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Warm-Up Problem! What volume of 0.175 M solution of KOH is needed to titrate 30.0 mL of 0.200 M H 2 SO 4 ?

Warm-Up Problem!

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Warm-Up Problem!. What volume of 0.175 M solution of KOH is needed to titrate 30.0 mL of 0.200 M H 2 SO 4 ?. Chapter 10 : Gases. Units of Pressure. mm Hg or torr These units are literally the difference in the heights measured in mm ( h ) of two connected columns of mercury. Atmosphere - PowerPoint PPT Presentation

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Page 1: Warm-Up Problem!

Warm-Up Problem!

What volume of 0.175 M solution of KOH is needed to titrate 30.0 mL of 0.200 M H2SO4?

Page 2: Warm-Up Problem!

Chapter 10 : Gases

Page 3: Warm-Up Problem!

Units of Pressure

• mm Hg or torrThese units are literally the difference in the heights measured in mm (h) of two connected columns of mercury.

• Atmosphere1.00 atm = 760 torr

Page 4: Warm-Up Problem!

Manometer

Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

Page 5: Warm-Up Problem!

Boyle’s LawThe volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

P1V1 = P2V2

Page 6: Warm-Up Problem!

Charles’s Law

• The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

A plot of V versus T will be a straight line.

V1 = V2

T1 T2

Page 7: Warm-Up Problem!

Avogadro’s Law

• The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.

Relates V and moles

Page 8: Warm-Up Problem!

1 mol gas = 22.4 L gas, D ∝ MW

Page 9: Warm-Up Problem!

Ideal Gas Equation

V 1/P (Boyle’s law)V T (Charles’s law)V n (Avogadro’s law)

nTPV

PV = nRT

Page 10: Warm-Up Problem!

Warm-Up!

A sample of gas occupies a volume of 1248 ft3 at 0.988 atm and 28 °C.

Calculate the pressure of the gas if its volume is decreased to 978 ft3 while its temperature is held constant.

Page 11: Warm-Up Problem!

Calculating Density and Molecular Weight!

nV

PRT

=

PRT

mV

=

(x M=)

PRT

d =

dRTP =

(m/v=d)

rearrangeto solve for MW

Page 12: Warm-Up Problem!

Cumulative Problem!

An organic compound had the following analysis: C, 55.8% by mass; H, 7.03%; O, 37.2%. A 1.500 g sample was vaporized and was found to occupy 530 cm3 at 100oC and 740 torr. What is the molecular formula of the compound?

Page 13: Warm-Up Problem!

AP Question 1972A 5.00 gram sample of a dry mixture of potassium hydroxide, potassium carbonate, and potassium chloride is reacted with 0.100 liter of 2.0 molar HCl solution.

(a) A 249 milliliter sample of dry CO2 gas, measured at 22°C and 740 torr, is obtained from the reaction. What is the percentage of potassium carbonate in the mixture?

(b) The excess HCl is found by titration to be chemically equivalent to 86.6 milliliters of 1.50 molar NaOH. Calculate the percentages of potassium hydroxide and of potassium chloride in the original mixture.

Page 14: Warm-Up Problem!

Warm –Up!

A Scuba divers tank contains 0.29 kg of oxygen compressed into a volume of 2.3 L.

Calculate the pressure inside the tank at 9 °C.

What volume would this oxygen occupy at 26 °C and 0.95 atm.

Page 15: Warm-Up Problem!

Agenda for Today

• Go back to 1972…• Partial Pressure discussion• Partial Pressure Practice Problems!

Page 16: Warm-Up Problem!

KOH + HCl KCl + H2OK2CO3 + 2HCl 2KCl + H2O + CO2

KCl + HCl no reaction1. Find % K2CO3 in original 5 g total solidUse PV=nRT to find moles of CO2

.249 L * .974 atm = 0.01 moles CO2

.0821Latm/molK(295K)0.01 moles CO2 x 1molK2CO3 = 0.01 mol K2CO3

1 mol CO2

0.01 mol K2CO3x 138g/mol = 1.38 gFind percent mass = 1.38g K2CO3/5 g total solid = 27.6%

2. Find moles of original HCl 0.1L x 2 mol/L = 0.2 mols HCl total

3. Find moles of HCl 0.1 moles K2CO3 x 2 mol HCl = 0.02 moles HCl reacted with K2CO3 1 mol K2CO3

4. Use NaOH titration to find unreacted HCl:0.0866L NaOH x 1.5 mol = 0.13 moles NaOH = 0.13 moles HCl (b/c 1:1)

L

5. Find moles of HCl that reacted with KOH from steps 2, 3, and 4:0.2 moles total – 0.02 moles reacted with K2CO3 - 0.13 moles excess = 0.05 moles HCl

6. Find grams of KOH: 0.05moles HCl x 1 mol KOH x 56.1 g KOH = 2.81g KOH 1 mol HCl 1 mol KOH

7. Find mass % KOH 2.81g KOH = 56.2 %5g total

8. Find % KCl 100 - 56.2 - 27.6 = 16.1 % KCl

Page 17: Warm-Up Problem!

