sign in sign up
<<
Home , Ideal gas law, Solubility, Solution

GAS LAWS &

KINETIC MOLECULAR THEORY (KMT) (pp. 426-429)

I. The kinetic molecular theory was developed to help predict the behavior of . It describes the behavior of gases at the molecular level.

2. This theory is based on assumptions about a theoretical known as an I1H;;)q L coAS

3. The assumptions are: a. particles are so small that the of the individual particles if they were at rest is essentially zero when compared with the total volume of the gas. b. Ideal gas particles are in constant, rapid, random motion, moving in straight lines in all directions until they collide with other particles. c. There are no attractive or repulsive between particles and collisions between particles are elastic. d. The average kinetic of the particles is directly proportional to the absolute (measured in ).

FOUR VARIABLES DESCRIBE A GAS

Temperature

I. The temperature of a gas determines the average kinetic energy of the particles.

2. While the average kinetic energy of a collection of gases at a given temperature will be same, the velocity at which they travel will not. This is because the values of the various gases are different.

3. Temperature is usually measured in either or . Both are related to one another:

Volume

I. A derived unit: ~ X ~S X \1

2. Gas are expressed four different ways: .3 fY\

Pressure

I. The that the gas particles exert over a unit area: FvR.C.G' fftr=ssu/(<= s: UNIT ~"'A 2. Pressure is a measure of the total force exerted by the moving particles of a gas as they collide with the walls of the container.

3. is measured in terms of: cd-IVt.

10(,.3 J:{

Standard Temperature and Pressure (STP)

1. In order to study the effects of changing temperature and pressure on a gas, one must have a standard for comparison.

2. STP represents a pressure and temperature that are fairly easy to reproduce:

T s: a 7.3 J( oCe A,.:JD /,00 a.t-~

373K l!l,.:J o 10/, 3 ~p~ f .:= Iol-Wl a13~ A.v 0 l\oO~ CJ-13K -4..)0 7~O,O I11nA WJ

d13 II: "\"-'D 7~(}. 0 .h.l\' r

Particles

I. Gases with equal volumes under the same conditions of pressure and temperature have equal numbers of particles.

2. This is Avo GA~ 's LA\,J and it holds true only for gases.

3. The amount of gas is measured in Wl 0 Ie

4. The volume of one of gas at STP is known as the MO LA l V()LlJ/VI~

I VV\~:- a.~.Lj L

GAS LAWS (pp. 403-415)

I. The gas laws are a series of mathematical relationships that relate the following variables:

~I - Tbl'V\P8Z,A 'NfL l;- V- vo t.vm E- f- Pt~s,ul-l? V\ - t'\10LE~ Charles' Law

1. Relates volume and temperature.

2. As the temperature decreases, the volume of a gas decreases. As temperature increases, the volume of a gas increases.

3. The volume of a gas at constant pressure is directly proportional to the absolute temperature.

4. Charles' Law helped prove the existence of .

5. Mathematically this is: V, Yz.. TJ.

***** A sample of gas occupies 24.0m3 at IOO.OK. Determine its volume at 400.0K. V, 'V-z. s: - Tz.- Vz.... z t{Oo]

***** Gas in a occupies 2.50L at 300.0K. At what temperature will the balloon expand to 7.50L?

VI \[ L ( 3elO, 0) (1 ..r0 ) .=: Ia. :: T, T2.. 01·50

Q,S-O L 7, S-o l ::. 9(Jo. K 30G. I~ OJ. 9,00 )((0 K

Boyle's Law

I. The volume of a gas at constant temperature is inversely proportional to the pressure.

2. Like Charles' Law, we can write:

P,V. ~ P,- VL

***** The gas in a balloon has a volume of 4.00L at IOO.OkPa. The balloon is released into the atmosphere and the gas in it expands to a volume of 8.00L. Determine the pressure on the balloon at the new volume.

PIV, =~ 1-

(4.00)( 100.0) .: P'2. (6-.0d)

( ~•0 0) ( I00 .0] (8' ·00 J ***** Ifthe pressure of a 2.50m3 sample of gas is 1.50 atm, what volume will the gas occupy if the pressure is changed to 7.50 atm?

