Hard Water (1959)

[Music] The three metals calcium, berium, and strontium occur in the same family on the periodic table and have many similarities and properties. They all form hydroxides with alkaline characteristics and have long been known as the alkaline earth metals. In these test tubes, we have samples of the carbonates of barerium, calcium and franchium respectively. Now in order to keep these three straight uh during the experiment remember that the tubes are arranged in alphabetical order from left to right with berium on the left calcium in the middle and stranchium on the right. We will now add 5 ml of dilute hydrochloric acid to each tube and note the reaction. Recall that carbonates usually react with hydrochloric acid with effrovescent All three carbonates have reacted similarly with the hydrochloric acid having dissolved. And this means that the chloride of berium, calcium, and stranchium are all soluble. Now to each tube, we'll add a few drops of dilute sulfuric acid to test the solubility of the sulfate. And here we discover a difference because berium and stranchium sulfates are very insoluble and formed almost immediately while calcium sulfate is much less soluble and only formed slowly and then in much smaller amounts. In the three test tubes, we had place samples of berium, calcium and strontium carbonate to which we added hydrochloric acid. The reactions in all three cases are entirely similar and we obtain solutions of the respective chlorides, water and carbon dioxide. Then to solutions of the chlorides, we added some dilute sulfuric acid as a source of sulfate ion. In the case of berium and stranchium sulfate, a precipitate was obtained immediately because the solubility product constants for these two sulfates are very small 10 - 10 10 - 7. In the case of calcium sulfate, the precipitate formed rather slowly since calcium sulfate is much more soluble having a solubility product constant of about 10us4. We will now uh demonstrate the different colored flames that berium calcium and strontium ions give when solutions containing them are placed on platinum wires and held in the flame of the Bunson burner. The lights have now been turned out so that we can see the characteristic colors of the flames due to each ion. First we will test a solution containing the barerium ion. You should notice the greenish yellow flame imparted to the Bunson burner flame by the berium ion. Next we will uh test the calcium ion. This flame which is rather brief in duration is generally described as orange red. And last we'll test the stranchium ion. This ion gives a longerlasting crimson flame. In the test tube, we've placed a solution of calcium hydroxide. The tube leading down into the liquid is attached to a carbon dioxide generator. We will now put the generator into operation and cause carbon dioxide gas to bubble through the liquid in the tube. In a few minutes, you will see the white precipitate of calcium carbonate uh beginning to form. Precipitation will continue until uh virtually all of the calcium has been converted to calcium carbonate which will remain suspended as the white solid in the liquid. Now, as we continue to bubble carbon dioxide into the tube, the calcium carbonate will slowly dissolve, forming calcium bicarbonate, and the solution in the tube will again become clear. In the test tube, we placed a solution of calcium hydroxide. Then we added carbon dioxide gas and precipitated calcium carbonate. And then we illustrated that when calcium carbonate solid is exposed to solutions containing dissolved carbon dioxide, the calcium carbonate dissolves with the formation of calcium bicarbonate. This is what happens when rainwater containing these substances trickles over limestone rock. And this is the source of much of the calcium ion found in hard waters. We will now examine the reaction of metallic potassium with water. In this beaker, we place some distilled water to which has already been added some phenylene. The test tube has been filled with this uh distilled water phenothalene mixture and then inverted in the tube. Now I'm going to drop in a small piece of metallic calcium and place the tube over the calcium. You'll notice that the solution immediately becomes red indicating that uh calcium hydroxide has been formed and the tube is rapidly filled with gas. The test tube is now filled with hydrogen and we will demonstrate that it is hydrogen by mixing it with air and igniting it. the ferland meer flask uh simply filled with air. We'll place the test tube mouth to mouth with the flask and then invert the setup so that the hydrogen uh will diffuse rather rapidly up into the flask and the air uh will pass down to the test tube. Now uh we'll light a wooden splint at the Bunson burner. place the tube in the flask and from the explosions that we obtained hydrogen was obviously formed when the calcium reacted with the water in the beaker along with calcium hydroxide. You will recall that in the beaker we placed water containing phenaline and then added a piece of metallic calcium. Hydrogen was evolved rapidly and the phenthaline turned red indicating the presence of calcium hydroxide. We collected the hydrogen over water then mixed it with air and ignited it. An explosion took place as a result of the reaction between the hydrogen and oxygen to form water. Industrially calcium oxide or quick line is prepared by heating calcium carbonate to a high temperature. In the lab uh we can accomplish this transformation uh with the Bunson burner on the wire gauze. We have placed a small marble chip. marble of course is calcium carbonate and we are heating chip in the flame of the burner. The chip is slowly being converted from calcium carbonate to calcium oxide. In the two test tubes we placed some distilled water and we have added some phenaline to each tube. To the tube on the right, I will drop or add a marble chip which has not been