Chapter 15 – Solutions

Chapter 15 – Solutions 15.1 Definitions related to Solutions           Solution - uniform mixture of two or more substances. A solution is composed of a solute(s) dissolved in a solvent. Solute – the substance present in lesser amount in a solution. Solvent – the substance present in greatest amount in a solution. Water is the most common solvent. Soluble – when a solid dissolves in a solvent. An example is salt is soluble in water. Insoluble – when a solid does not dissolve in a solvent. An example is gold metal in water. Gold is insoluble in water. Miscible – when two liquids mix together. An example is tonic and vodka. Immiscible – when two liquids do not mix but rather form layers. An example is oil and vinegar. Dilute – when there is not a lot of solute in the solvent we say the solution is dilute. Concentrated – when there is a lot of solute in the solvent we say the solution is concentrated. An example would be very strong coffee. Concentration – a measurement that gives a number and unit as to how concentrated or dilute a solution is. Concentration can be measured in mass mass percent (nursing) or molarity (chemistry). The physical states of the solute and solvent can be solid, liquid or gas: CO2 gas dissolved in liquid soda (gas in liquid), gin and tonic (2 liquids), Brass is Cu and Zn (2 solids), salt water is NaCl dissolved in water (solid in liquid), air is N2 and O2 gases primarily with hundreds of trace gases (gas in gas). 15.2 Like Dissolves Like The “Like dissolves like” rule says that    Polar solutes dissolve in polar solvents Nonpolar solutes dissolve in nonpolar solvents Nonpolar and polar substances don’t mix or dissolve in one another. Ionic solids often dissolve in polar solvents – check solubility rules on your periodic table to see if the solid dissolves in water. But ionic solids do not dissolve in nonpolar solvents because ions have charges while nonpolar solvents have no charges at all. So they are not alike and therefore do not mix. 1 polar solvent nonpolar solvent polar liquid miscible immiscible nonpolar liquid immiscible miscible polar solid soluble insoluble nonpolar solid insoluble soluble ionic solid Check Solubility Rules! insoluble 15.3 The Dissolving Process As an ionic solid solute is dropped into a liquid solvent such as water, the solvent molecules begin to attack and pull apart the solute ions. The water molecules align themselves so that opposite charges attract. In the picture to the right we see how the water aligns with the - oxygens pointing in towards the Na+1 ion because opposite charges attract. YouTube videos that shows how a salt dissolves in water and a second one. 15.4 Mass Mass Percent The concentration of a solution may be expressed quantitatively as a mass/mass %: Mass Percent  g solute g solute  100%   100% g solution g solute  g solvent  Note that the mass of solution = mass of solute plus mass of solvent YouTube videos on Mass Mass Percent 1 and Mass Mass Percent 2 2 15.5 Molarity The concentration of a solution can also be expressed quantitatively as molarity: Molarity M  moles solute Liters of solution Units : M  mol Liter Quite often the problem with give you grams of solute so you have to convert grams to moles using the molar mass. YouTube video on Molarity Practice Problems Ex. 1: Ex. 2 Indicate whether the following are soluble / insoluble in or miscible / immiscible with water: vegetable oil: AgI(s): NaCl(s): HBr (l): C5H12(l): grease(s): Br2 (l): Mg(OH)2(s): Circle all of the following that would be soluble in or miscible with vegetable oil: I2 (s) KCl(s) P4 (s) grease(s) HBr(l) MgBr2(s) 1) A solution is prepared by dissolving 12.3 g of Na2SO4 in 71.6 g of water. a) The solute is ___________, and the solvent is _____________. b) What is the mass percent for this solution? 2) If a normal IV glucose solution is 5.00 % glucose, what is the mass of solution that contains 15.0 g of glucose? 3 3) What mass of solute is required to prepare 150.0 g of a 2.50% KCl solution? 4) What mass of water is required to dissolve 5.00 g of NaOH to prepare a 10.0% NaOH solution? 5) Find the molarity of a solution prepared by dissolving 0.250 moles of NaOH in 100.0 mL of solution: 6) Find the molarity of a solution prepared by dissolving 30.6 g of NaOH in 350 mL of solution. Answers Ex. 1: vegetable oil: immiscible AgI: insoluble NaCl: soluble HBr (l): miscible C5H12 (l): immiscible grease (s): insoluble Br2 (l): immiscible Mg(OH)2: insoluble Ex. 2 I2 (s) KCl(s) P4 (s) grease (s) HBr(l) MgBr2(s) 4 Smith, Clark (CC-BY-4.0) GCC CHM 130 Chapter 15: Solutions 1) The solute is Na2SO4 , and the solvent is H2O . mass % = 12.3 g 12.3 g x 100% = x 100% = 14.7 % 12.3g  71.6 g 83.9 g 2) 5.00 = (15.0 / x ) * 100 5.00x = 15.0 * 100 x = 3.00x102 g solution 3) 2.50 = (x / 150.0 ) * 100 2.50 * 150.0 = x * 100 x = 3.75 g KCl 4) 10.0 = (5.00 / x ) * 100 10.0x = 5.00 * 100 x = grams solution total = 50.0 g = g solute + g solvent Now 50.0 g solution – 5.00g solute = 45.0 g solvent water 5) Find the molarity of a solution prepared by dissolving 0.250 moles of NaOH in 100.0 mL of solution:  1L   = 0.1000 L 100.0mL  1000 mL  M= 0.250 mol NaOH mol solute = = 2.50 M NaOH L solution 0.1000 L 6) Find the molarity of a solution prepared by dissolving 30.6 g of NaOH in 350 mL of solution.  1 mol NaOH   = 0.765 mol NaOH 30.6 g   40.0 g NaOH   1L   = 0.350 L 350 mL   1000 mL  M= 0.765 mol NaOH mol solute = = 2.19 M NaOH L solution 0.350 L 5