More on Solutions Solubility Curves, Molarity, Dilutions

Solubility How many grams of solute will dissolve in a solvent depends on temperature and amount of solvent based on a saturated solution

Solubility Curve Solubility vs. Temperature 140 KI 130 120 gases solids NaNO3 110 Solubility Curve 100 KNO3 90 80 HCl NH4Cl shows the dependence of solubility on temperature 70 Solubility (grams of solute/100 g H2O) 60 NH3 KCl 50 “Solubility Curves for Selected Solutes”   Description: This slide is a graph of solubility curves for 10 solutes. It shows the number of grams of solute that will dissolve in 100 grams of water over a temperature range of 0cC to 10 cC. Basic Concepts The maximum amount of solute that will dissolve at a given temperature in 100 grams of water is given by the solubility curve for that substance. When the temperature of a saturated solution decreases, a precipitate forms. Most solids become more soluble in water as temperature increases, whereas gases become less soluble as temperature increases. Teaching Suggestions Use this slide to teach students how to use solubility curves to determine the solubilities of various substances at different temperatures. Direct their attention to the dashed lines; these can be used to find the solubility of KClO3 at 50 cC (about 21 g per 100 g of H2O). Make sure students understand that a point on a solubility curve represents the maximum quantity of a particular solute that can be dissolved in a specified quantity of solvent or solution at a particular temperature. Point out that the solubility curve for a particular solute does not depend on whether other solutes also are present in the solution (unless there is a common-ion effect; this subject usually is covered at a later stage in a chemistry course). Questions Determine the solubilities (in water) of the following substance at the indicated temperatures: NH3 at 50 oC; KCl at 90 oC; and NaNO3 at 0 oC. Which of the substances shown on the graph is most soluble in water at 20 oC? Which is lease soluble at that temperature? For which substance is the solubility lease affected by changes in temperature? Why do you think solubilities are only shown between 0 oC and 100 oC? In a flask, you heat a mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has just been dissolved. At what temperature does this occur? You then allow the flask to cool. When you examine it later, the temperature is 64 oC and you notice a white powder in the solution. What has happened? What is the mass of the white powder? Compare the solubility curves for the gases HCl, NH3, and SO2) with the solubility curves for the solid solutes. What generalizations(s) can you make about the relationship between solubility and temperature? According to an article in an engineering journal, there is a salt whose solubility in water increases as the water temperature increases from 0 oC to 65 oC. The salt’s solubility then decreases at temperatures above 65 oC, the article states. In your opinion, is such a salt likely to exist? Explain your answer. What could you do to verify the claims of the article? 40 30 NaCl KClO3 20 10 SO2 0 10 20 30 40 50 60 70 80 90 100

Gas Solubility Higher Temperature …Gas is LESS Soluble CH4 O2 CO 2.0 O2 Higher Temperature …Gas is LESS Soluble CO Solubility (mM) 1.0 The general rule of thumb is that the solubility of gases decreases when temperature increases. He 10 20 30 40 50 Temperature (oC)

READING A SOLUBILITY CURVE KNO3 (s) KCl (s) Solubility (g/100 g H2O) Temp. (oC) Solubility (g/100 g H2O) KNO3 (s) KCl (s) HCl (g) unsaturated: solution could hold more solute; below line saturated: solution has “just right” amt. of solute; on line supersaturated: solution has “too much” solute dissolved in it; above the line

Practice Problems Solubility vs. Temperature 140 KI 130 How many grams of potassium nitrate can dissolve in 100 g of water at 50C? At 20C, a solution contains 120 g of NaNO3 in 100 g of water. Is this solution saturated, unsaturated, or supersaturated? You need to make a solution containing 150 g of potassium chloride in 300 g of water. What temperature is required? 120 gases solids NaNO3 110 ~85 g 100 KNO3 90 80 HCl NH4Cl 70 Solubility (grams of solute/100 g H2O) 60 NH3 KCl supersaturated 50 “Solubility Curves for Selected Solutes”   Description: This slide is a graph of solubility curves for 10 solutes. It shows the number of grams of solute that will dissolve in 100 grams of water over a temperature range of 0cC to 10 cC. Basic Concepts The maximum amount of solute that will dissolve at a given temperature in 100 grams of water is given by the solubility curve for that substance. When the temperature of a saturated solution decreases, a precipitate forms. Most solids become more soluble in water as temperature increases, whereas gases become less soluble as temperature increases. Teaching Suggestions Use this slide to teach students how to use solubility curves to determine the solubilities of various substances at different temperatures. Direct their attention to the dashed lines; these can be used to find the solubility of KClO3 at 50 cC (about 21 g per 100 g of H2O). Make sure students understand that a point on a solubility curve represents the maximum quantity of a particular solute that can be dissolved in a specified quantity of solvent or solution at a particular temperature. Point out that the solubility curve for a particular solute does not depend on whether other solutes also are present in the solution (unless there is a common-ion effect; this subject usually is covered at a later stage in a chemistry course). Questions Determine the solubilities (in water) of the following substance at the indicated temperatures: NH3 at 50 oC; KCl at 90 oC; and NaNO3 at 0 oC. Which of the substances shown on the graph is most soluble in water at 20 oC? Which is lease soluble at that temperature? For which substance is the solubility lease affected by changes in temperature? Why do you think solubilities are only shown between 0 oC and 100 oC? In a flask, you heat a mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has just been dissolved. At what temperature does this occur? You then allow the flask to cool. When you examine it later, the temperature is 64 oC and you notice a white powder in the solution. What has happened? What is the mass of the white powder? Compare the solubility curves for the gases HCl, NH3, and SO2) with the solubility curves for the solid solutes. What generalizations(s) can you make about the relationship between solubility and temperature? According to an article in an engineering journal, there is a salt whose solubility in water increases as the water temperature increases from 0 oC to 65 oC. The salt’s solubility then decreases at temperatures above 65 oC, the article states. In your opinion, is such a salt likely to exist? Explain your answer. What could you do to verify the claims of the article? 40 30 NaCl KClO3 20 10 SO2 750C 0 10 20 30 40 50 60 70 80 90 100 6

the term “molar” represents molarity substance being dissolved mol L M total combined volume the term “molar” represents molarity

Concentration = # of moles volume (L) V = 1000 mL V = 1000 mL n = 8 moles [ ] = 32 molar V = 1000 mL V = 1000 mL V = 5000 mL n = 2 moles n = 4 moles n = 20 moles Concentration = 2 molar [ ] = 4 molar [ ] = 4 molar

Molarity Example Problem #1 How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L sol’n 0.25 mol NaCl 1 L sol’n 58.44 g NaCl 1 mol NaCl = 7.3 g NaCl

Molarity Example Problem #2 Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g NaF 1 mol NaF 41.99 g NaF 0.25 L sol’n = 0.95M NaF

Dilution Preparation of a desired solution by adding water to a concentrate. Moles of solute remain the same.

Concentrated vs. Dilute Dilution Add Solvent Moles of solute before dilution after dilution =

Dilution Example Problem What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M) V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

Dilution Problem #2 Example #1: 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make?

Solution (1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x) x = 100. mL Using the dilution equation, we write: (1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x) x = 100. mL Notice that the volumes need not be converted to liters. Any old volume measurement is fine, just so long as the same one is used on each side. (However, as mentioned above, If you are calculating how many moles of solute are present, you need to have the volume in liters.)

Your turn. Example #2: 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results?

Solution (2.500 mol/L) (100.0 mL) = (0.5500 mol/L) (x) x = 454.5454545 mL (oops, my fingers got stuck typing. Bad attempt at humor, really bad!) x = 454.5 mL