STUDY GUIDE SOLUBILITY GRAPH   When using a solubility graph it's important to understand how it is created in the first place. To start with,the solubility of a substance for a given temperature must be determined. This is done by finding out what mass of solute is needed to make a saturated solution in 100 cm3 of water for that temperature(referred to as the solubility). This is repeated for each of the temperatures from 0ºC to 100ºC. The data is then plotted on a temperature/solubility graph,and the points are connected. These connected points are called a solubility curve. To see the actual graph mentioned click on Solubility Graph. A solubility graph can be used to solve a variety of questions. The first type is simply to identify a substance when you are given the solubility in g/100 cm3 of water and the temperature. All you do is see which solubility curve the solubility and temperature intersect at. For example,what substance has a solubility of 90 g/100 cm3 of water at a temperature of 25ºC ? The only substance whose solubility curve is located at the intersection of 90 g/100 cm3 and 25ºC is sodium nitrate. Sometimes you're given the substance name and either the temperature or the solubility. Then you have to find out what the solubility or temperature is. Locate the solubility curve needed and see for a given temperature,which solubility it lines up with and visa versa. For example,what is the solubility of potassium nitrate at 80ºC ? Locate where 80ºC is on the solubility curve for potassium nitrate. Extend a line over to the solubility axis that is level with this point. If you did it correctly,the solubility should be 170 grams. If you need to find what temperature a substance has for a given solubility,locate where the solubility intersects the curve for the substance,and drop a vertical line down from that point to the temperature below. Example: at what temperature will sodium nitrate have a solubility of 95 g/100 cm3 ? You'll need to more carefully estimate where 94 g will be,but if you drop a vertical line down from this intersection point,it should be about 30ºC. These types of situations will only work if the amount of water is the same as the graph,namely 100 cm3. If it is not,you'll need to set up a proportion. For example,what is the solubility of sodium chloride at 25ºC in 150 cm3 of water ? From the solubility graph we see that sodium chlorides solubility is 36 g.Place this in the proportion below and solve for the unknown solubility. Solve for the unknown quantity by cross multiplying.   Solubility in grams = unknown solubility in grams 100 cm3 of water other volume of water ___36 grams____ = unknown solubility in grams 100 cm3 of water 150 cm3 water The unknown solubility is 54 grams. You can use this proportion to solve for the other volume of water if you're given the other solubility. You can also use a solubility graph to determine if a solution is saturated,unsaturated,or supersaturated. If the solubility for a given substance places it anywhere on it's solubility curve it is saturated. If it lies above the solubility curve,then it's supersaturated,and if it lies below the solubility curve it's an unsaturated solution. Remember though,if the volume of water isn't 100 cm3 to use a proportion first as shown above. Sometimes you'll need to determine how much additional solute needs to be added to a unsaturated solution in order to make it saturated. For example,30 grams of potassium nitrate has been added to 100 cm3 of water at a temperature of 50ºC. How many additional grams of solute must be added in order to make it saturated? From the graph you can see that the solubility for potassium nitrate at 50ºC is 84 grams. If there are already 30 grams of solute in the solution,all you need to get to 84 grams is 54 more grams ( 84g-30g ). Lastly you can use a solubility graph to determine at what temperature a precipitate(crystals) will begin forming when an unsaturated solution cools down. Suppose you have a sodium nitrate solution at 100ºC containing 120g of dissolved solute. Locate where 120g intersects the curve for sodium nitrate,and the temperature at this point is when crystals can start forming. In this example the temperature will be about 56ºC. Remember though,if the volume of water isn't 100 cm3 to use a proportion first.