking vigorously to wet the inside and so avoid trapped air bubbles. 2. Completely fill the bottle under water with the sample, making sure that there are no trapped air bubbles. 3. Using a dropper placed well below the surface of the water, add 1 cm^3^ (approximately) each of manganese(II) sulfate solution and of alkaline potassium iodide solution to the bottle. 4. Stopper the bottle so that no air is trapped - a few cm^3^ of solution will overflow at this point. 5. Invert the bottle repeatedly for about a minute, and then allow the brown precipitate to settle out. 6. In order to dissolve the precipitate, carefully add 1 cm^3^ of concentrated sulfuric acid to the bottle, by running the acid down the side of the bottle. 7. Restopper the bottle - avoiding trapping any air. Invert repeatedly to redissolve the precipitate. If all the precipitate has not dissolved at this point add 0.5 cm^3^ of acid, invert repeatedly, and allow to stand for one minute. Continue repeating the process of inverting repeatedly followed by addition of acid until all of the precipitate has dissolved. Iodine should now be released resulting in a golden brown solution. 8. Wash the pipette, burette and conical flask with deionised water. Rinse the pipette with the iodine solution and the burette with the sodium thiosulfate solution. 9. The free iodine can now be estimated by means of the thiosulfate titration. Measure out 50 cm^3^ samples into clean conical flasks and titrate each of these with 0.005 M thiosulfate solution in the usual way. 10. Add about 1 cm^3^ starch indicator as the end point approaches (when the solution becomes pale yellow in colour). Titrate until the blue colour has just disappeared. 11. Record the results in the usual way taking the average of two accurate titration results, i.e. two titres within 0.1 cm^3^ of each other. 12. Calculate the results in (i) moles of oxygen per litre (ii) grams of oxygen per litre (iii) dissolved oxygen in p.p.m. **[Sewage Treatment]** \- Sewage treatment is the term used to describe the material delivered to waste treatment plants of towns and cities. Sewage treatment is essential to prevent waste getting into the water system as serious health consequences could arise otherwise. **[Primary Treatment\ ]**\*\*Definition: Primary treatment of sewage is a mechanical process in which large solids are removed by screening and some suspended solids are removed by settlement (sedimentation).\*\* **[Secondary Treatment\ ]**\*\*Definition: Secondary treatment of sewage is a biological oxidation process in which the levels of suspended and dissolved organic materials are reduced i.e., the sewage is decomposed by means of bacteria which use the nutrients from the sewage together with oxygen from the air to break down the sewage.\*\*\ - Using bacteria is called the activated sludge process.\ - Any sludge removed is used as nutrients for fertilisers or possible as a fuel **[Tertiary Treatment\ ]**\*\*Definition: Tertiary treatment of sewage is a process involving the removal of phosphorous compounds by precipitation and the removal of nitrogen compounds by biological and ion-exchange methods.\*\*\ - Prevents Eutrophication **[Instrumental Methods of Water Analysis]** \- pH Analysis: pH can be measured using a pH sensor.\ - Atomic Absorption Spectroscopy (AAS): Used to detect the concentration of heavy metals in water. It works on the principal that each element has its own unique absorption spectrum, so that the element absorbs specific frequencies of light. The amount of light absorbed tells us the concentration of that element, e.g., lead.\ - Colorimetry: Used to measure the concentration of coloured substances. Works on the principle that the more light the coloured solution absorbs, the higher its concentration \*\*Definitions: The principle of colorimetry is that the amount of absorbance of light by a coloured solution is proportional to the concentration of the solution.\*\* **[Mandatory Experiment: To estimate the concentration of free chlorine in swimming pool water of bleach using (a) a comparator or (b) a colorimeter]** 1. Add 2.5 cm^3^ of Milton Sterilising fluid to a 250 cm^3^ volumetric flask and dilute to the mark with deionised water. 2. To a series of five 50 cm^3^ volumetric flasks add 5 cm^3^ of 5% ethanoic acid solution. 