General Chemistry: Concentration and Molarity Lecture Notes PDF

Lecture General Chemistry: Concentration and Molarity Joseph M. Hayes...

Lecture General Chemistry: Concentration and Molarity Joseph M. Hayes 1 © 2017 Pearson Education, Inc.  Covered in Today’s Lecture  The Mole  Concentrations  Changing the Concentration: Dilutions  Titration Calculations 2 © 2017 Pearson Education, Inc.  The Mole Mass of 1 H atom = 1 Da (or 1 a.m.u.) Mass of 1 H2O molecule = (2 ×(atomic mass H)) + (1×(atomic mass O)) = (2 ×1 Da) + (1 ×16 Da) = 18 Da But the atomic mass number in periodic table also tells you the molar mass (in grams) of that element Atomic mass (Da) Molar mass (g) A mole is Avogadro’s number of particles = 6 ×1023 1 mole of H = 6 ×1023 atoms = 1 g [molar mass = 1 g/mol] 1 mole of H2O = 6 ×1023 molecules = 18 g [molar mass = 18 g/mol] 3 © 2017 Pearson Education, Inc.  The Mole The molar mass of a compound is the sum of the molar masses of each atom from which the compound is composed Example 1: What is the molar mass of glucose, C6H12O6? Molar mass = (6 ×(molar mass C)) + (12 ×(molar mass H)) + (6 ×(molar mass O)) = (6 ×12 g mol-1) + (12 ×1 g mol-1) + (6 ×16 g mol-1) = 180 g/mol Example 2: Calculate the amount (in moles) of glucose in a 540 g sample? Amount in moles = = = 3 moles 4 © 2017 Pearson Education, Inc.  Concentrations General Properties of Aqueous (Water) Solutions  A solution is a homogenous mixture of two or more substances.  The substance present in the largest amount (moles) is referred to as the solvent.  The other substances present are called the solutes.  A substance that dissolves in a particular solvent is said to be soluble in that solvent.  Concentration relates to the amount of solute per unit volume of solution.* *Important to note we also measure concentrations of substances other than chemical compounds (e.g. number of bacteria present, number of red blood cells) but in all cases it relates the amount of that entity to a particular volume 5 © 2017 Pearson Education, Inc.  Concentrations  Concentration is often expressed as the molarity of a solution  The molarity (M) is the number of moles of a chemical substance present in 1 L of solution*  A 3 M solution of hydrochloric acid (HCl) has a concentration of 3 mol L-1  [HCl] = 3 mol L-1  If we know the concentration of a solution, we can then calculate the amount of a substance in a sample of that solution 6 * 1 L is the same as 1 dm3 (1 decimeter3) © 2017 Pearson Education, Inc.  Concentrations  Example 3: Suppose we have 100 mL (i.e. 0.1 L) of a NaCl solution whose concentration is 2 M. What amount of NaCl (in moles) do we have in the 100 mL sample? Answer: 2 M = 2 moles in 1 L 1 L = 2 moles 1000 mL = 2 moles 1 mL = 0.002 moles 100 mL = 0.2 moles 7 * 100 mL (mls) = 100 cm3 © 2017 Pearson Education, Inc.  Concentrations Preparing a solution of known concentration  Example 4: To prepare 500 mL of a 1 M NaCl solution, what amount of NaCl do we need in moles and in grams? Answer: 1 M = 1 mole in 1 L 1 L = 1 mole 1000 mL = 1 mole 500 mL = 0.5 mole So, we need 0.5 moles 1 mole NaCl = molar mass NaCl = (1×molar mass Na) + (1×molar mass Cl) = (1 ×23.0 g mol-1) + (1×35.5 g mol-1) = 58.5 g mol-1 0.5 mole = 0.5 × molar mass NaCl 8 © 2017 Pearson Education, Inc. = 0.5 × 58.5 g = 29.25 g  Concentrations Preparing a solution of known concentration  Volumetric flasks 9 © 2017 Pearson Education, Inc.  Concentrations Preparing solutions according to % by weight  Example 5: How would you prepare a 1.5% w/v nutrient agar solution for growing bacterial colonies if you need to prepare 250 mL? Answer: 1.5 % w/v = 1.5 % weight/volume, means mass of agar (in g) = 1.5% of the volume of the solution (250 mL)  so we need to calculate 1.5% of 250 mL  1.5% of 250 mL = 0.015×250 = 3.75 g of agar needed to prepare a solution with the desired concentration 10 © 2017 Pearson Education, Inc.  Concentrations Calculating the concentration of a solution  Example 6: 2 moles of glucose are dissolved in water to make a 500 mL solution. What is the concentration of the solution? Answer: = 4 mol L-1 = 4 M 11 © 2017 Pearson Education, Inc.  