solubility review

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Chemistry 11 Review Solutions Hebden Chemistry 12 Workbook: p.73-81

Terms you should know: : 

Terms you should know: Homogeneous mixture: Mixtures in which the components are uniformly distributed. (Typically these are clear and may be coloured or colourless). Heterogeneous mixture: Mixtures in which the components are segregated when at rest. (Typically these are cloudy before they are at rest and have two distinct phases when they are at rest. They may be coloured or colourless). Solution: A combination of two or more substances that exist as a homogeneous mixture. Example : Sodium Chloride in water. Solute: The substance of lesser quantity in a homogeneous mixture. Example: The Sodium Chloride portion of the Sodium Chloride in water solution.

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Solvent: The substance of greater quantity in a homogeneous mixture. Example: The water portion of the Sodium Chloride in water solution Solubility: A measure of the amount of solute that will dissolve in a given amount of solvent at a specified temperature. Usually measured in units of moles/liter, (M), or g/100mL. Concentrated: A relatively large amount of solute to solvent. Dilute A relatively small amount of solute to solvent. .

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Saturated solution: A solution in which a maximum quantity of solute has been dissolved. Precipitate: An insoluble product that may form when a chemical reaction occurs in solution. (ie. The solid that causes a cloudy appearance and eventually settles to the bottom of the container). Soluble: Any solution whose concentration is greater than or equal to 0.100 M. Insoluble: Any solution at equilibrium whose concentration is less than 0.100 M

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Prescription drugs in the correct concentration make you better. In higher concentration they can kill you.

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Chemists need to make solutions that have precise concentrations

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Pesticides must be in proper concentrations. Food additives must be in correct concentrations. A driver is legally impaired at 0.08 mg/mL blood alcohol content. A new driver is legally impaired at 0.00 mg/mL blood alcohol content. Chemists control the concentration of chemicals using the concepts we will develop in this unit.

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The concentration of a solution tells you how much solute that is dissolved in a given amount of solvent (water). The molarity is the concentration of a solution. Molarity = Moles Litre 1. 2.255 moles of NaCl is dissolved in 4.0 L of water, calculate the molarity. Molarity =

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The concentration of a solution tells you how much solute that is dissolved in a given amount of solute (water). The molarity is the concentration of a solution. Molarity = Moles Litre 1. 2.255 moles of NaCl is dissolved in 4.0 L of water, calculate the molarity. Molarity = 2.255 moles 4.0 L

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The concentration of a solution tells you how much solute that is dissolved in a given amount of solute (water). The molarity is the concentration of a solution. Molarity = Moles Litre 1. 2.255 moles of NaCl is dissolved in 4.0 L of water, calculate the molarity. Molarity = 2.255 moles 4.0 L = 0.56 moles/litre = 0.56 M (Molar) where M means moles per 1 litre

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2. 5.00 g KOH is dissolved in 250. mL of water, calculate the molarity. Molarity =

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2. 5.00 g KOH is dissolved in 250. mL of water, calculate the molarity. Molarity = 5.00 g

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2. 5.00 g KOH is dissolved in 250. mL of water, calculate the molarity. Molarity = 5.00 g x 1 mole 56.11 g

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2. 5.00 g KOH is dissolved in 250. mL of water, calculate the molarity. Molarity = 5.00 g x 1 mole 56.11 g 0.250 L

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2. 5.00 g KOH is dissolved in 250. mL of water, calculate the molarity. Molarity = 5.00 g x 1 mole 56.11 g 0.250 L = 0.356 M

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3. How many moles are there in 205. mL of a 0.172 M solution?

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3. How many moles are there in 205. mL of a 0.172 M solution? 0.205 L

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3. How many moles are there in 205. mL of a 0.172 M solution? 0.205 L x 0.172 moles 1 L

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3. How many moles are there in 205. mL of a 0.172 M solution? 0.205 L x 0.172 moles = 0.0353 moles 1 L

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4. How many grams NaCl are there in 250.0 mL of 0.500 M solution?