heated. And you will notice that nothing much happened. Calcium carbonate then does not react with water containing phenalate. Now to the tube on the left we will add the marble chip which has been heated. And if as we suspect calcium oxide has been formed during the heating it should react with the water to form calcium hydroxide which should turn the phenaline pink. The instant appearance of the pink color indicates that calcium oxide was formed when the calcium carbonate chip was heated. Calcium carbonate here represents the marble chip. which we heated to a very high temperature with Bunson burner causing some decomposition into calcium oxide and carbon dioxide. After the chip had cooled, we then dropped it into a little distilled water containing phenaline. And immediately we saw the phenylene turn red as the water we reacted with the calcium oxide forming calcium hydroxide. The O ions of which caused the phenthalene to change from its colorless to its red color. When the ions of calcium or magnesium uh appear in ordinary tap water, the water is said to be hard. Chemists distinguish two kinds of hardness. However, permanent hardness which is caused by calcium and or magnesium ions present as their sulfate or chloride and temporary hardness which contains calcium and/or magnesium ions as the bicarbonate. The reason for these names um is that permanent hardness is not removed by boiling the water whereas most of the temporary hardness is removed by boiling. Calcium and magnesium ions react with soap to form a precipitate. Hard water therefore requires a considerable amount of soap because much of the first portions of soap combines the calcium and magnesium are removed from the solution. We will demonstrate the reaction of these ions with soap in the two tubes in the setup. The tube on the left contains tap water and the tube on the right distilled water. To each tube, we will add several drops of soap solution. You notice that no precipitate appears with the distilled water, but in the tube containing the tap water, a milky substance is formed. This is the precipitate caused when the calcium and magnesium ions in the tap water react with the soap. Now we'll place a stopper in each tube and shake vigorously. In the tube containing the distilled water, we have lost the soap. So when we shake the tube containing the tap water, practically no suds is obtained. This is because the calcium and magnesium in the tap water precipitated most of the soap. Now, if we added more and more and more soap to this tube, we could eventually obtain suds. And this is in effect what you do when you wash in hard water. One way to remove the offending ions of calcium and magnesium from hard water is through the use of materials called called ion exchange resin. Let this uh heavy line represent a very large organic molecule polystyrene which has been sulfenated at uh various positions give sulonic acid groups and then the hydrogen associated with the sulfonic acid group has been replaced with sodium ion. The large polystyrene backbone makes the whole molecule insoluble, but the sodium uh sulfonic acid groups sort of stick out into the solution. And when water containing calcium ion trickles down over such a resin, the sodium ions are displaced. The calcium ion becomes associated with the two sulfonic acid groups and the sodium ions are released into solution and since sodium ions don't react with soap uh the water has been effectively softened. Now, we're going to demonstrate this phenomena uh in the next sequence on the film, but we're not going to use calcium ion in our ion exchange experiment because the calcium ion is colorless and the sodium ions are colorless and it would be very difficult to follow the course of the reaction. Instead, we're going to use the copper ammonia complex ion which is dark blue in color. The color permits us to identify this ion and which is in equilibrium with a small amount of copper ion and ammonia. Now the copper ion will react here in exactly the same way uh that the calcium ion did and will become associated with the sulonic acid groups and sodium ions will be released. Therefore, when a solution containing the equilibrium mixture, copper ammonia complex and copper ions and ammonia flows down through an ion exchange resin tube, the copper ions should be removed from the equilibrium. This of course will cause the complex ion to dissociate and the filtrate or liquid being produced at the lower end of the tube should contain sodium ions and ammonia. The ammonia can be tested for because of its basic properties. We will now demonstrate one of these ion exchange columns in action. As we have just uh discussed, we've seen that it's possible to remove offending positive ions from solution using a cation exchange resin. In this tube we have some cation exchange resin essentially a sulfonated polystyrene in which the hydrogen ions on the sulfonic acid groups have been replaced with sodium ions. In this bottle we have a solution of coupric ammonium sulfate and the only part that we're interested in is the copper ammonia complex ion. We're using this ion because of its blue color, which permits us to follow the uh series of events as they take place. Now, we'll add some of this solution to the ion exchange resin in the tube. I think you can clearly see the blue color of the solution in the upper part of the tube, but notice that the emerging liquid at the bottom is colorless. Now in the tube we have the equilibrium between the coupriic ion and the copper ammonia complex and the ion exchange resin is removing the copper ions replacing it with two sodium ions. the ammonia which is uh another product of the reaction and the two sodium ions should pass into the collecting tube. Now the sodium ions we can't test for but the ammonium ions should give a pink color with phenaline. So as we add phenthalene to this tube, we see the immediate formation