3. Transfer the diluted Milton solution to a burette. To the first 50 cm^3^ flask add nothing at this stage; to the second 50 cm^3^ flask add 1.0 cm^3^ of this solution. To the third, fourth and fifth flasks, add 2.0 cm^3^, 4.0 cm^3^ and 8.0 cm^3^ of the solution respectively. 4. Using a burette, transfer 5.0 cm^3^ of 2% potassium iodide solution to each flask and dilute to the mark with deionised water. Stopper each flask, mix thoroughly, and allow about five minutes for the colour to develop. These solutions are now the working standards. Label the flasks A, B, C, D and E respectively. 5. Switch on the colorimeter and place a 440 nm wavelength filter in the filter slot. Pour each working standard into a cuvette, rinsing each cuvette first with the solution it is to contain. 6. Using the operating procedure for the colorimeter, as per the manufacturer's instruction book, zero the instrument. 7. Obtain the absorbance for each standard, starting with the most dilute. Rinse the sample cells with deionised water after each sample has been used. 8. Plot a graph of absorbance versus concentration (in terms of chlorine) for the series of standards. 9. To a 50 cm^3^ volumetric flask add 5 cm^3^ of 5% ethanoic acid solution, and then 5.0 cm^3^ of 5% potassium iodide solution. Fill the flask up to the mark with the swimming pool water or diluted bleach. Allow about five minutes for the colour to develop. Label this flask as flask F. 10. Obtain the absorbance for the solution in flask F. 11. From the graph, obtain the concentration of NaOCl in the sample. 12. Multiply by 71 / 74.5 to calculate the concentration of free chlorine in the sample (b) 1. Rinse the compartments of the comparator with portions of the water sample under test and discard the rinsings. 2. Fill each compartment to their graduation marks with fresh portions of the sample. 3. Add a DPD No.1 tablet to each appropriate compartment, and crush each tablet with a stirring rod. 4. Fit the lid and shake until the tablets have dissolved completely. 5. Compare the colour produced in the sample with the pre-calibrated standards, using daylight rather than artificial light as an aid. 6. Choose the best colour match, and read the free chlorine concentration of the selected standard in p.p.m. (mg l^-1^). **[Exam Questions]** [2012 -- HL -- Section B -- Question 7] 7\. (a) Explain how hard water is caused and how it wastes soap.\ - Caused by calcium (magnesium) hydrogencarbonate.\ - Soap used up in reaction with Ca and Mg ions to give scum.\ How can hard water be softened by ion exchange so that it is suitable for use as deionised water in the laboratory?\ - Pass through resin to replace positive ions with hydrogen ions (3) and negative ions with hydroxyl ions\ (b) In water treatment, what is the purpose of adding each of the following:\ flocculating agent: Clumping of fine particles.\ Chlorine: Kills pathogens.\ Fluorine-containing compound: Prevents tooth decay.\ Calcium hydroxide: Raises\ Sulfuric acid: Lowers pH.\ State the problem that would arise when each of any two of these substances is added in excessive quantity.\ Chlorine: Water tastes of chlorine.\ Sulfuric acid: Corrodes pipes\ (c) Why is water pollution by heavy metal ions, e.g. Hg^2+^ or Pb^2+,^ a cause of concern? It is a danger to health\ Name an instrumental technique that could be used to detect and measure the concentration of a heavy metal ion in a water sample. Atomic Absorption Spectroscopy\ Explain how Hg^2+^ or Pb^2+^ ions can be removed from a water supply. Precipitation\ (d) Describe a test for the presence of chloride ion (Cl^--^ ) in water.\ - Add silver nitrate and dilute nitric acid.\ - White precipitate forms. [2011 -- HL -- Section B -- Question 11] 11\. Answer any two of the parts (a), (b) and (c)\ (a) What is meant by the biochemical oxygen demand (BOD) of a water sample?\ ppm of oxygen consumed when sample kept in the dark for 5 days at 20^o^C\ The BOD of a raw sewage sample was 350 ppm and the BOD of the same sample after treatment was about 25 ppm.\ (i) Describe how the BOD was reduced by about 30% in primary sewage treatment. Removing of solids by screening and settlement\ (ii) Explain the processes by which the BOD was further reduced in secondary sewage treatment.\ - Biological.\ - Aerobic.\ - Decomposition of waste using activated sludge process