~ V I ~ P2. \f 2-

( I •50 ')( Ol ' S-O) z: ( 1.~o 1V d- (I. so) (0I.$1:J)_ 7. )"0

Gay-Lussac's Law

I. The pressure of a given gas varies with the Kelvin temperature when the volume remains constant.

2. This is expressed as: P, P-z.. - T, TL ***** The pressure of a gas in a tank is 324.24kPa at 295.0K. Determine the gas pressure if the temperature is raised to 333.0K. Pr P2- :: - T, /2.-

{J-z... = 33.3

***** A gas in a sealed container has a pressure of 125.0kPa at 30.0oC. Determine the temperature in the container ifthe pressure is increased to 201.0kPa. P, P; T _(z.ul)(303) --I .., ~ -, ILJ T, do

'leQ lol- 0 :: 303.0 -I~

Dalton's Law (pp. 420 - 425)

I. The total pressure in a gas is the sum of the partial of the individual components.

2. The M~n '"' L P..'2k--:s.sUIZl.?' of a gas is the pressure of an individual gas in a gas mixture that contributes to the total pressure of the mixture.

3. This is expressed mathematically as:

PrDTAL:: PT:: ~ of p~-f Ps -I •• , 4. 's law is often used to determine the pressure of a gas collected over .

***** gas is collected over water at a total pressure of 95.0kPa. The volume of gas collected is 28.0mL at 25.0oe. Determine the of the hydrogen gas if the pressure is 3.17kPa.

p·rt)"; ;! P,..j'L- + P;.fz.o

q S .0 llc(. :: (J,-h + .3.' 1)c~

-Rf 2-:: q 1 ,If k Ib.

***** A gas is collected over water at 50.0De and a barometric pressure of I05.00kPa. Determine the pressure of the gas if the water is 12.34kPa.

PruT ~ ~aJ ·f PI-h. 0

10 S. o0 .kPtIl " 10J-I I;), a if

fjaS .: q;;J. & (p:: q ;). 7)c ~

COMBINED GAS LAW

1. The combined gas law states the relationship among pressure, volume and temperature of a fixed amount of gas:

i~ V,

2. The combined gas law allows one to out problems involving more variables that change.

***** A gas at 11O.0kPa and 30.0oe fills a flexible container with an initial volume of2.00L. Ifthe temperature is raised to so.o-c and the pressure increased to 440.0kPa, what is the new volume? T,!' e;l1,3.,. 3.0 :: .30 .3

T2'= Jli3+ 81J:- 3.53

(110) ( t.ooo') (Ifl(O) V l :: 3.5 3

:58';) L ***** At O.OOoCand 1.00 atm pressure, a sample of gas occupies 30.0mL. If the temperature is increased to 30.0oC and the entire gas sample is transferred to a 20.0mL container, what will be the gas pressure inside the container? T, ~ d-/3 -10 s: ';)'73 (( .;173 /2.. ~ ~ 7"3 -+ 30 s: :lO 3. J{. P,V, :: f2- v'z Tz-

IDEAL GAS LAW (pp. 415-420)

1. The number of moles is the fourth variable that can be used along with pressure, volume and temperature to describe a gas.

2. The is: N;:: n f2... T

p ~ N'-£SSlJe..(;-- T; nll1 Pt:--.eA7VR.-l::-

\/: VO LU /l'\ (; {l : MVLE"j

3. Ifpressure is expressed in atmospheres, then: L'JYY\ wwl /(

4. If pressure is expressed in kPa, then:

R.:: [,,3/4

***** Determine the number of moles of gas in a 3.00L container at 300.0K and a pressure of 1.50 atm. Pv= fl Q./ py (1,.$)(3) (\: ~ ~T :: C. oill) ( 300)

***** Determine the Celsius temperature of2.49 moles of gas contained in a l.OOL container at a pressure of 143kPa. N= nQ.. T PV (P/3)(1' T :: .- :: flfL (~,l(q)( 4', 3/ LI) s: ~,q, K

0 il.:- Ci" 313 SOLUTIONS

WHAT ARE SOLUTIONS?

Characteristics of Solutions

1. Solutions are homogeneous containing two or more substances called the solute and the .

2. The ,sOLUTF is the substance that dissolves.

3. The SoLV&.J T is the dissolving medium.

4. A may exist as a gas, or depending on the state of its solvent.

5. Solutions can contain more than one solute.

6. Some combinations of substances can readily form solutions and others do not.

7. A substance that dissolves in a solvent is said to be SCJI..v&L/.;;' in that solvent.

8. A substance that does not dissolve in a solvent is said to be ,bJSuLlJ ,g, L/;; in that solvent.

9. Two that are not soluble in each other are said to be rmfl1I'sC., &rL t:

10. Two liquids that are soluble in each other are said to be M J sc, 6 L(;;'

Solvation in Aqueous Solutions

1. To form a solution, solute particles must separate from one another and the solute and solvent particles must mix.

2. Forces of attraction exist between the pure solute particles, between the pure solvent particles, and between the solute and solvent particles.

3. When a solid solute particle is placed in a solvent, the solvent particles completely surround the surface of the solid solute.

4. If the forces of attraction between the solvent and solvent particles are greater than the attractive forces holding the solute particles together, the solvent particles pull the solute particles apart and surround them.

