Slide 1. Slide 2. Slide 3. Hardness. Concentration is. What s the concentration of red triangles? What s in your pipes? 500 ml

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Slide 1 Hardness What s in your pipes? Slide 2 What s the concentration of red triangles? 500 ml 1 g 1 g 1 g A. 10 B. 10 C. D. 1 g 1 g It s all of the above! Slide 3 Concentration is any statement of the relationship between the amount of stuff ( solute ) dissolved in a solvent/solution.

Slide 4 Could be ANYTHING Slide 5 UNITS! UNITS! UNITS! The units are your friend ALWAYS! The units tell you how to measure your stuff. So, if I ve got Molarity (M), I probably want to measure the volume Slide 6

Slide 7 If I have Then I want to measure GRAMS of solution! Slide 8 Concentration is always just a conversion factor between the way you measured the solution and how much solute you ve got! The SOLUTE is almost always the thing you care about. The solvent/solution is just the carrier. Slide 9 It s a question of what they DO! All I really see is what they do. I never actually see them. This is really the take home lesson for waste water analysis: how specific is your test? Mg 2+ Mg 2+ Mg 2+ Mg 2+

Slide 10 Alkalinity We saw this with alkalinity. We don t really know what the base is, we only know how much acid it eats. For alkalinity, this is the only thing that matters. And from a site standpoint, that s all you ll care about. Slide 11 Sometimes, what you don t know will kill you. The difference between OH - and HCO 3- is unimportant. What about the difference between iron (Fe) and lead (Pb)? Pb could shut your site down if you find it. Pb could get your butt sued down the line if you don t find it. Slide 12 Spend the bucks on what counts Some tests are more specific than others. Spend your testing budget on the things that matter. Total alkalinity is usually enough. Total metals is usually NOT enough.

Slide 13 Hardness We experience hardness of water directly in several ways: 1. A slimy feel to our water when bathing. 2. Reduced lather or foaming in soaps. 3. Formation of scale in pipes and near drains. Slide 14 Chemical Identity of Hardness Hardness is caused by dissolved metal ions. These ions can form precipitates (with things like soap) which result in water-insoluble scale. Slide 15 Every Cation has its Anion Metal Cations Most common anion Ca 2+ HCO - 3 Mg 2+ SO 2-4 Sr 2+ Cl - Fe 2+ NO - 3 Mn 2+ SiO 2-3 Do you recognize these species?

Slide 16 Every Cation has its Anion Metal cations Most common anion Ca 2+ (calcium) HCO 3- (bicarbonate) Mg 2+ (magnesium) SO 2-4 (sulfate) Sr 2+ (strontium) Cl - (chloride) Fe 2+ (iron) NO 3- (nitrate) Mn 2+ (manganese) SiO 2-3 (silicate) What happens when they meet? Slide 17 Every Cation has its Anion Metal cations Most common anion Ca 2+ (calcium) HCO 3- (bicarbonate) calcium bicarbonate - Ca(HCO 3 ) 2 Mg 2+ (magnesium) SO 2-4 (sulfate) magnesium sulfate MgSO 4 Sr 2+ (strontium) Cl - (chloride) strontium chloride SrCl 2 Fe 2+ (iron) NO 3- (nitrate) iron nitrate Fe(NO 3 ) 2 Mn 2+ (manganese) SiO 2-3 (silicate) manganese silicate MnSiO 3 Slide 18 And the problem is all of the compounds are water-insoluble solids.

Slide 19 How do you make a precipitate? How do I make a water-insoluble precipitate with water? I need two sources of ions could even be two water sources. I need to decrease the water and increase the concentration of the ions until I am below the solubility. Slide 20 What is solubility? Quick Review It is the MAXIMUM amount of a substance that will dissolve in a liquid. If I decrease the volume of water to increase the concentration, eventually I have a supersaturated solution and the solid precipitates. Slide 21 Determining Hardness If you are looking for hardness, what are you actually searching for? Metal ions! What s the easiest way to quantify the amount of metal ions? TITRATE THEM!

Slide 22 Titrations you can t escape em (ethylenediaminetetraacetic acid) is a chemical compound that binds to most metal ions, especially divalent species (charges of 2+). In any titration, what do you need? Slide 23 Titrations Balanced chemical equation Indicator of equivalence Slide 24 Balanced equation M 2+ + 4- [M-] 2- + 2H + (the H + comes from the ) The important point is that the reaction is 1:1

Slide 25 It s a question of what they DO! Any of the divalent metals (not to mention a few others) will bind to That makes the NON-SPECIFIC! Mg 2+ Ca 2+ Mg 2+ Pb 2+ Slide 26 Indicator, M 2+, and M- are all soluble and colorless. So, you won t see any change We need a secondary indicator a second chemical reaction that will result in some visible change. Slide 27 A couple of possible indicators Calmagite or Eriochrome Black T are blue dyes when alone in water. When it is complexed with a Metal ion, it turns red. How does this help you? What would you see?

