Properties of Solutions Glenn V. Lo, Ph.D. Department of Physical Sciences Nicholls State University.

Properties of Solutions Glenn V. Lo, Ph.D. Department of Physical Sciences Nicholls State University

Properties of Solutions LESSON OBJECTIVES. Students will be able to Express concentration in various ways Calculate amount of solute present in samples of solutions of known concentration Determine amounts of solutes and solvent needed to prepare solutions of desired concentrations Describe and calculate concentrations of molecular and ionic solutes Describe and calculate colligative properties of solutions (vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure) Determine molar mass of solute particles from colligative properties Use Henry’s Law and Raoult’s Law to describe solutions with volatile solutes

Solution Concentration Concentration: amount of solute (or solvent) amount of solvent or solution Type 1: denominator is volume of solution; depends on temperature. Ex. Molarity, mg/dL, %(w/v), %(v/v) Type 2: denominator is mass or moles of solvent or solution; independent of temperature. Ex. Molality, mass percent or %(w/w), mole fraction, ppm, ppb

Type 1 Concentration Units Molarity = moles solute / Liters solution = mmol solute / mL solution %(w/v) = [ grams solute / mL solution ] x 100% = [ grams solute / dL solution ] Ex. Vinegar is 5%(w/v) acetic acid in water I.V. glucose solution is 5%(w/v) glucose Physiological saline solution is 0.9%(w/v) NaCl %(v/v) = [ mL solute / mL solution ] x 100% = [ mL solute / dL solution ] Ex. “100 proof” liquor is 50%(v/v) ethanol mg/dL = [milligrams solute / dL solution] Ex. blood glucose and cholesterol levels

Test Yourself Suppose 5.844 g of NaCl is mixed with enough water to make a 200.0 mL solution. Calculate molarity and %(w/v).

Test Yourself Which of these has the highest molarity? A. 0.10 mol NaOH in 100.0 mL solution B. 0.05 mol NaOH in 100.0 mL solution C. 0.10 mol NaOH in 50.0 mL solution

Test Yourself Which has a higher %(w/v)? A. 5.0 g NaOH in 100.0 mL solution B. 5.0 g NaCl in 100.0 mL solution C. neither (same)

Test Yourself Which has a higher molarity? A. 5.0 g NaOH in 100.0 mL solution B. 5.0 g NaCl in 100.0 mL solution C. neither (same)

Type 2 Concentration Units Type 2: denominator is mass or moles of solvent or solution Molality = moles solute / kg solvent Percent by mass or %(w/w) = [ grams solute / grams solution ] x 100% ppm = [ grams solute / grams solution ] x 10 6 = [ mg solute / kg solution ] = [  g solute / grams solution ] ppb = [ grams solute / grams solution ] x 10 9 = [  g solute / kg solution ] mole fraction = moles component / total moles x A = n A / (n A + n B + …)

Test Yourself Suppose 18.0 g of glucose (C 6 H 12 O 6 ) is mixed with 200.0 g of water. Calculate Molality %(w/w) or mass percent ppm ppb mole fractions

Amount of Solute If Type 1 concentration unit is known, Amount of Solute = (known concentration) times (Volume of solution) grams solute = %(w/v) x (V in dL) mg solute = (Conc in mg/dL) x (V in dL) moles solute = M x (V in L) mmol solute = M x (V in mL)

Test Yourself A diabetic’s blood glucose level is found to be 225 mg/dL two hours after ingesting 75 grams of glucose. How much glucose can be found in a 5-cc blood sample drawn from this person? [Note: 1 cc = 1 mL]

Test Yourself A commonly used reagent in the laboratory is concentrated HCl, which is 12.0 M HCl(aq). How much HCl is present in a 5.0 mL sample of concentrated HCl?

Test Yourself Which contains more moles of solute? A. 10.0 mL of 0.50M NaCl B. 20.0 mL of 0.25M NaOH C. neither (same)

Test Yourself Which contains more grams of solute? A. 10.0 mL of 0.50M NaCl B. 20.0 mL of 0.25M NaOH C. neither (same)

Amount of Solute If Type 2 concentration unit is known, Amount of Solute = (known concentration) times (Amount of solvent or solution) Moles solute = molality x (kg of solvent) Moles component = mole fraction x (total moles) Grams solute = mass percent x (grams of solution) Milligrams solute = ppm x (kg of solution) Micrograms solute = ppb x (kg of solution)

Test Yourself A solution was prepared by mixing 2.00 g of an unknown compound with 20.0 g of water. The molality of the unknown aqueous solution was found to be 0.50 mol/kg. The number of moles of solute in the sample is… A. 10 mol, B. 0.010 mol, C. 1.0 mol What is the molar mass of the unknown compound ?