5. These surrounded solute particles then move away from the solid solute out into the solution.

6. The process of surrounding solute particles with solvent particles to form a solution is called ~LvA T7(nJ . When it occurs in water it is called 1-y1D{lj:\TION

7. Lit-IE t>ISSOLVES Ll\(G" is the general rule used to determine whether salvation will occur in a specific solvent.

8. Water will dissolve ionic compounds such as sodium and polar compounds such as , but not nonpolar substances such as oil. Factors That Affect Rate of

I. For salvation to occur the solute and solvent particles must come in contact with one another.

2. There are three ways to increase the collisions between solute and solvent particles which then increases the rate of solvation:

a. A e, ITA 'n ,,-' 6 fH E m I 'f. .,1/-eJ:

b . .:t"NC~""'f\ s I ~ G n+G ScJW'"i"E"'5 SuZi==l-\q; Ml.::.~

Heat of Solution

1. Energy is needed to overcome the forces of attraction of the solute particles for each other and solvent particles for each other.

2. This process is l:"'N DO n-t Gilm I.C

3. When the solute and solvent particles mix, they are attracted to one another and energy is released. This process is ic""!

4. The overall energy change that occurs during solution formation is called the t-ltt\T of ~LV n .,J

Solubility

1. Only a limited amount of solute can dissolve in a solvent at a given set of conditions.

2. .)OwS ILI T'1 refers to the maximum amount of solute that will dissolve in a given amount of solvent at a specified temperature and pressure.

3. It is expressed in terms of: J SOLVn::

/DOj SOI..Y/;1J I

4. When a solute dissolves, its particles mix in with the solvent particles. As time passes, the spaces available to the solute particles decrease and they collide with one another and the original crystal. Some of the solvated particles will re-attach themselves to the crystal.

5. As long as the rate of salvation remains greater than the rate of , salvation will continue.

6. Depending on the amount of solute present, the two rates may eventually equalize and no more solute appears to dissolve.

7. A state of dynamic equilibrium is reached. As one particle is solvated, another crystallizes. The overall amount of dissolved solute remains constant.

8. Such a solution is said to be a Sl.\nJe4Tb---n SOLI) 77U,J ; it contains the maximum amount of dissolved solute for a given amount of solvent at a specific temperature and pressure. 9. An U~eA n l) SOLUT] VAlis one that contains less dissolved solute for a given temperature and pressure.

Factors That Affect

1. Pressure affects the solubility of gases. The greater the pressure, the more gas that can be dissolved in a liquid.

2. Solubility also depends on the nature of the solute and the solvent.

3. Temperature affects the solubility of all substances. For most substances, increasing the temperature of the solvent increases the solubility of the solute.

4. A SllP/':/l!i.AnJeA-n-t:> can be formed using temperature effects. This type of solution contains more dissolved solute than a saturated solution at the same temperature.

5. Pressure affects the solubility ofa gas. The solubility ofa gas in any solvent increases as its external pressure increases.

Solubility Curves

I. Solubility curves can be used to predict the mass of a solute that will dissolve in IOOgof water at various .

***** Using the solubility curves provided, answer the following questions:

a. Determine the solubility of (KN03) at 60°C. II S J-

b. Determine which , chloride (NH4CI) or (KCI) has a greater solubility at OOC.

c. Determine which of the following salt's solubility in water is most temperature dependent: (NaN03), potassium chloride (KCI) or (NaCl).

t--.I~JJ03

***** A saturated solution of NaN03 in IOOgof water at 60°C is cooled to 10°C. Determine the mass, in grams, of solute that will precipitate out of solution.

***** A solution ofNH4CI in IOOgof water at 60°C contains 52.0g of solute. This solution is vtJ S4TVRAW D

***** A solution of HCI in lOOg of water at 40°C contains 61.0g of solute. This solution is ..5,+nJt

***** A solution of KCI03 in IOOgof water at 80°C contains 45.0g of solute. This solution is ~uP8Z.~Tk~ SOLUTION

1. The totJCk~'TlO,J ofa solution is a measure of how much solute is dissolved in a specific amount of solvent or solution.

2. Concentration may be described qualitatively as eitherCUI\lCk'")J~ 11::J:> or D IL u 1l::: but these are not useful to a chemist.