Slide 28 Initially (before is added): M 2+ + dye M 2+ -dye blue red When you begin to add : M 2+ + M- M 2+ + dye M 2+ -dye blue red At equivalence ([]=[M]): M 2+ + M- Dye (blue) Mg 2+ Ca 2+ Mg 2+ Pb 2+ Slide 29 Initially, there is NO Initially (before is added): M 2+ + dye M 2+ -dye blue red The indicator is the ONLY thing that binds to the metal. Then you start titrating.. Mg 2+ Ca 2+ Mg 2+ Pb 2+ Slide 30 Now there IS Initially (before is added): M 2+ + dye M 2+ -dye blue red When you begin to add : M 2+ + M- M 2+ + dye M 2+ -dye blue red The can bind to the metal also. Eventually, every metal has either an or an indicator then Mg 2+ Ca 2+ Mg 2+ Pb 2+

Slide 31 Something has to win the competition Initially (before is added): M 2+ + dye M 2+ -dye blue red When you begin to add : M 2+ + M- M 2+ + dye M 2+ -dye blue red Mg 2+ Ca 2+ If the indicator is a better binder than the, I m done for I ve got a mix of binders but I have no way to know when I reach the endpoint. Mg 2+ Pb 2+ Slide 32 Something has to win the competition Initially (before is added): M 2+ + dye M 2+ -dye blue red When you begin to add : M 2+ + M- M 2+ + dye M 2+ -dye blue red Mg 2+ Ca 2+ My solution will get purplish. I ve got some red complex and I ve got some free blue dye. At equivalence Mg 2+ Pb 2+ Slide 33 Something has to win the competition My solution will get purplish. I ve got some red complex and I ve got some free blue dye. At equivalence all the metal has and the dye is all free. At equivalence ([]=[M]): M 2+ + M- Dye (blue) Mg 2+ Ca 2+ Mg 2+ Pb 2+

Slide 34 This is a tricky endpoint Your solution will start red (all bound metal-indicator complex) As you add, it eventually gets purple (mix of red metal-indicator complex and then free blue indicator dye) At the endpoint it goes from purplish to straight blue. You are looking for the end of any red color. Slide 35 An example 10.00 ml of a waste water sample is dilute to 50 ml total volume. Titration with a 0.2150 M solution shows a Calmagite endpoint after addition of 36.23 ml. What is the total hardness of the water sample? Slide 36 What is total hardness? Total hardness means that we are not differentiating the different metals present. Generally, total hardness is taken as the sum of calcium hardness and magnesium hardness. (Other metals are just lumped into those 2)

Slide 37 An example 10.00 ml of a waste water sample is dilute to 50 ml total volume. Titration with a 0.02150 M solution shows a Calmagite endpoint after addition of 36.23 ml. What is the total hardness of the water sample? (10.00 ml) X = (36.23 ml) (0.02150 M) X = 0.07789 M Metals Slide 38 Why 10.00 ml and not 50.00 ml? Dilution does not change the amount of anything present! 1 L of water + 100 grams of sugar Add another L of water Slide 39 Why 10.00 ml and not 50.00 ml? 100 grams of sugar in both! Concentration is different, but we don t care. Why? Because the diluted sample is NOT my waste water.

Slide 40 Reactions are between molecules Reactions happen because 2 (or more) molecules stick together. It is only the number of molecules that count. Instead of 100 g of sugar, pretend I have 5 metal molecules. Slide 41 Reactions are between molecules If I react them with Slide 42 Reactions are between molecules 5 metal ions react with 5 ions no matter how much water.

Slide 43 An example 10.00 ml of a waste water sample is dilute to 50 ml total volume. Titration with a 0.02150 M solution shows a Calmagite endpoint after addition of 36.23 ml. What is the total hardness of the water sample? (10.00 ml) X = (36.23 ml) (0.02150 M) X = 0.07789 M Is Molarity a good unit? Molarity of what? Slide 44 Depends on what you mean by good Hardness is usually expressed in mg/l of CaCO 3 equivalents. Since, in this case, all the metals are lumped together, they are taken to be Ca 2+ equivalents Slide 45 0.07789 mol Ca 2+ * 1 mol CaCO 3 * 100.09 g CaCO 3 * 10 3 mg = 7796 mg/l L solution 1 mol Ca 2+ 1 mol CaCO 3 g 7796 mg/l as CaCO 3 would be how you would express this number. NOTE: There may be no Calcium carbonate in the sample at all!!! But we are expressing it as an equivalence.