Preparing Solutions Decide on desired concentration and value of numerator (amount of solute) or denominator (amount of solvent or solution) Use defining equations to calculate denominator or numerator Convert denominator and numerator to measurable lab quantities (grams) Liquid solute or solvent – if you know density, convert to mL and use “TD” glassware to dispense required volume If desired concentration is Type 1, put solute in “TC” glassware and add enough solvent to get desired total volume. (Why? Volumes of components do not necessarily add up to the volume of solution.)

Test Yourself How would you prepare 500.0 mL of 0.500M NaCl(aq)? Desired Concentration = Desired Denominator Volume of solution = Calculate Numerator Moles of solute = Convert denominator and numerator to measurable quantities in the lab Grams of solute = Amount of solvent =

Preparing Solutions: Dilution Dilution = adding more solvent Why? Useful for preparing very low concentrations (very dilute solutions) Typical laboratory balances are not sensitive enough for small masses. Uncertainty of commonly available “analytical balance” is + 0.2 mg. Some substances are more readily available in solutions (example: HCl)

Test Yourself Solution A is 10.0 mL of 0.10M NaOH(aq). Solution B is prepared by adding enough water to solution A to get a total volume of 100.0 mL. Which of the following is true? A. Solution A contains more NaOH B. Solution B contains more NaOH C. Solution B has a higher molarity D. none of the above

Test Yourself Which of these most likely needs to be prepared by dilution? A. 100 mL of 0.10 M NaCl(aq) B. 2.0 L of 1.00x10 -6 M NaCl(aq)

Test Yourself In the laboratory, you have a stock solution of 12.0 M HCl(aq). You want to prepare 500.0 mL of 0.50M HCl. How much stock solution would you need to dilute to 500.0 mL?

Test Yourself Rubbing alcohol is typically 70%(v/v) isopropanol in water. How would you prepare 2.00 L of 0.100M isopropanol (C 3 H 7 OH) in water, how much of the rubbing alcohol do you need? Density of pure isopropanol is 0.786 g/mL. Strategy: calculate moles of isopropanol, convert to grams, convert to mL pure, convert to mL sample.

Converting concentration units Strategy (same type): assume any value for the denominator (amount of solution or solvent) use defining equations to calculate the numerator (amount of solute) Convert denominator and numerator to desired units

Test Yourself Vinegar is 5.0%(w/v) CH 3 COOH in water; calculate molarity

Test Yourself Calculate %(w/v) of 0.15M NaCl solution.

Test Yourself According to the American Association for Clinical Chemistry: (http://www.labtestsonline.org/understanding/analytes/glucose/test.html) Glucose concentration in blood collected after an 8-10 hour fast is considered normal if it is within 70-99 mg/dL. A diabetic person will have fasting blood glucose level higher than 125 mg/dL. If a person’s fasting glucose concentration is 8.0 mM or 8.0x10 -3 M, is the person diabetic?http://www.labtestsonline.org/understanding/analytes/glucose/test.html

Test Yourself Concentrated HCl is 37.0%(w/w) HCl in water, calculate molality, ppm, ppb, x HCl, x water

Test Yourself Suppose in a mixture of benzene (C 6 H 6 ) and toluene (C 6 H 5 CH 3 ), the mole fraction of benzene is 0.250. Calculate x toluene, %(w/w) toluene, %(w/w) benzene, molality, ppm, ppb

Converting concentration units Strategy (Type 1 to Type 2, or vice versa) If Type 1 concentration is known, assume any volume of solution, then use density to calculate the mass of solution. If Type 2 concentration is known, assume any value for the denominator (amount of solution or solvent), then figure out mass of solution and use density to calculate the volume of solution. Convert known quantities to desired quantities for numerator and denominator

Test Yourself Concentrated HCl is 37.0%(w/w) HCl in water; density of solution is 1.20 g/mL. Calculate molarity.

Test Yourself A 5.08 M NaOH(aq) solution has a density of 1.22 g/mL. Calculate the molality.