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4. How many grams NaCl are there in 250.0 mL of 0.500 M solution? 0.250 L

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4. How many grams NaCl are there in 250.0 mL of 0.500 M solution? 0.250 L x 0.500 moles 1 L

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4. How many grams NaCl are there in 250.0 mL of 0.500 M solution? 0.250 L x 0.500 moles x 58.5 g 1 L 1 mole

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4. How many grams NaCl are there in 250.0 mL of 0.500 M solution? 0.250 L x 0.500 moles x 58.5 g = 7.31 g 1 L 1 mole

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5. How many grams of NaCl must be used to prepare a 100.0 mL of 0.250 M solution.

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5. How many grams of NaCl must be used to prepare a 100.0 mL of 0.250 M solution. 0.1000 L

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5. How many grams of NaCl must be used to prepare a 100.0 mL of 0.250 M solution. 0.1000 L x 0.250 moles 1 L

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5. How many grams of NaCl must be used to prepare a 100.0 mL of 0.250 M solution. 0.1000 L x 0.250 moles x 58.5 g 1 L 1 mole

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5. How many grams of NaCl must be used to prepare a 100.0 mL of 0.250 M solution. 0.1000 L x 0.250 moles x 58.5 g = 1.46 g 1 L 1 mole

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6. How many millilitres of a 0.200 M solution of CoCl2 will contain 25.0 g?

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6. How many millilitres of a 0.200 M solution of CoCl2 will contain 25.0 g? 25.0 g

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6. How many millilitres of a 0.200 M solution of CoCl2 will contain 25.0 g? 25.0 g x 1 mole 129.9 g

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6. How many millilitres of a 0.200 M solution of CoCl2 will contain 25.0 g? 25.0 g x 1 mole x 1 L 129.9 g 0.200 mol

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6. How many millilitres of a 0.200 M solution of CoCl2 will contain 25.0 g? 25.0 g x 1 mole x 1 L x 1000 mL 129.9 g 0.200 mol 1 L

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6. How many millilitres of a 0.200 M solution of CoCl2 will contain 25.0 g? 25.0 g x 1 mole x 1 L x 1000 mL = 962 mL 129.9 g 0.200 mol 1 L

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7. Describe how you would prepare 100.0 mL of a 0.200 M solution of CoCl2. 0.1000 L x 0.200 moles x 129.9 g = 2.60 g 1 L 1 mole Weigh 2.60 g of CoCl2 Dissolve in water Transfer to a 100 mL volumetric flask and fill to the line

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Diluting Solutions

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Diluting Solutions Concentrated solutions have a relatively high molarity. Dilute solutions have a relatively low molarity. It is often faster to prepare a number of standard solutions by diluting a more concentrated solution. The following equation can be used to solve dilution problems – when water is added or removed from a solution. M1V1 = M2V2 M1= the initial molarity M2 = the final molarity V1 = the initial volume V2 = the final volume

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NaCl(aq) NaCl(aq) Concentrated solution Diluted solution Is there a difference in the number of moles of solute in each beaker? NO, there is the same number in each beaker. water

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1. 25.0 mL of 0.10 M solution is diluted by adding 75.0 mL of water. Calculate the molarity of the new solution. M1V1 = M2V2 Initial Solution M1= 0.10 M M2 = ? V1 =25.0 mL V2 = 100.0 mL 25.0 mL + 75.0 mL (0.10 M)(25.0 mL) = M2(100.0 mL) M2 = 0.025 M (100.0 mL) (100.0 mL)

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What volume of 0.250 M CoCl2 solution can be diluted to 100.0 mL in order to make a 0.0150 M solution? Initial Solution M1V1 = M2V2 M1= 0.250 M M2 = 0.0150 M V1 = ? V2 = 100.0 mL (0.250 M) V1 = (0.0150 M)(100.0 mL) V1 = 6.00 mL

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3. 50.0 mL of 0.500 M CuSO4 solution is diluted to 250.0 mL. Calculate the new concentration and the number of grams CuSO4 in the new solution. Initial Solution M1V1 = M2V2 M1= 0.500 M M2 = ? V1 = 50.0 mL V2 = 250.0 mL (0.500 M)(50.0 mL) = M2(250.0 mL) M2 = 0.100 M 0.0500 L 0.500 mol 1 L x x 159.6 g 1 mol = 3.99 g

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4. 1.53 g of NaCl is dissolved in 100.0 mL of water. Calculated the molarity. The solution is concentrated by evaporating off 60.0 mL of water. Calculate the new molarity.

