kg = 10 3 g eg. 1.6 kg = 1.6 x 10 3 g mg = 10 -3 g 6 mg = 6 x 10 -3 g

1. Review of Formulas kg = 103 g eg. 1.6 kg = 1.6 x 103 g mg = 10-3 g 6 mg = 6 x 10-3 g μg = 10-6 g 0.6 mg = 6 x 10-4 g

2. Making Molar Solutions From Liquids (More accurately, from stock solutions)

3. What are molar solutions? A molar solution is one that expresses “concentration” in moles per volume Molar solutions are prepared using: • a balance to weigh solids (in grams) • a pipette to measure small liquid volumes(μL/mL) • a volumetric flask to measure large volumes (L) and for mixing Molar Volume is measured in mol/L, ∴ we can use the equation c = n/V • mol/L can be abbreviated as M or [ ]

4. x 0.10 mol NaCl x 0.10 mol NaCl x 58.44 g NaCl 1 mol NaCl 1 L 1 L Calculations with molar solutions M=n/L, n = 0.10 M x 7.5 L = 0.75 mol Q: How many moles of NaCl are required to make 7.5 L of a 0.10 M solution? # mol NaCl = 7.5 L = 0.75 mol But in the lab we weigh grams not moles, so … Q: How many grams of NaCl are required to make 7.5 L of a 0.10 M solution? # g NaCl = 7.5 L =43.83 g

5. Practice Questions 63 g 101 g • How many grams of nitric acid are present in 1.0 L of a 1.0 M HNO3 solution? 2. Calculate the number of grams needed to produce 1.00 L of these solutions: a) 1.00 M KNO3 b) 1.85 M H2SO4 c) 0.67 M KClO3 3. Calculate the # of grams needed to produce each: a) 0.20 L of 1.5 M KCl b) 0.160 L of 0.300 M HCl c) 0.20 L of 0.09 mol/L AgNO3 d) 250 mL of 3.1 mol/L BaCl2 4. Give the molarity of a solution containing 10 g of each solute in 2.5 L of solution: a)H2SO4 b)Ca(OH)2 181 g 82 g a) 22 g b) 1.75 g c) 3 g d)0.16kg 5. Describe how 100 mL of a 0.10 mol/L NaOH solution would be made. a) 0.041 mol/L b) 0.054 mol/L

6. C, V, n, m  if these variables are in the question (or required for your answer) use the following steps e.g. Given c and V. Find m. • Find # of moles (c=n/v) • Convert to grams (n= m/M)

7. Preparing Standard Solutions  This is a volumetric flask • It is used for preparing solutions (but not for storing them) • A standard solution is one with an accurate, known concentration. This is also known as a stock solution. • These are used as reactant solutions • They usually have a higher concentration than is needed for creating solutions and therefore must be diluted

8. Preparing Standard Solutions After diluting a solution, the concentration of the solution changes. To calculate the new molar concentration, we use the equation: c1V1 = c2V2 C = concentration; V = volume 1 = initial (concentrated); 2 = final (diluted) Don’t forget the equation for molar concentration! (c = n/V)

9. Reading a pipette Identify each volume to two decimal places (values tell you how much you have expelled) 4.48 - 4.50 4.86 - 4.87 5.00

10. The Dilution Formula E.g. if we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L? M1 = 3 mol/L, V1 = 1 L, V2 = 6 L M1V1 = M2V2, M1V1/V2 = M2 M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M V1 = 1 L M1 = 3 M V2 = 6 L M2 = 0.5 M M1V1 = 3 mol M2V2 = 3 mol

11. Examples Q – What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl? M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 L M1V1 = M2V2, M1V1/M2 = V2 V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L Q – 1 L of a 3 M HCl solution is added to 0.5 L of a 2 M HCl solution. What is the final concentration of HCl? (hint: first calculate total number of moles and total number of L) # mol = (3 mol/L)(1 L) + (2 mol/L)(0.5 L) = 3 mol + 1 mol = 4 mol # L = 1 L + 0.5 L = 1.5 L # mol/L = 4 mol / 1.5 L = 2.67 mol/L

12. Dilution problems • How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution? • You have 200 mLof 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L? 3. 100 mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M?

13. 1. M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M) V1 = 0.03125 L = 31.25 mL 2. M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 L M2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L) M2 = 1.2 M 3. M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ? V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M) V2 = 0.4 L or 400 mL

14. Dilution problems (cont’d) • What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl? • What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of 0.2 M NaCl? • 6. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water?

15. Mixing two solutions together, need to find new concentration • Calculate the amount of moles in the final solution (n=m/M) n1+n2=n(final) • Calculate the final volume after mixing the two solutions together • From #1 & 2  calculate final concentration

16. 4. # mol = (2.0 mol/L)(0.4 L) + (3.0 mol/L)(0.6 L) = 0.8 mol + 1.8 mol = 2.6 mol # L = 0.4 L + 0.6 L # mol/L = 2.6 mol / 1 L = 2.6 mol/L 5. # mol = (0.5 mol/L)(3 L) + (0.2 mol/L)(2 L) = 1.5 mol + 0.4 mol = 1.9 mol # mol/L = 1.9 mol / 5 L = 0.38 mol/L 6. # mol = (0.5 mol/L)(3 L) + (0 mol/L)(2 L) = 1.5 mol + 0 mol = 1.5 mol # mol/L = 1.5 mol / 5 L = 0.3 mol/L Or, using M1V1 = M2V2, M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L

17. Dilution problems (cont’d) • Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution? • 8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out,whatistheconcentrationoftheremaining HF?

18. 7. M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ? V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M) V2 = 16 L 8. The concentration remains 0.2 M, both volume and moles are removed when the solution is poured out. Remember M is mol/L. Just like the density of a copper penny does not change if it is cut in half, the concentration of a solution does not change if it is cut in half.

19. Homework • Read Table 8.7 and Table 8.8 thoroughly to prepare for Friday’s lab • MANDATORY!!