Very Dilute Aqueous Solutions For very dilute aqueous solutions: molarity and molality are almost equal. ppm (mg/kg is approx. same as mg/L) ppb (  g/kg is approx. same as  g/L) Why? Density of water is 1.00 kg/L. Small amount of solute hardly affects the density of a large amount of water. Therefore, we can say that, in very dilute solutions, 1 L soln = 1 kg soln = 1 L solvent = 1 kg solvent

Test Yourself For which of the following is the molality of HCl essentially equal to the molarity? A. 12M HCl (aq), B. 0.10M HCl (aq)

Test Yourself For which of the following is the concentration essentially equal to 5 mg/L. A. 5 ppm benzene in water B. 5 ppm benzene in ethyl alcohol Note: the density of ethyl alcohol is 0.79 g/mL

The Solution Process SOLVATION: surrounding of a solute particle by solvent particles. HYDRATION: solvation with water molecules

Ionic Solutes When an ionic compound dissolves in water, all ions are dispersed and individually hydrated. Example, in a 0.2M Ca(NO 3 ) 2 (aq): [Ca 2+ ] = 0.2 mol/L [NO 3 - ] = 2x0.2 mol/L, or 0.4 mol/L [Ca(NO 3 ) 2 ] = 0

Test Yourself A 1-L aqueous solution contains 0.10 mol NaCl and 0.20 mol Na 2 SO 4. What is the concentration of Na + ions in the solution? A. [Na + ] = 0.30 mol/L, B. [Na + ] = 0.40 mol/L C. [Na + ] = 0.50 mol/L

Test Yourself Which figure best represents the following solutions? NaCl(aq) CaCl 2 (aq) KNO 3 (aq)

Molecular Solutes Strong acids (HCl, HBr, HI, HNO 3, HClO 4, H 2 SO 4 ) are essentially 100% ionized in water. Example. HCl(aq)  H + (aq) + Cl - (aq) In 0.50M HCl(aq), [H + ] = 0.50 mol/L, [Cl - ] = 0.50 mol/L [HCl] = 0 mol/L

Molecular Solutes Weak acids and bases are very slightly ionized. All other molecular compounds are not ionized. Weak acid example: in 0.50M HNO 2 (aq), HNO 2 (aq) H + (aq) + NO 2 - (aq) [H + ] = trace, [NO 2 - ] = trace [HNO 2 ]  0.50 mol/L (essentially)

Test Yourself In 0.50M HC 2 H 3 O 2 (aq), which of the following species has a concentration closest to 0.50 mol/L? A. H +, B. C 2 H 3 O 2 -, C. HC 2 H 3 O 2

Molecular Solutes Weak acids and bases are very slightly ionized. Weak base example: in 0.10M NH 3 (aq), NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) [NH 4 + ] = trace, [OH - ] = trace [NH 3 ]  0.10 mol/L (essentially) Note: base yields OH - ions in water

Test Yourself Which of the following has a higher total concentration of solute particles? A. 0.10 M HCl B. 0.10 M HF

Test Yourself For which of these is the concentration of solute molecules closest to 0.2 mol/L? A. 0.2M HNO 3 (aq) B. 0.2M HF(aq) C. 0.2M NH 3 (aq) D. 0.2M C 6 H 12 O 6 (aq) E. 0.1M NaCl(aq)

Evidence of ionization Pure water is a poor electrical conductor; aqueous solutions of ionic compounds and strong acids conduct electricity [“electrolytes”]. Most molecular compounds are either nonelectrolytes or weak electrolytes. Colligative properties: can be used to determine total concentration of solute particles. Example: a solution containing 0.001 mol NaCl appears to have 0.002 mol of solute particles based on colligative properties

Osmolality and Osmolarity Osmolality = total moles of solute particles per kilogram of solvent Osmolarity = total moles of solute particles per Liter of solution With respect to a given solute: osmolality = (i )(molality) osmolarity = (i )(molarity) i = van’t Hoff factor

Test Yourself If the molality of NaCl in a solution is 0.50 mol/kg, what is the osmolality of the solution ? Assume NaCl is completely dissociated under these conditions. A. 0.50 m, B. 1.00 m, C. 0.25 m

Test Yourself What is the osmolarity of a solution which is 0.20 M with respect to Na 2 SO 4 and 0.50M with respect to glucose (C 6 H 12 O 6 )? A. 0.70M, B. 1.10 M, C. 0.90 M, D. 1.70 M

Test Yourself Suppose “MX” is 90% ionized in water MX(aq) = M + (aq) + X - (aq) What is the osmolarity of 0.20M MX(aq)? A. 0.20 M B. between 0.2 and 0.4 M C. between 0.4 and 0.6 M HINT: i = (1-f) + nf

Test Yourself Suppose “AB” is partially ionized in water AB(aq) = A + (aq) + B - (aq) If the total solute concentration in 0.50M AB(aq) is 0.60M, what fraction of AB molecules are ionized? A. 10%, B. 20%, C. 50%

Colligative Property depends on the total concentration of solute particles depends on the nature of solvent, not the solute. C.P. = (constant) times (total concentration of solute particles) where constant depends on nature of solvent

Colligative Properties Vapor pressure lowering, P*-P = (P* solvent )(x solute ) Freezing point depression, T f -T f * = - K f m Boiling point elevation, T b -T b * = K b m Osmotic pressure,  = MRT NOTE: x, m and M in these formulas refer to total concentrations of solute particles.