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4. 1.53 g of NaCl is dissolved in 100.0 mL of water. Calculated the molarity. The solution is concentrated by evaporating off 60.0 mL of water. Calculate the new molarity. [ NaCl ] = 1.53 g 58.5 g 1 mol x 0.1000 L M1V1 = M2V2 M1= 0.262 M M2 = ? V1 = 100.0 mL V2 = 40.0 mL = (0.262 M)(100.0 mL) = M2(40.0 mL) M2 = 0.654 M 100.0 mL – 60.0 mL 0.262 M

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5. What volume of 0.200 M CuSO4 solution must be diluted to 500.0 mL to produce a 0.150 M solution? Initial Solution M1V1 = M2V2 M1= 0.200 M M2 = 0.150 M V1 = ? V2 = 500.0 mL (0.200 M)(V1) = (0.150 M)(500.0 mL) V1 = 375 mL

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6. What volume of water must be added to 150.0 mL of 0.200 M NaCl solution to change its concentration to 0.0250 M? Initial Solution M1V1 = M2V2 M1= 0.200 M M2 = 0.0250 M V1 = 150.0 mL V2 = ? (0.200 M)(150.0 mL) = (0.0250 M)(V2) V2 = 1200 mL = 1.20 L Final Total Volume Volume of H2O added = 1.20 L – 0.1500 L = 1.05 L

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Ion Concentration

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1. What is the concentration of each ion in a 0.300 M AlCl3 solution? Al3+ Cl- Cl- Cl- AlCl3 Al3+ 3 Cl- + 0.300 M 0.300 M 0.900 M

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2. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl2 in 600.0 mL of water? [ CaCl2 ] = 80.0 g 111.1 g 1 mol x 0.6000 L = 1.20 M CaCl2 Ca2+ 2 Cl- + 1.20 M 1.20 M 2.40 M

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3. If 1.25 L of 0.560 M Aluminum sulphate solution is added to 4.75 L of water, what is the resulting concentration of each ion? M1V1 = M2V2 M1 = 0.560 M V1 = 1.25 L V2 = 6.00 L M2 = ? (0.560 M)(1.25 L) = M2(6.00 L) M2 = 0.117 M Al2(SO4)3 2 Al3+ 3 SO42- + 0.117 M 0.234 M 0.351 M

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4. If 40.0 mL of 0.400 M Potassium chloride solution is added to 60.0 mL of 0.600 M Calcium nitrate, what is the resulting concentration of each ion. KCl K+ Cl_ + Ca(NO3)2 Ca2+ + 2 NO3_ 0.160 M 0.160 M 0.160 M 0.360 M 0.360 M 0.720 M M1V1 = M2V2 (0.400 M)(40.0 mL) = M2(100.0 mL) M2= 0.160 M For KCl For Ca(NO3)2 M1V1 = M2V2 (0.600 M)(60.0 mL) = M2(100.0 mL) M2= 0.360 M

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5. If the [Cl-] = 0.400 M, calculate the number of grams of AlCl3 that would be dissolved in 3.00 L of water. AlCl3 Al3+ 3 Cl- + 0.400 M 0.133 M 3.00 L 0.133 mol 1 L x x 133.5 g 1 mol = 53.3 g

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6. If the [SO42-] = 0.100 M in 20.0 mL of Ga2(SO4)3, determine the [Ga3+] and the molarity of the solution. Ga2(SO4)3 2 Ga3+ 3 SO42- + 0.100 M 0.0667 M 0.0333 M

Activities: : 

Activities: Workbook Exercises: Do Questions from pages 73 – 81. Check your own work