Slide 46 Analytical Methods You can also determine metal concentrations using advanced instrumentation like atomic absorption spectroscopy (AAS) and inductively coupled plasma (ICP). Slide 47 Determining Ca and Mg separately With advanced techniques (other than titration), you can determine the Ca 2+ and Mg 2+ concentrations separately. These could be reported separately, or they could be combined into CaCO 3 equivalents. Slide 48 Sample problem AAS analysis of a water sample determined the Ca 2+ hardness to be 36 mg/l and the Mg 2+ hardness to be 16 mg/l. What is the total hardness expressed as CaCO 3 equivalents?

Slide 49 Units! Units! Units! This is really just a unit conversion problem. You need to recognize the stoichiometry is 1:1. MgCO 3 CaCO 3 There is 1 metal ion for each carbonate ion. Slide 50 36 mg Ca 2+ * 1 g * 1 mol Ca 2+ * 1 mol CaCO 3 *100.1 g CaCO 3 * 10 3 mg = 1 L 10 3 mg 40.1 g Ca 2+ 1 mol Ca 2+ 1 mol CaCO 3 1 g = 90 mg/l as CaCO 3 Similarly for Mg: 16 mg Mg 2+ * 1 mmol Mg * 1 mmol Ca 2+ * 1 mmol CaCO 3 *100.1 mg CaCO 3 = 1 L 24.3 mg Mg 2+ 1 mmol Mg 2+ 1 mmol Ca 2= 1 mmol CaCO 3 = 66 mg/l as CaCO 3 Total hardness as CaCO 3 = 90 mg/l + 66 mg/l = 156 mg/l Slide 51 Notice it s just the masses: 36 mg Ca 2+ * 1 g * 1 mol Ca 2+ * 1 mol CaCO 3 *100.1 g CaCO 3 * 10 3 mg = 1 L 10 3 mg 40.1 g Ca 2+ 1 mol Ca 2+ 1 mol CaCO 3 1 g = 90 mg/l as CaCO 3 Because the stoichiometry is 1:1, it s just the ratio of the masses: 36 mg Ca 2+ * 100.1 g CaCO 3 = 90 mg/l as CaCO 3 1 L 40.1 g Ca 2+

Slide 52 Good old carbonate You can also look at the hardness in terms of the anions. In this case: Total hardness = carbonate hardness + non-carbonate hardness Carbonate includes both bicarbonate and carbonate. This is really alkalinity they are kindred spirits! Cation (Ca 2+ et al) + anion (CO 3 2- et al) = CaCO 3 Hardness + alkalinity = CaCO 3 Slide 53 Why is carbonate special? CO 2 carbon dioxide from the air CaCO 3 - limestone Slide 54 Carbonate is singled out because it s nasty! Bicarbonate hardness: Ca 2+ (aq)+ 2 HCO 3 - (aq) CaCO 3 (s) + CO 2 (g) + H 2 O (l) Bicarbonate hardness in the presence of softeners!: Ca 2+ (aq)+ 2 HCO 3 - (aq) + Ca(OH) 2 (s) 2 CaCO 3 (s) + 2 H 2 O (l)

Slide 55 When CaCO 3 is not CaCO 3 NOTE that both hardness and alkalinity are measured in CaCO 3 equivalents but that doesn t mean they will ever be the same number. In one case, I m looking at metals. In the other case, I m looking at bases. Slide 56 Consider I ve got a total alkalinity of 100 mg CaCO 3 /L. What does that mean? It means that I ve got enough base to neutralize the same amount of acid as 100 mg CaCO 3 in each liter of my waste water. Suppose the actual species present is ammonia (NH 3 ). The ammonia is NOT CaCO 3 and has no metal ion at all. So the total hardness might be 0 mg CaCO 3 /L. Slide 57 On the flip side Suppose I have a hardness that is 100 mg CaCO 3 /L. That means I have as much metal ions as 100 mg of CaCO 3 in each liter of waste water. If the actual metal species present is Mg(NO 3 ) 2 there is NO base present. The total alkalinity will be 0 mg CaCO 3 /L!