Vapor Pressure Vapor pressure = partial pressure of gas in equilibrium with liquid. At “equilibrium,” rates of vaporization and condensation are equal. Vapor pressure … is constant at a given temperature for a given liquid increases with temperature reaches 760 mm Hg, or 1 atm at the normal boiling point of a liquid (“normal” means at sea level, with a barometric pressure of 1 atm)

Test Yourself Consider two identical closed containers on the right with different amounts of water. At equilibrium (same temperature), in which container is the partial pressure of H 2 O(g) higher? A. Container A B. Container B C. Neither (same)

Test Yourself Consider two identical closed containers on the right with different amounts of water. At equilibrium (same temperature), which container has more water molecules in the gas phase? A. Container A B. Container B C. Neither (same)

Test Yourself The vapor pressure of water at room temperature is …. A. Less than 1 atm B. Equal to 1 atm C. Higher than 1 atm

Test Yourself Ethyl alcohol boils at 78.4 o C. Which of these two liquids has a higher vapor pressure at 78.4 o C? A. water, B. ethyl alcohol

Vapor Pressure Lowering If nonvolatile solutes are added, vapor pressure is lowered. Explanation: sites on liquid surface where solvent molecules can leave are blocked by solute molecules. Raoult’s Law: P A = P A *x A (A = solvent) P A *– P A = P A * - P A *x A = P A *(1-x A ) = P A *x B P A *– P A is the “vapor pressure lowering” Nonvolatile solutes = ordinarily solids in pure form

Test Yourself The vapor pressure of saltwater at 100 o C is… A. less than 1 atm B. equal to 1 atm C. higher than 1 atm

Test Yourself Which of the following has a higher vapor pressure of water at the same temperature? A. 0.1 mol NaCl + 99.8 mol water B. 0.1 mol C 6 H 12 O 6 + 99.9 mol water

Test Yourself What is the vapor pressure of water in a solution at 85 o C, containing 18.0 g glucose, C 6 H 12 O 6, and 5.844 g NaCl dissolved in 500.0 grams of water? The vapor pressure of pure water at 85 o C is 433.6 mm Hg.

Boiling Point Elevation Vapor pressure lowering  elevation of boiling point. A higher temperature is needed to raise the vapor pressure to 1 atm.  T b = T b -T b * = K b m K b depends on solvent; for water is 0.512 o C/m or K/m K b is called the ebullioscopic or boiling point elevation constant Note: m is total molality of solute particles

Test Yourself Which boils has at a higher temperature? A. 0.100 m NaCl(aq) B. 0.100 m C 12 H 22 O 11 (aq)

Test Yourself What is the boiling point of a solution containing 18.0 g glucose, C 6 H 12 O 6, and 5.844 g NaCl dissolved in 500.0 grams of water?

Freezing Point Depression  T f = T f -T f * = -K f m K f depends on solvent; for water is 1.86 o C/m or K/m K f is called the cryoscopic or freezing point depression constant m is total molality of solute particles

Test Yourself On a winter day, sprinkling salt on an icy road will … A. cause the ice to melt B. cause any liquid water in contact with the ice to freeze

Freezing Point Depression What is the freezing point of a solution containing 18.0 g glucose, C 6 H 12 O 6, and 5.844 g NaCl dissolved in 500.0 grams of water?

Osmotic Pressure Osmosis: passage of solvent molecules through a semi- permeable membrane Natural tendency: from high solvent concentration to low solvent concentration Osmotic pressure = additional pressure on solution to prevent a net flow of solvent molecules from the pure side.  = MRT At higher additional pressure, reverse osmosis occurs. (Application: purification of water)

Test Yourself Suppose the left compartment in the figure shown below contains pure water and the right compartment contains saltwater. Assuming the barrier allows only water molecules to pass. What is the net direction of water flow? A. left-to-right B. right-to-left

Test Yourself Osmotic pressure of seawater is equivalent to that of an aqueous solution containing 27 g NaCl per liter. What pressure can be used to desalinate seawater by reverse osmosis at room temperature (298K)? A. 20 atm, B. 25 atm

Test Yourself An intravenous solution is 5.00%(w/v) glucose. Calculate its osmotic pressure. This is an example of an “isotonic” solution. What is the %(w/v) of an isotonic saline solution, NaCl(aq)?

