Chemistry 101: Solutions, Concentrations, and Molarity, Lecture notes of Chemistry

Solutions and explanations for various chemistry concepts related to homogeneous mixtures, solvents and solutes, concentration, molarity, titration, and dilution. It includes examples, formulas, and calculations. Students can use this document as study notes, summaries, or cheat sheets to help them understand these topics.

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Solutions and concentration
Solution: a homogeneous mixture of two or
more substances. Example: water, sugar, flavor
mixture (Coke). The substances are physically
combined, not chemically combined or bonded
to each other.
Nanoscale pictures: Figs. 5.1, 5.4, and 5.5
Solvent: usually the substance in the greater
amount. The substance used to dissolve the
solute or solutes. Example: water.
Solute: usually the substance in the lesser
amount. The substance dissolved by the solvent.
Example: sugar.
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Solutions and concentration Solution: a homogeneous mixture of two or more substances. Example: water, sugar, flavor mixture (Coke). The substances are physically combined, not chemically combined or bonded to each other. Nanoscale pictures: Figs. 5.1, 5.4, and 5. Solvent: usually the substance in the greater amount. The substance used to dissolve the solute or solutes. Example: water. Solute: usually the substance in the lesser amount. The substance dissolved by the solvent. Example: sugar.

Concentration: the ratio of the amount of solute to the total amount of solution. Examples: 80 proof alcohol, 3% hydrogen peroxide, 12 M HCl. In chemistry, we use molarity (M) because it’s based on moles and the mole ratio concept (Ch 3 and 4). Molarity = moles of solute = mol = M Volume of solution L Example: What is the molarity of a sugar solution of 3. g sugar dissolved in a total volume of 100* mL? Sugar = C 12 H 22 O 11 = 342 g/mol. Msugar = [C 12 H 22 O 11 ] =?

Titration example: moles OH

  • = moles H + 17.85 mL of 0.104 M NaOH solution is needed to titrate 20.00 mL of H 2 SO 4 solution. What is the concentration of the H 2 SO 4 solution? Stoichiometry (mole ratio) in H 2 SO 4 H 2 SO 4 has two acidic protons (H

). When titrating H 2 SO 4 with OH

, keep in mind that 2 OH

are needed for every 1 H 2 SO 4!

More practice with molarity Dissolve 1.75 g (NH 4 ) 2 SO 4 in enough water to make 250* mL of solution. What is the concentration (in M) of all ions present in the solution? MM of (NH 4 ) 2 SO 4 = 132.139 g/mol.

Dilution: a process in which the concentration (molarity) of a solution is lowered. The amount of solute (atoms, moles, grams, etc.) remains the same, but the volume is increased by adding more solvent. Example: orange juice from frozen concentrate. molesstock = molesdiluted Stock solution: the more concentrated solution that is used to make the less concentrated solution. A sample of the stock solution is diluted by adding more solvent to the sample. Mstock x Vstock= molesstock= molesdiluted = Mdiluted x Vdiluted moles x L = moles L molesstock = molesdiluted

Example: molesstock = molesdiluted A stock solution of urea (NH 2 ) 2 CO has a concentration of 2.94 x 10

  • 2 M. How many milliliters (mL) of the stock solution are needed to prepare 500* mL of a solution with a concentration of 1.35 x 10
  • 3 M? Vstock =?

Conservation of energy Energy can neither be created nor destroyed. The total energy of the universe is constant. (Law of Conservation of Energy, First Law of Thermodynamics) Example: A hot bar of copper at 100 °C is placed into cold water at 4.0 °C. What is the final temperature of the bar and water? Internal energy of each substance (kinetic and thermal energy) Heating (heat): the transfer of thermal energy between substances with different temperatures when they are in contact with each other.

Need to know how much energy is needed to change the temperature of each substance (the copper and the water). Heat capacity: the amount of thermal energy gained or lost when the temperature of a substance changes by one degree. For water, 4.184 J of energy for every 1°C change for each gram of water that changes temperature. For copper, 0.385 J per 1 °C per 1 gram. Written as specific heat capacity, c : cwater = 4.184 J = 4.184 J g

  • 1 °C - 1 g °C ccopper = 0.385 J = 0.385 J g
  • 1 °C
  • 1 g °C

Substance 1: q 1 = c 1 x m 1 x ΔT 1 Substance 2: q 2 = c 2 x m 2 x ΔT 2 Direction of energy transfer and signs: q 1 = - q 2 (same as q 2 = - q 1 ) c 1 x m 1 x ΔT 1 = - (c 2 x m 2 x ΔT 2 ) Or m 1 c 1 ΔT 1 = - (m 2 c 2 ΔT 2 ) Can be asked to solve for any of the 8 variables: m 1 , m 2 , c 1 , c 2 , Ti,1, Ti,2, Tf,1, Tf, m = mass in grams, and must use specific heat capacity = c = s. c = J or cal g-°C g-°C

More on precipitation reactions Precipitation: a process in which a solid forms from a solution. Soluble: a term that describes a solid that easily dissolves in a solvent. Dissolve = soluble. Example: NaCl in water. Insoluble: a term that describes a solid that does not easily dissolve in a solvent. Not dissolve = insoluble. Example: CaCO 3 (chalk, seashells, etc.) in water. Soluble and insoluble are actually the two extremes on a scale of solubility. We oversimplify when we classify a solid as either extreme.