Solution Chemistry Notes Solution Chemistry Notes

Definition of a “Solution”  A solution is a homogeneous (uniform) mixture of a  A solution is a homogeneous (uniform) mixture of a solute and a solvent.  Everyday solutions: CO 2, sugar dissolved in water Salt dissolved in water

Relationships  Solute:present in smaller amount  Solvent: present in larger amount

Solution = Solute + Solvent Important Note: Unless otherwise specified, assume that the solvent for all solutions discussed in this course is WATER.

Importance of Solutions  In nature, it is rare to find pure solids, liquids or gases; most substances are mixtures

Using Solutions to Speed Up Chemical Reactions  Most elements and most compounds of interest are solids at RT; however the rxn rate between solids is very slow due to low surface area which limits contacts between reacting particles  In the lab, most chemical reactions are carried out by dissolving solids in liquid solvents to increase the # of collisions and hence the rxn rates

Concept of Concentration as Ratio  Concentration = Moles of Solute  Volume of Solution 1 drop in 100 mL __ drops in 200 mL == 100 mL 200 mL 400 mL __ drops in 400 mL

Concept of Concentration as Ratio  Concentration = Moles of Solute  Volume of Solution 1 drop in 100 mL 2 drops in 200 mL == 100 mL 200 mL 400 mL __ drops in 400 mL

Concept of Concentration as Ratio  Concentration = Moles of Solute  Volume of Solution 1 drop in 100 mL 2 drops in 200 mL == 100 mL 200 mL 400 mL 4 drops in 400 mL

Molarity Definition  Molarity (M) = moles of solute Liter of solution 58.5 g 1 M NaCl solution: 1 Mole (58.5 g) of NaCl dissolved in a total aqueous solution volume of 1 Liter.

What is the Molarity of each solution if represents 1 mole? (Recall 1 L = 1000 mL)  Molarity = Moles of Solute 1 Liter of Solution 1 M ? M == 1000 mL ? M

What is the Molarity of each solution if represents 1 mole? (Recall 1 L = 1000 mL)  Molarity = Moles of Solute 1 Liter of Solution 1 M 2 M == 1000 mL ? M

What is the Molarity of each solution if represents 1 mole? (Recall 1 L = 1000 mL)  Molarity = Moles of Solute 1Liter of Solution 1 M 2 M == 1000 mL 4 M

Example Problem: What is the concentration of the following solution?  Answer: 1 Molar (1 M) – there is 1 mole in a liter of solution 1 Liter 1 Mole

Although the definition of Molarity is the moles of solute in 1 liter of solution, as long as the ratio of moles to volume is unchanged any volume of solution at the same concentration can be prepared.

What is the Molarity of each solution if represents 1 mole? (Recall 1 L = 1000 mL)  Molarity = Moles of Solute Liters of Solution ? M == 500 mL 1000 mL 2000 mL ? M

What is the Molarity of each solution if represents 1 mole? (Recall 1 L = 1000 mL)  Molarity = Moles of Solute Liters of Solution 2 M == 500 mL 1000 mL 2000 mL 2 M

Example: Calculating Molarity If 9.45 g of CsBr is dissolved in enough water to give 250. mL of solution, what is the molarity of the solution? Molarity = moles of solute Liters of solution What is the solute, solvent and solution in this problem? Ans: solute = CsBr, solvent = water, solution = CsBr dissolved in water

Calculating Molarity: Comparing Given to Defn of Molarity Molarity = moles of solute 9.45 g CsBr Liters of solution 250 mL Solution How would you solve this problem? Ans: 1) Convert g → moles 2) Convert mL → L 3) Divide moles by L

Calculating Molarity: Comparing Given to Defn of Molarity Molarity = moles of solute 9.45 g CsBr Liters of solution 250 mL Solution 1)g CsBr → moles CsBr: 9.45 g CsBr ( 1 mole CsBr) = moles (212.8 g CsBr)

Calculating Molarity: Comparing Given to Defn of Molarity Molarity = moles of solute 9.45 g CsBr Liters of solution 250 mL Solution 2) mL solution → L solution 250. mL ( 1 Liter) = L (1000 mL)

Molarity = moles of solute Liters of solution M = moles CsBr = moles = M L L = M

Complete the following table: Grams of KBr Moles of NaCl Volume of solution Molarity of Solution 119 g 1.00 moles1000. mL 1.00 M moles500. mL 476 g2000. mL Recall 1000 mL = L

Complete the following table: Grams of KBr Moles of KBr Volume of solution Molarity of Solution 119 g 1.00 moles1000. mL 1.00 M 59.5 g moles500. mL 1.00 M 238 g 2.00 moles2000. mL 1.00 M Recall 1000 mL = L

Homework 7-1 p. 476 #32  To prepare 500. mL of 1.02 M sugar solution, which of the following would you need?  A) 500. mL of water and 1.02 mole of sugar  B) 1.02 mol of sugar and enough water to make a total volume of 500. mL  C) 500. g of water and 1.02 mol of sugar  D) 0.51 mol of sugar and enough water to make the total volume 500. mL

Homework 7-1 p. 476 #32  To prepare 500. mL of 1.02 M sugar solution, which of the following would you need?  A) 500. mL of water and 1.02 mole of sugar  B) 1.02 mol of sugar and enough water to make a total volume of 500. mL  C) 500. g of water and 1.02 mol of sugar  D) 0.51 mol of sugar and enough water to make the total volume 500. mL

HW 7-1, p. 476, #34- Calculate Molarity of each solution 34a) 0.50 mol KBr: 250 mL Solution Ans: 0.50 mol/.250 L = 2.0 M 34c) 0.50 mol KBr: 750 mL Solution Ans: 0.50 mol/.750 L = 0.67 M

p. 476, #36a- Calculate Molarity  36a) 1.25 g KNO 3 ; 115 mL  Ans:  K =  N =  O = 3 x =  Molar Mass =  1.25 g KNO 3 ( g KNO 3 1 mole KNO 3 ) = moles KNO 3

p. 476, #36a- Calculate Molarity  115 mL ( 1000 mL 1 Liter ) = L moles KNO L Molarity = = M

p. 476, #36c Calculate Molarity  Don’t forget to convert mg → g (1000 mg = 1 g)  Final Answer: M

HW 7-1: p. 480 #119  10. g AgNO 3  Ag: N: O: 3 x =  Molar Mass AgNO 3 = g/mole  10 g/ g/mole = moles  M = moles/ L;  LM = moles  L = moles/M = / 0.25 = 0.24 L