Chapter 18 Solutions

Chapter 18Solutions

Liquids • Miscible means that two liquids can dissolve in each other • water and antifreeze, water and ethanol • Partially miscible- slightly • water and ether • Immiscible means they can’t • oil and vinegar

In general the principle concerning whether substances will mix is "Like dissolves Like"; that is, polar (and ionic) substances will dissolve in polar solvents and non-polar substances will dissolve in non-polar solvents, while non-polar and polar substances will not mix.

+ Solubility in Water : The simple rule to remember is ‘like dissolves like 2- + Water is a polar solvent and will dissolve polar molecules and substances that contain charged particles. • Substances that dissolve in H2O are said to be soluble • E.g. Sugar, ethanol which are polar • most ionic compounds Site of polarity can form H-bonding sucrose

Water is polar, the different ends are attracted to the charged ions.  + 2-  + Cl- Na+

Substances that don’t dissolve are called insoluble E.g. Petroleum (crude oil), which are non-polar So if you want to dissolve grease which is non-polar, you need to use a non-polar solvent. Petroleum in a non-polar organic molecule

Solution formation • Nature of the solute and the solvent • Whether a substance will dissolve • How much will dissolve • Factors determining rate of solution... • stirred or shaken (agitation) • particles are made smaller • temperature is increased

Temperature and Solutions • Higher temperature makes the molecules of the solvent move around faster and contact the solute harder and more often. • Speeds up dissolving. • Usually increases the amount that will dissolve (exception is gases)

How Much? • Solubility- The maximum amount of substance that will dissolve at a specific temperature (g solute/100 g solvent) • Saturated solution- Contains the maximum amount of solute dissolved • Unsaturated solution- Can still dissolve more solute • Supersaturated- solution that is holding more than it theoretically can; seed crystal will make it come out;

In a saturated solution an equilibrium is reached between dissolving and recrystallization.

Concentration is... • a measure of the amount of solute dissolved in a given quantity of solvent • Aconcentrated solution has a large amount of solute • Adilute solution has a small amount of solute • thus, only qualitative descriptions • But, there are ways to express solution concentration quantitatively

Concentration = amount of solute amount of solvent (or solution)

Molarity (M) = amount of solute in moles amount of solution in liters

What is the molarity of a solution with 2.0 moles of NaCl in 250 mL of solution? • Molarity= moles/liter • = 2.0 mol/ .250 L • = 8.0 M

Calculating molarity • Calculate the molarity of Na2SO4 (s) if you have 4.2 moles in 500. L of solution. • Molarity = moles/liter • = 4.2 mol/ 500. L • = 8.4 x 10-3 M • Could also have mol/l or molar as units

How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution? • Moles = (0.75 mol/l)(6.0 l) • = 4.5 mol

Calculate the mass of Na2 SO4 (s) in 1.6 L of a 2.5 M solution. • Molarity = moles/volume • Moles = (molarity)(volume) • = ( 2.5 mol/l) ( 1.6 l) • = 4.0 mol moles M x V Then Find mass Mole x molar mass mass

Need molar mass: 142.02 g/mol Mass =568.08g = 570 g

Making solutions • 10.3 g of NaCl are dissolved in a small amount of water, then diluted to 250 mL. What is the concentration? • How many grams of sugar are needed to make 125 mL of a 0.50 M C6H12O6 solution?

Alternate Measures of Concentration • Molar concentration is the accepted method of determining the amount of substance dissolved in a solvent. • parts per million (ppm), • parts per billion (ppb) • parts per trillion (ppt).

Part per million • ppm = mass of solute x 106 mass of solution

Part per billion • ppb = mass of solute x 10 9 mass of solution

Part per trillion ppt = mass of solute x 10 12 mass of solution

Molality (independent of temperature) 1 kg of water = 1 liter of water Molarity = moles of solute kg of solvent

Molality • - What is the molality when 0.750 mol is dissolved in 2.50 L of solvent? • Molality = 0.750 mol 2.50 kg = 0.300m

Molality • Ex- Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 kg of pure water (the solvent). What would be the molality of the solution?

Step One: convert grams to moles. • Step Two: divide moles by kg of solvent to get molality. • 58.44 grams/mol is the molecular weight of NaCl • 58.44g x 1 mol = 1.00 mol. • 58.44 g Molality = 1.00 mol 2.0 kg =0.500 mol/kg (or 0.500 m). or 0.500-molal.

1) Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water. • 2) 80.0 grams of glucose (C6 H12 O6, mol. wt = 180. g/mol) is dissolved in1.00 kg of water. Calculate the molality. • 3) Calcuate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent.

Mass Percent 1. We can describe the concentration of a solution by using the mass percent. Mass percent is the mass of solute present in a given mass of solution. • Mass Percent = mass of solute  100% • Mass of solution • = grams of solute  100% • (grams of solute) + (grams of solvent)

Example 1: We make a solution by dissolving 1.0g of sodium chloride in 48g of water. What is the mass percent of the solute? Mass Percent = 1 .0 g solute x100% 49 g solution = 2.0% NaCl

Example 2: A solution is prepared by dissolving 1.00g of ethanol, C2H5OH, with 100.0g of water. What is the mass percent of ethanol in this solution? Mass Percent = 1 .00 g ethanol 100% 101 g solution = 0.990% C2H5OH

Example 3: Cow’s milk typically contains 4.5% by mass of the sugar lactose, C12H22O11. Calculate the mass of lactose present in 175 g of milk. Mass of solution (milk) = 175g Mass percent of solute (lactose) = 4.5% Mass Percent = grams of solute  100% grams of solution 4.5% = grams of solute  100% 175 g Grams of solute = (4.5%)(175 g) = 0.045  175g 100% Grams of solute = 7.9g lactose

Dilution M1V1 = M2V2 Adding water to a solution

Dilution • The number of moles of solute doesn’t change if you add more solvent! • The # moles before = the # moles after • M1 x V1 = M2 x V2 • M1 and V1 are the starting concentration and volume. • M2 and V2 are the final concentration and volume. • Stock solutions are pre-made to known Molarity

Ex:1 - 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make? M1V1 = M2V2 (1.50 mol/L)(53.4 mL) = (0.800 mol/L) (x) • (1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x) (0.800 mol/L) (0.800 mol/L) • 100. mL = x

Ex:2 - 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results? • M1 V1 = M2 V2 (2.500 mol/L)(100.0 mL)=(0.5500mol/L)(x) • 454.55 mL = x • 454.6 mL = x

Ex: 3 - 53.4 mL of a 1.50 M (mol/l) solution of NaCl is on hand, but you need some 0.500 M solution. How many ml of 0.500 M can you make? • M1V2 = M2V2 • (1.50 mol/L) (53.4 ml) = (0.500 mol/L) (x) • 160. ml = x

Cb Vb = Ca Vb • This is the same as the dilution equation, concentration before and volume before is equal to the concentration after and the volume after.

Practice a. 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity? b. You have 150 mL of 6.0 M HCl. What volume of 1.3 M HCl can you make? c. Need 450 mL of 0.15 M NaOH. All you have available is a 2.0 M stock solution of NaOH. How do you make the required solution?

Chapter 18 Solutions