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Solutions. STOICHIOMETRY, ACIDS AND BASES, THE LITMUS TEST, AND THE WOOD SPLINT TEST. Solutions. Quick definitions: Solution: is a homogeneous mixture composed of two or more substances.
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Solutions STOICHIOMETRY, ACIDS AND BASES, THE LITMUS TEST, AND THE WOOD SPLINT TEST
Solutions • Quick definitions: • Solution: is a homogeneous mixture composed of two or more substances. • In such a mixture a solute is dissolved in another substance, known as a solvent. The solvent does the dissolving. • Ex.: salt is a solute, and water is the solvent. • Homogeneous mixture is a solution that is uniform in composition. If you divide the solution in half, an equal amount of solute and solvent will be found in both halves. • Example.
Simple stoichiometry • Concentrations = g/ml • C1 V1 = C2 V2
Concentrations are typically represented as mass per volume, e.g.: g/ml or g/L. • When liquids or solids are mixed with other liquids, we can also represent the concentration as a percent: • 1 g/ml NaCl = 100% solution of NaCl • Remember, when you see a percentage, the compound is usually dissolved in water (but could be other liquids)
Also, 1g = 1ml so, • 20% solution is equal to 20g/100ml or 0.2g/ml
Also, 1g = 1ml so, • 20% solution is equal to 20g/100ml or 0.2g/ml • Convert the following salt solutions into g/ml: • 80% , 50% , 10% , 1% , and 0.1%.
Problem • In the laboratory, a student is given a quantity of NaCl. He must prepare 1250ml of an NaCl solution with a concentration of 12g/L • What mass of NaCl should the student use?
Problem • You need to make a 250 ml solution of sodium hydroxide with a concentration of 1.5 g/L. • How much sodium hydroxide will you need?
C1 V1 = C2 V2 • C1V1 = C2V2 • Where: • C1 = Concentration of the stock solution • V1 = Volume of the stock solution • C2 = Concentration of the working solution • V2 = Volume of the working solution • NOTE: • The relationship between the volumes and concentrations involved is that of an inverse proportion;that is, when the volume of one solution is greater, the concentration of the solution of the greater volume is less.
C1 V1 = C2 V2 • a. Basic Rules for Solving C1V1 = C2V2 Problems. • Three of the 4 values must be known. • The units of volume and concentration must be the same respectively. • It’s crucial that the volume and concentration that relate to one another are identified. • Any unit of volume or concentration may be used.
Demonstration • I need to make a 10 mg/ml sugar solution. • How many grams in how much water? • Then I to dilute the stock concentration to a 5mg/ml sugar solution with a volume of 200ml. • What do I do?
Problem • You have a solution of HCl with a concentration of 4 g/l and some water. As part of an experiment, you must prepare 200 ml of a diluted solution with a concentration of 0.5 g/l. • Determine the volume of concentrated solution required. • Determine the volume of water you must add to prepare this solution.
Problem • You found a 2.0% solution of sodium hydroxide on the shelf. You need 500 ml of a 0.4 % solution of sodium hydroxide. What volume of the 2.0 % solution will you need?
Tough Problem • How much 0.100 mol/L Ca(OH)2 can be prepared from 10.0 mL of 6.00 mol/L Ca(OH)2?
Practice • A reaction involving 51.0g of aluminum oxide (Al2O3) and 109.5g of hydrochloric acid (HCl) produces 133.5g of aluminum chloride (AlCl3) and a certain amound of water (H2O). • Al2o3 + HCl AlCl3 + H2O • 1.Balance the equation. • 2. Determine the mass of water (H2O) produced during this reaction.
Balance the following: • AlBr3 + Cl2 → AlCl3 + Br2You have 130.5g of aluminum bromide (AlBr3), 145.5g of chloride (Cl2), and 137g of aluminum chloride (AlCl3). How much bromide (Br2) do you end up with?
How much caffeine in a 450 ml coffee which has a concentration of 7g of caffeine per 1L (7g/L)?
If I need to dilute a solution of hydrogen sulfide with a concentration of 6g/L to a 300ml solution with a concentration of 0.7g/L, how much stock solution do I need? • How much water do I need to add?