Test Yourself A hypertonic solution is more concentrated and has higher osmotic pressure than an isotonic solution. If 0.3M glucose is isotonic, which of these is hypertonic? A. 0.3M NaCl(aq), B. 0.1 M CaCl 2 (aq)

Determining molar mass Prepare mixture. (Know mass of solute and solvent) Measure colligative property Determine total concentration of solute particles Calculate total moles of solute particles Average Molar mass = mass of solute divided by total moles of solute particles If van’t Hoff factor = 1, molar mass equals average molar mass.

Determining molar mass Grams soluteGrams solvent Moles solute ? Kilograms solvent  T f or  T b Osmolality ? x = 

Test Yourself A 2.000 g sample of unknown solid is mixed with 20.00 g of benzene. Pure benzene freezes at 5.533 o C. The mixture freezes at –0.87 o C. K f for benzene is 5.12 o C/m. Calculate the molar mass of the unknown.

Test Yourself The molar mass of NaCl is 58.44 g/mol. If we were to attempt to determine the molar mass of solute particles in NaCl(aq) from colligative property data, what average molar mass would we get? (assume complete ionization) A. 106.88 g/mol, B. 29.22 g/mol

Solutions with volatile solutes Volatile solutes = ordinarily exist as liquids or gases in pure form Ideal liquid mixture: each component obeys Raoult’s Law. For a mixture of liquids “A” and “B” P = P A * x A + P B * x B Note: most components deviate from Raoult’s Law when their mole fraction is significantly less than 1.

Test Yourself At a given temperature, pure liquid “A” has a vapor pressure of 80 Torr, while pure liquid “B” has a vapor pressure of 100 Torr. If 0.20 mol of liquid B is mixed with 0.80 mol of liquid A, assuming ideal behavior, the vapor pressure of the resulting solution is… A. between 80 and 100 Torr B. greater than 100 Torr C. less than 80 Torr

Test Yourself At a given temperature, pure liquid “A” has a vapor pressure of 80.0 Torr, while pure liquid “B” has a vapor pressure of 100.0 Torr. In a liquid mixture of A and B, the mole fraction of B is 0.20. Assuming ideal behavior, the mole fraction of B in the vapor is… A. larger than 0.20 B. less than 0.20 C. still 0.20

Solutions with volatile solutes Volatile solutes typically obey Henry’s Law. The concentration of volatile solute “A” is directly proportional to the partial pressure of “A” in the vapor above the solution. Unit for proportionality constant depend on concentration and pressure units used. If k H unit is pressure, P A / x A = kH What is the unit for k H if it is equal to [A]/P A ? What is the unit for k H if it is equal to P A /m, where m is the molality?

Test Yourself If you open a bottle of soda, carbon dioxide bubbles out of the solution. This suggests that the pressure of CO 2 in the sealed bottle is _________ than the partial pressure of CO 2 in the atmosphere. A. higher, B. lower

Test Yourself Estimate the molar concentration of O 2 in water if it is saturated with O 2. The Henry’s law constant, k H, for O 2 in water is 1.3x10 -3 M/atm, and air is about 21% O 2

Test Yourself Suppose a scuba diver goes down to a depth of 100 ft. Eventually, by approximately how much will the concentration of dissolved gases change in his bloodstream? Hint: air pressure at the surface is 1 atm; underwater it’s higher; 1 Torr or 1 mm Hg is equivalent to 13.6 mm H 2 O since mercury is 13.6 times more dense than water. A. 2x, B. 3x, C. 4x

Example If solute obeys Raoult’s law, then it obeys Henry’s law: If P A = P A * x A, then k H = P A *, However, k H is general not necessarily equal to P A * Acetic acid (HC 2 H 3 O 2 ) @298K: P*=15.73 Torr, In water @298K:k H = 1.3 Torr Calculate the vapor pressure of acetic acid in 1-molal acetic acid solution? Hint: First calculate mole fraction of acetic acid Express k H in Torr/(mol kg) Express k H in molality/atm

Properties of Solutions Glenn V. Lo, Ph.D. Department of Physical Sciences Nicholls State University.

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Properties of Solutions Glenn V. Lo, Ph.D. Department of Physical Sciences Nicholls State University.

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