of the pink color indicating that ammonia has emerged from the bottom of the ion exchange column. We will now proceed to show how we can calculate the total hardness in a sample of hard water. In this experiment, both the calcium and magnesium present will be calculated in parts per million of calcium carbonate. In this flask, we've placed 50 ml of Akran tap water and we will add this to the small erlin meer flask. Next, we will add some indicator. The indicator we're using here is called Univer and forms a purplish red colored complex with calcium and magnesium ions. We'll swirl to dissolve the indicator. The color of the solution indicates that calcium and/or magnesium are present. The bureete, which this time is a 10 ml bureete instead of the usual 50, has been filled with a solution of versine and the level adjusted to zero. Versine is a chemical compound which itself forms a complex with calcium and magnesium ion. Now as we add the versine solution in small batches to the flask, the versine will form a complex with the calcium and magnesium and remove those ions from their complex with the indicator. The end point is signaled by the appearance of a blue color which is the color given by the indicator in the presence of very small amounts of calcium and magnesium present in equilibrium with the versine. This particular versine solution has been made up in such a way that when we use a 50 ml sample of tap water, the parts per million of calcium carbonate are given by multiplying the milliliters of versine used by 20. I think you can see that the original pink color of the indicator is beginning to disappear and the end point appears when the color of the indicator changes to a purplish blue color. This is the end point of this titration. As I said before, this is a 10 ml bureete. This is the 4.5 ml mark. This the 5 ml mark. The level of the liquid in the bureete uh can be carefully noted and determined to be 4.80 milliliters. To calculate total hardness we need to know the volume of our versine solution used and the strength or tighter of the versine. Now this particular versine has been made up in such a way that each milll of versine used represents 20 parts per million of hardness in water figured as calcium carbonate. when we deal with a 50 ml sample of hard water and this was the size of the sample that we used. So we need only multiply the volume of the versine used by 20 [Applause] to find that the total hardness in a tap water in this experiment is about 96 parts per million. This means that um if we had a million pounds uh of Akran tap water, we would expect to find dissolved therein about 96 pounds of calcium carbonate or its equivalent in calcium bicarbonate. We will now calculate the permanent hardness in soft water uh by this device. With the Versine uh univer mixture, we titrated total hardness. Now in this bureete, we've placed some hydrochloric acid. We'll titrate a sample of tap water with the hydrochloric acid, which will titrate the carbonate. And assuming that all the carbonates are calcium carbonate, we can calculate temporary hardness. and permanent hardness may then be calculated by subtracting the temporary hardness from the total hardness. In the graduated cylinder, we've again placed 50 ml of Akran tap water and we will add that to the Earl Meyer flask. The indicator of choice in this experiment is methyl red indicator which has a yellow color on the basic side and a reddish pink color on its acid side. In the bureete we have some 015 normal hydrochloric acid. The bureete has been filled and adjusted to zero. You can see the pink color of the indicator beginning to appear. and the appearance of the permanent pink color signals the end point of the titration. At the end of the titration, we can read the volume of acid used. Again, we are using the 10 ml bureete. This is the 2.5 milll line. This the 3ml line. The reading when taken is found to be 2.87 ml of HCl used. From our titration data, we can now proceed to calculate the temporary hardness of our sample of tap water. And from this data, then the permanent hardness. Recall our old titration experiment uh rule that at the end point the equivalence of acid used always equal equivalence of base used. Now in this experiment we know that we used 2.87 ml or 00287 L of 0015 normal acid. The equivalence of base then equals the product of these two numbers or 4043 or 4.3 * 10us 5 equivalence of base. Now we know that this much base was present in a volume of 50 ml of tap water which was the quantity we measured out and we need to know how many grams of base not equivalent are present in 1 million milliliters of tap water. This will be our hardness in parts per million. Well, the equivalent weight of calcium carbonate is 50. So, we multiply the gram equivalent weight 50 times the number of equivalents present. This will give us the number of grams of calcium carbonate present in 50 milliliters of the solution. And we want to know how many grams are present in the million milliliters. And solving this equation for X, we find that 43 g of calcium carbonate would be present in 1 million milliliters of this solution or that the hardness then on the calcium carbonate scale is 43 parts per million. Now this figure permits us to calculate permanent hardness because our univer uh versine titration told us that the total hardness was 96 parts per million. We've now found temporary hardness to be 43 parts per million. Permanent hardness must be the difference or 53 parts per million. In this film then we have examined several of the important reactions of calcium, berium and stranchium ions. We've seen how hard water uh is caused by the presence of these ions and we've examined a method or two for removing them from water. Finally, we've examined methods by which the chemist can tell how hard a sample of water is both with respect to its total permanent and temporary hardness. [Music]