Presentation on theme: "Review of Formulas kg = 103 g eg. 1.6 kg = 1.6 x 103 g"— Presentation transcript:
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 gμg = 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 volumeMolar 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 mixingMolar Volume is measured in mol/L, ∴ we can use the equation c = n/Vmol/L can be abbreviated as M or [ ]
4 Calculations with molar solutions Q: How many moles of NaCl are required to make 7.5 L of a 0.10 M solution?M=n/L, n = 0.10 M x 7.5 L = 0.75 mol# mol NaCl =7.5 Lx 0.10 mol NaCl1 L= 0.75 molBut 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?Misslucilaresources.wordpress.com# g NaCl =7.5 Lx 0.10 mol NaCl1 Lx g NaCl1 mol NaCl=43.83 g
5 Practice Questions 63 g 101 g 181 g 82 g a) 22 g b) 1.75 g c) 3 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 KNO3b) 1.85 M H2SO c) 0.67 M KClO33. Calculate the # of grams needed to produce each:a) 0.20 L of 1.5 M KCl b) L of M HClc) 0.20 L of 0.09 mol/L AgNO3d) 250 mL of 3.1 mol/L BaCl24. Give the molarity of a solution containing 10 g of each solute in 2.5 L of solution: a)H2SO4 b)Ca(OH)263 g101 g181 g82 ga) 22 gb) 1.75 gc) 3 gd) 0.16 kg5. Describe how 100 mL of a 0.10 mol/L NaOH solution would be made.a) mol/Lb) mol/L
6 C, V, n, m if these variables are in the question (or required for your answer) use the following stepse.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 flaskIt 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 solutionsThey 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 = c2V2C = concentration; V = volume1 = initial (concentrated); 2 = final (diluted)Don’t forget the equation for molar concentration! (c = n/V)
9 Reading a pipetteIdentify each volume to two decimal places(values tell you how much you have expelled)5.00
10 The Dilution FormulaE.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 LM1V1 = M2V2, M1V1/V2 = M2M2 = (3 mol/L x 1 L) / (6 L) = 0.5 MV1 = 1 LM1 = 3 MV2 = 6 LM2 = 0.5 MM1V1 = 3 molM2V2 = 3 mol
11 ExamplesQ – 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 LM1V1 = M2V2, M1V1/M2 = V2V2 = (12 mol/L x 1 L) / (0.5 L) = 24 LQ – 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 L = 1.5 L# mol/L = 4 mol / 1.5 L = 2.67 mol/L
12 Dilution problemsHow many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution?You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L?mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M?
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 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 togetherFrom #1 & 2 calculate final concentration
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 L of this is poured out, what is the concentration of the remaining 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 L8. 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 HomeworkRead Table 8.7 and Table 8.8 thoroughly to prepare for Friday’s labMANDATORY!!