Dalton’s Law of Partial Pressures• The total pressure of a mixture of gases

equals the sum of the pressures that each would exert if it were present alone.

Ptotal = P1 + P2 + P3 + …

P1 = ntRTV

nt = n1 + n2 + n3 +...

Page 18: Warm-Up Problem!

Partial Pressure Practice (say that 3 times fast)

What is the total pressure exerted by a mixture of 2.00 g of hydrogen and 8.00 g of nitrogen at 273 K in a 10.0-L vessel?

Page 19: Warm-Up Problem!

Partial Pressure = Mole Fraction x Ptot

• What’s a mole fraction?– The relationship between the moles of each gas to

the total moles of the gas mixture

n1/ntotal = 1

Use the mole fraction and total pressure to find partial pressures!

P1 = 1 (Ptotal)

What is the mole fraction of oxygen in air?Air = 78% nitrogen, 21% oxygen, 1% water

vapor, 0.9% argon, and .04% carbon dioxide

Page 20: Warm-Up Problem!

Partial Pressures

• A mixture of 9.00 g of oxygen, 18 g of argon, and 25 g of carbon dioxide exert a pressure of 2.45 atm. What is the partial pressure of Argon in the mixture?

• What is the partial pressure of oxygen in air? (Atmospheric pressure = 760 torr)

Page 21: Warm-Up Problem!

AP Multiple Choice Warm Up!!• A student pipetted five 25.00-milliliter samples of hydrochloric acid and

transferred each sample to an Erlenmeyer flask, diluted it with distilled water, and added a few drops of phenolphthalein to each. Each sample was then titrated with a sodium hydroxide solution to the appearance of the first permanent faint pink color. The following results were obtained.

Volumes of NaOH Solution First Sample..................35.22 mL

Second Sample..............36.14 mLThird Sample.................36.13 mLFourth Sample..............36.15 mLFifth Sample..................36.12 mL

Which of the following is the most probable explanation for the variation in the student's results?

(A) The burette was not rinsed with NaOH solution.(B) The student misread a 5 for a 6 on the burette when the first sample was titrated.

(C) A different amount of water was added to the first sample.(D) The pipette was not rinsed with the HCl solution.(E) The student added too little indicator to the first sample.

Page 22: Warm-Up Problem!

Collecting Gases over Water

Ptotal = Pgas + PH2O

2KClO3(s) 2KCl(s) + 3O2(g)

The volume of gas collected is 0.250 L at 26°C and 765 torr total pressure. (a) How many moles of O2 are collected? (b) How many grams of KClO3 were decomposed? (The pressure of water vapor at 26°C = 25 torr .)

Page 23: Warm-Up Problem!

Kinetic – Molecular Theory• Gases consist of large numbers of molecules

that are in continuous, random motion.• The combined volume of all the molecules of

the gas is negligible relative to the total volume in which the gas is contained.

• Attractive and repulsive forces between gas molecules are negligible.

• The collisions between molecules are perfectly elastic.

• The average kinetic energy of the molecules is proportional to the absolute temperature.

Page 24: Warm-Up Problem!

• Pressure is caused by collisions of molecules with the walls of the container!– Magnitude of pressure is related to how

often and how forcefully the particles strike the walls.

• Temperature of a gas is a measure of the average kinetic energy of its molecules!– Motion increases with increasing

temperature.

• Why does pressure increase with increasing temperature?

Page 25: Warm-Up Problem!

r1 = M2

r2 M1√t1 = M1

t2 M2√

KE = 1/2mu2

Page 26: Warm-Up Problem!

Effusion Problem• At a particular T and P, neon gas effuses at a

rate of 16 mol/s.

• (a.) What is the rate at which Ar effuses under the same conditions?

• (b.) Under a different set of conditions, 3.0 mol of Ar effuse in 49.0 seconds. How long will it take an equal amount of helium to effuse?

Page 27: Warm-Up Problem!

Deviation from Ideal Behavior

Real gases at high pressure arelike the mall at christmas.

Too many people in a small space,people are on top of one another!Compared to the space in the checkout line, the people in front of you and behind you have significant volume. Ugh…

If molecules are crowded in a container, as pressure increases attractiveforces take over. Also the volume of the gas particles are significant compared tothe volume of the container. NOT IDEAL!

Page 28: Warm-Up Problem!

Deviation from Ideal BehaviorAt low temperature there is notenough energy to overcome attractive forces and the gas isforced to condense into a liquid.

Ice is NOT an ideal gas…

The van der Waals Equation – corrects for non-ideal behavior

) (V − nb) = nRTn2aV2(P +

Page 29: Warm-Up Problem!

Non-Ideal Practice

Arrange the following gases in order of increasing deviation from ideality:

H2O, CH4, Ne

WHY???