Concentration

C11-4-16 CONCENTRATION AND DILUTION OUTCOME QUESTION(S): C11-4-16 CONCENTRATION AND DILUTION Differentiate among the various representations of concentration, and solve concentration problems. Solve problems involving the dilutions and combinations of common solutions. Include: stock solutions Vocabulary & Concepts  Molarity M

Solution with a lot of solute compared to solvent. Concentrated: Solution with a lot of solute compared to solvent. Just like soluble/insoluble: dilute and concentrated are relative terms Dilute: Solution with a little solute compared to solvent. Solution A 500 g of solute in 100 mL of solvent Solution B 100 g of solute in 100 mL of solvent Solution C 1 g of solute in 100 mL of solvent

There are many ways to express the solute/solvent proportions (concentration): Parts per …(ppm / ppb) Number of particles of solute for every million / billion particles of solution. Seen in water/air quality and pollution levels Grams per Litre (g/L) Number of grams of solute for every litre of solution. Seen in medical testing – cholesterol or sugar levels in the blood/urine Percentage (%w/v : %v/v : %w/w) Percentage of weight/volume of solute in total weight/volume of solution. Seen in industrial and household solutions. These are all useful depending on the situation, but we use a different form for concentration

Consider the equation: units of molarity (M) are obviously mol/L Concentration in Chemistry: Molarity (mol/L) Number of moles of solute for every litre of solution. Seen in chemical solutions and laboratory preparations moles of solute litres of solution = Molarity n V M = Consider the equation: units of molarity (M) are obviously mol/L

0.80 M is phonetically pronounced “0.8 mol-er” Calculate the concentration of a NaCl solution if 0.200 moles is dissolved in 250 mL of solution. Watch your units 250 mL 1000 mL = 0.25 L 1 L n V M = 0.25 L = 0.80 mol/L 0.200 mol 0.80 M is phonetically pronounced “0.8 mol-er”

n V M = Depending on your strength in Stoichiometry and your comfort level with dimensional analysis, there are two ways you can solve concentration/molarity questions: 1) Use the formula and some dimensional analysis 2) Use ALL dimensional analysis

What volume of a 1.25 M solution contains 5.0 moles of solute? FORMULA: n M = V = 5.0 mol 1.25 M V = 4.00 L DIMENSIONAL ANALYSIS: This is not a gas at STP, so can’t use 22.4L/mol. Concentration is used as the conversion factor between moles and litres in solutions. 5.0 mol 1 L 4.00 L = 1.25 mol

n n = (0.4)(0.225) M = V = = 0.09 mol 1 L 0.225 mol 0.09 mol 1 L How many moles of solute needed to make 400 mL of a 0.225 M solution? Watch your units FORMULA: Have to convert to litres first n n = (0.4)(0.225) M = V = 0.09 mol DIMENSIONAL ANALYSIS: 400 mL 1 L 0.225 mol = 0.09 mol 1 L 1000 mL

50.0 g 1 mol n = 0.313 mol M = V 159.6 g = 0.5 M 0.6 L 50.0 g 1 mol = What is the molar concentration of a solution of copper (II) sulfate if 50.0 g is dissolved in 600 mL of solution? CuSO4 = 159.6 g/mol FORMULA: 50.0 g 1 mol n V M = = 0.313 mol 159.6 g 0.6 L = 0.5 M 0.313 mol DIMENSIONAL ANALYSIS: 50.0 g 1 mol -- = 0.5 mol/L 159.6 g 0.6 L The top can be left empty because 0.6 L is not part of a ratio – we just want to divide the moles by that volume

What mass of sodium chloride is needed to make 0. 250 L of a 0 What mass of sodium chloride is needed to make 0.250 L of a 0.100 M solution? FORMULA: NaCl = 58.5 g/mol n M = = (0.250)(0.100) V = 0.0250 mol 58.5 g = 1.46 g 1 mol DIMENSIONAL ANALYSIS: 0.250 L 0.100 mol 58.5 g = 1.46 g 1 mol 1 L

Means weigh it out on a scale... Making A Solution: Means weigh it out on a scale... 1. Calculate the mass of the solute needed and “mass out.” 2. Add the solid to the clean and empty volumetric flask. 3. Pour in some of the solvent and swirl to dissolve. 4. Add more solvent up to the correct total volume. 5. Mix again. solvent solute

How much in grams will you need? Describe the steps needed to make 500.0 mL of a 0.50 M solution of sodium hydroxide. How much in grams will you need? NaOH = 40.0 g/mol FORMULA: n M = = (0.500)(0.50) V = 0.25 mol 1 40.0 mol g = 10.0 g solvent solute 1. Mass out 10.0 g of NaOH 2. Add the solid to the clean and empty volumetric flask. 3. Pour in some of the solvent and swirl to dissolve. 4. Add more solvent up to 500 mL. 5. Mix again.

What is the [Cl-] in 0.200 M solution of aluminum chloride? […] means “concentration of … Dissociation equation: AlCl3 (s) Al+3(aq) + Cl-1(aq) 3 0.200 mol AlCl3 3 mol Cl- = 0.600 M Cl- 1 L 1 mol AlCl3 What you know Add conversion factor Units to cancel on bottom Units in need on the top Multiply top, divide bottom Let the units guide you…

Cu(NO3)2 (s) Cu+2(aq) + NO3-1(aq) 2 2.5 L NO3- 0.1 mol 1 187.5 g 2 = What mass of copper (II) nitrate is needed for 2.50 L of a solution with a nitrate ion concentration of 0.100 M? Cu(NO3)2 = 187.5 g/mol What’s the plan? Dissociation equation: Cu(NO3)2 (s) Cu+2(aq) + NO3-1(aq) 2 ? g 2.50 L mol mol DIMENSIONAL ANALYSIS: 2.5 L NO3- 0.1 mol 1 mol Cu(NO3)2 187.5 g 1 L 2 mol NO3- 1 mol Cu(NO3)2 1. What you know 2. Add conversion factor 3. Units to cancel on bottom 4. Units in need on the top 5. Multiply top, divide bottom = 23.4 g Cu(NO3)2

C11-4-16 CONCENTRATION AND DILUTION CAN YOU / HAVE YOU? C11-4-16 CONCENTRATION AND DILUTION Differentiate among the various representations of concentration, and solve concentration problems. Solve problems involving the dilutions and combinations of common solutions. Include: stock solutions Vocabulary & Concepts  Molarity M