Changing the Concentration: Dilutions When we dilute a solution, the amount of solute remains unchanged, but the concentration decreases moles of solute before dilution = moles of solute after dilution A stock solution is a solution that is prepared in bulk in one concentration – 12 dilutions © 2017 Pearson Education, Inc. are used to prepare other lower concentration solutions  Changing the Concentration: Dilutions Formula we use for dilutions: Cinit ×Vinit = Cfinal ×Vfinal  Example 7: 100 mL of a 2 M solution of HCl is diluted with 900 mL of water to produce 1 L solution. What is the concentration of the diluted HCl solution? Vinit = volume of initial  Answer: Cinit ×Vinit = Cfinal ×Vfinal (stock) solution used to make up final solution 2 mol L-1 × 100 mL = Cfinal × 1000 mL Vfinal = volume after dilution Cfinal = = 0.2 mol L-1 = 0.2 M 13 © 2017 Pearson Education, Inc.  Changing the Concentration: Dilutions Dilution problems can be solved in an intuitive manner. Let’s look again at the previous example (alternative approach)  Example 7: 100 mL of a 2 M solution of HCl is diluted with 900 mL of water to produce 1 L solution. What is the concentration of the diluted HCl solution?  Answer: Intuitive approach 100 mls of stock is diluted to 1000 mL Stock solution is therefore diluted by Stock solution is therefore diluted by = factor of 10 i.e. final solution is 10 times less concentrated = = 0.2 M is the diluted concentration 14 © 2017 Pearson Education, Inc.  Changing the Concentration: Dilutions A serial dilution reduces the concentration of a solution in a series of steps, such that the concentration gradually falls from step to step. Why serial dilutions? If we have a solution that we need to dilute by a significant amount, it can be impractical to dilute it “in one go”, e.g. 10 mL to 500 L Example 8: We have a stock solution of concentration 2 M (2 mol L-1). If we take 1 mL and add it to 9 mL solvent and carry out 3 more subsequent dilution step using the same volumes, what would be the final concentration? Etc. 15 © 2017 Pearson Education, Inc.  Changing the Concentration: Dilutions Example 8: We have a stock solution of concentration 2 M (2 mol L-1). If we take 1 mL and add it to 9 mL solvent and carry out 3 more subsequent dilution steps using the same volumes, what would be the final concentration? Answer: Dilution Step 1 Intuitive method Step 1 Cinit ×Vinit = Cfinal ×Vfinal 1 mL diluted to 10 mL 2 mol L-1 × 1 mL = Cfinal × 10 mL Diluted by factor of 10 Cfinal = = 0.2 mol L-1 10 times less concentrated 2 M initial now 0.2 M after step 1 1:10 dilution of stock solution: conc. of new solution 10 times less than stock solution Dilution Step 2 Intuitive method Step 2 Cinit ×Vinit = Cfinal ×Vfinal 1 mL diluted to 10 mL Diluted by factor of 10 0.2 mol L-1 × 1 mL = Cfinal × 10 mL 10 times less concentrated Cfinal = = 0.02 mol L-1 0.2 M from step 1 now 0.02 M after step 2 Overall, 1:100 dilution of original stock solution (2 M): conc. of new solution (0.02 M) 100 times less than 2 M stock solution. [Original Conc./New Conc. = 2 M / 0.02 M = 100] © 2017 Pearson Education, Inc. 16  Changing the Concentration: Dilutions Answer (continued): Step 3 and Step 4 are calculated in the very same way (and therefore not shown). Overall results for Steps 1-4 serial dilutions are as follows: Stock solution 2 mol L-1 Step 1 0.2 mol L-1 1: 10 dilution Step 2 0.02 mol L-1 1:100 dilution Step 3 0.002 mol L-1 1:1000 dilution Step 4 0.0002 mol L-1 1:10000 dilution 17 [Original Conc./New Conc. = 2 M / 0.0002 M = 10000] © 2017 Pearson Education, Inc.  Titrations: Using Chemical Reactions to Measure Concentrations Titrations use chemical reactions to allow us measure directly the concentration or amount of compound in a solution. Consider the reaction: A+B C A few drops of indicator added that changes colour when reaction has gone to completion Know concentration of either A or B, can determine concentration of other. 18 © 2017 Pearson Education, Inc.  Titrations: Using Chemical Reactions to Measure Concentrations Example 9: A titration yields a colour change when 23.5 mL of a 0.5 M solution of HCl is added to 50 mL of a solution of sodium carbonate (Na2CO3) of unknown concentration. The reaction occurs as follows: 2HCl + Na2CO3 2NaCl + CO2 + H2O What is the concentration of the Na2CO3 solution? Answer: So, HCl and Na2CO3 react in the ratio 2HCl:1Na2CO3 That is, 2 moles of HCl react with 1 mole of Na2CO3 In this example, 23.5 mL of a 0.5 M solution of HCl have reacted. We work out the corresponding number of moles in 23.5 mL 0.5 M = 0.5 moles in 1 L (1000 mL) 1000 mL = 0.5 moles 1 mL = 0.0005 moles 23.5 mL = (0.0005 X 23.5) moles = 0.01175 moles HCl reacted 19 2HCl:1Na2CO3 0.01175 moles HCl = 0.005875 moles Na2CO3 50 mL = 0.005875 moles 1 mL = 0.0001175 moles © 2017 Pearson Education, Inc. So, Conc Na2CO3 = 0.1175 M