Click here to load reader

View

11Download

2

Tags:

Embed Size (px)

DESCRIPTION

materi kimdas

SolutionSolution

Mass PercentageMass Percentage

Mass % of A = mass of A in solutiontotal mass of solution 100

Parts per Million andParts per Billion

ppm = mass of A in solution

Parts per Million (ppm)ppm mass of A in solution

total mass of solution 106

Parts per Billion (ppb)

ppb = mass of A in solutiontotal mass of solution 109

Mole Fraction (X)moles of AX

Mole Fraction (X)moles of A

total moles in solutionXA =

In some applications one needs the mole In some applications, one needs the mole fraction of solvent, not solutemake sure you find the quantity you need!you find the quantity you need!

Molarity (M)Molarity (M)

mol of soluteL of solutionM =

You will recall this concentration measure from Chapter 4.from Chapter 4.

Because volume is temperature dependent molarity can change withdependent, molarity can change with temperature.

Molality (m)Molality (m)

mol of solutekg of solventm =

Because neither moles nor mass changeBecause neither moles nor mass change with temperature, molality (unlike molarity) is not temperature dependentis not temperature dependent.

Changing Molarity to MolalityChanging Molarity to Molality

If we know the densityIf we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations

(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. What is the mass (a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4 g of Zn2+What is the concentration of Zn2+ in parts per million?

PRACTICE EXERCISE(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3 62 mass % sodium hypochlorite NaOCl What is the mass of NaOClcommercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?

PRACTICE EXERCISEA commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the molality and (b) the mole fraction of NaOCl in the solution.

SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations

(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. What is the mass (a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4 g of Zn2+What is the concentration of Zn2+ in parts per million?

Solution(a) Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent(a) Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent (0.100 kg = 100 g). From this we must calculate the mass percentage of solute.Plan: We can calculate the mass percentage by using Equation 13.5. The mass of the solution is the sum of the

mass of solute (glucose) and the mass of solvent (water).Solve:

Comment: The mass percentage of water in this solution is (100 11.9)% = 88.1%. (b) Analyze: In this case we are given the number of micrograms of solute. Because 1 g is 1 106 g, (b) a y e t s case we a e g ve t e u be o c og a s o so ute. ecause g s 0 g,5.4 g = 5.4 106 g. Plan: We calculate the parts per million using Equation 13.6.Solve:

PRACTICE EXERCISE(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?

Answers: (a) 2.91%, (b) 90.5 g of NaOCl Answers: (a) 0.505 m, (b) 9.00 103

Colligative PropertiesColligative Properties Colligative properties depend only onColligative properties depend only on

the number of solute particles present, not on the identity of the solutenot on the identity of the solute particles.

Among colligative properties are Among colligative properties areVapor pressure lowering Boiling point elevationBoiling point elevationMelting point depressionO tiOsmotic pressure

Vapor PressureVapor Pressure

As solute molecules are added to a solution, the solvent become less volatile (=decreased vapor(=decreased vapor pressure).

Solute solventSolute-solvent interactions contribute to this effect.o s e ec

Vapor PressureVapor Pressure

Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

Raoults LawRaoult s Law

P = X PPA = XAPAwhere XA is the mole fraction of compound A PA is the normal vapor pressure of A at A

that temperature

NOTE: This is one of those times when you want to make sure you have the vapor pressure of the solvent.

SAMPLE EXERCISE 13.8 Calculation of Vapor-Pressure Lowering

Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25C. Calculate the vaporGlycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 C. Calculate the vapor pressure at 25C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25C is 23.8 torr (Appendix B).

PRACTICE EXERCISEThe vapor pressure of pure water at 110C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110C. Assuming that Raoults law is obeyed, what is the mole fraction of ethylene p g y , yglycol in the solution?

SAMPLE EXERCISE 13.8 Calculation of Vapor-Pressure Lowering

Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25C. Calculate the vaporGlycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 C. Calculate the vapor pressure at 25C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25C is 23.8 torr (Appendix B).

SolutionAnalyze: Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute.Plan: We can use Raoults law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, XA, is the ratio of the number of moles of solvent (H2O) to total solution ( l C H O l H O)(moles C3H8O3 + moles H2O).

Solve: To calculate the mole fraction of water in the solution, we must determine the number of moles of C3H8O3 and H2O:

SAMPLE EXERCISE 13.8 continued

We now use Raoults law to calculate the vapor pressure of water for the solution:

The vapor pressure of the solution has been lowered by 0.6 torr relative to that of pure water.

We now use Raoult s law to calculate the vapor pressure of water for the solution:

PRACTICE EXERCISEThe vapor pressure of pure water at 110C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110C. Assuming that Raoults law is obeyed, what is the mole fraction of ethylene glycol in the solution?

Answer: 0.290

Boiling Point Elevation and Freezing Point Depression

Solute-solvent interactions also cause solutions to have higher boiling points and lowerpoints and lower freezing points than the pure solventthe pure solvent.

Boiling Point ElevationBoiling Point ElevationThe change in boiling

i t i ti l tpoint is proportional to the molality of the solution:solution:

Tb = Kb x m

where Kb is the molal boiling point elevationboiling point elevation constant, a property of the solvent.T is added to the normalTb is added to the normal

boiling point of the solvent.

Freezing Point DepressionFreezing Point Depression The change in freezing

i t b f dpoint can be found similarly:

T KTf = Kf x m

H K i th l l Here Kf is the molal freezing point depression constant ofdepression constant of the solvent.

Tf is subtracted from the normal ffreezing point of the solvent.

Boiling Point Elevation and Freezing Point Depression

In both equations, T T = K x mIn both equations, Tdoes not depend on what the solute is, but

Tb = Kb x monly on how many particles are di l d Tf = Kf x mdissolved. Tf Kf m

Colligative Properties of Electrolytes

Because these properties depend on the number of ti l di l d l ti f l t l t ( hi hparticles dissolved, solutions of electrolytes (which

dissociate in solution) show greater changes than those of nonelectrolytes.

e.g. NaCl dissociates to form 2 ion particles; its limiting vant Hoff factor is 2.

Colligative Properties of Electrolytes

However, a 1 M solution of NaCl does not showHowever, a 1 M solution of NaCl does not show twice the change in freezing point that a 1 Msolution of methanol does.

It doesnt act like there are really 2 particles.

vant Hoff Factorvan t Hoff Factor

One mole of NaCl inOne mole of NaCl in water does not really give rise to two moles of ions.

vant Hoff Factorvan t Hoff FactorSome Na+ and Clreassociate as hydrated ion pairs, so the true concentrationthe true concentration of particles is somewhat less than two times the concentration of NaClNaCl.

The vant Hoff FactorThe van t Hoff Factor

Reassociation is moreReassociation is more likely at higher concentration.

Therefore, the number of particles present is concentration dependentdepend