Using Percent to Describe Concentration

I. Concentration expressed as a percent is a ratio of a measured amount of solute to a measured amount of solution.

2. Percent by mass is the ratio of the solute's mass to the solution's mass: MASS SOLvTl:;- (100) MASS SoLuno,.J

MAS~ o~ SoLVn ON .: W\A Sso;:: ScJLu IE" + flIIA.n CFF !oLV cxr r= ***** What is the percent by mass ofNaHC03 in a solution containing 20.0g NaHC03 dissolved in 600.0 mL of water? I./.1 OJ

j I~l-O .: £toO.O n1 L 't Id~)O'>1 L z: ~OO. OJ

~O. 0 :: & io. OJ

;/0.00 (IUU) .: l.t LtJ, 0

***** A student has 1500.0g of a bleach solution. The percent by mass of the solute, sodium hypochlorite (NaOCI) is 3.62%. Determine the mass ofNaOCI in the solution. X

x t: s tf. ~J-

3. Percent by volume usually describes solutions in which both solute and solvent are liquids. "'OLuM G .sou.) 113"' (IIJD) SOLUnIJ,J

VVL\)YYlLE- af=' S()L-vnoA!.:- \/OLUm~ SOLlJn: i "0 LV M~ SO Lv'l:?J 'T ***** Determine the percent by volume of in a solution that contains 35.0 mL of ethanol dissolved in 115.0 mL of water.

3$,0 - ---- (H'S-+ 3 s-) ***** A student has 100.0 mL of30.0% of oil in . Determine the volumes of the oil and the gasoline present in the solution. ')( .:: 30, OM L 0 I L ('0 (») /00,0 loa. 0 - .ie:!. () == 70,0 w. L

l 3.00 z

Molarity

1. MOLA~ \ T'-f is the number of moles of solute dissolved per liter of solution.

2. M

3. To calculate a solution's molarity, one must know the volume of the solution and the amount of dissolved solute: (11OLl? s 0 F -SOL.v n=

***** Determine the molarity of an aqueous solution containing 40.0g of (C6H1206) in 450.0 mL of solution.

+se.o WI L:- O.l.{SO L

1V\u--Q -to, 0J CeJI,z. O~ \ I",~ I-I'L O,~ Ci. I~IL0 (, = c.. • ;J:J-:J I I PO. 1"-1 Ct, !l,LO(; - h1J • ol..Ol ;). l\.loLI~ \ 'r-( -- - .4S0 - O. lfq3 11f..

***** A bleach solution ofNaOCI has a concentration ofO.128M. Determine the mass ofNaOCI in the solution if its volume is 1.00 L.

. rz 8 r.o o t,

y- 128 MuJ

d L ~ 11'uJ Jr", a cJ 171/,1./ tfs tJ~oC/ In-,,.f «l« Preparing Molar Solutions

I. Many lab solutions are prepared by diluting stock solutions purchased from a supply company.

2. To determine the volume of stock solution that must be diluted use: MJ\j, .;:.M z, \/2- (Yj a.: IV(OLAe. t rf DiLVTE ~ ,YloiAlLn't .).TCCl-\c:.. SuL.\JDOrJ

I/OWWlG DC S:m(¥ .5:OLV·noIJ \]a. I' VOLU"ftI,e D1Lun; ,sQLUno"'-' ***** Determine the volume, in mL, of a 3.00M KI stock solution needed to make 0.300 L ofa 1.25M KI solution. M ' I M V , '\I r::: z, t: {~,oo)\J, : (l'dS)(,30o)

(I. J.S)(. 3(0) ./;).S L iss ; l

***** Determine the molarity ofa dilute solution made by starting with 20.0 mL ofa 3.50M stock solution and diluting it to 100.0 mL. M/ VI :- M2- v; (3 • .5)( so. 0): 1\1d ( 100.0")

(~.s) (1.0, 0") z: 0, "70 fY/ (IOO,O) OF SOLUTIONS

1. Solutes affect some of the physical properties of their .

2. Most of these effects are caused by the number of particles in the solution rather than the type of particle in solution.

3. Physical properties of solutions that are affected by the number of particles, but not the identity of dissolved solute particles are called WLLI~A'n v(; fMPI.:<-a:(n;;,5

4. Colligative properties include:

Electrolytes and Colligative Properties

I. Ionic compounds are called because they dissociate in water to form a solution that conducts an electric .

2. Some molecular compounds also ionize in water.

3. Electrolytes that produce many are called strong electrolytes while those that produce only a few ions in solution are called weak electrolytes.

4. Compare the dissolution of 1 mole ofNaCI and I mole ofCaCh in water:

.IONS Vapor Pressure Lowering

I. Vapor pressure is he pressure exerted in a closed container by liquid particles that have escaped the liquid's surface and entered the gaseous state.

2. Adding a nonvolatile solute (one that has little tendency to become a gas) to a solvent lowers the solvent's vapor pressure.

3. This is because the solvent's surface is now a combination of solvent particles and solute particles. There are fewer solvent particles available to escape from the surface so fewer particles enter the gaseous state and the vapor pressure is lowered.

4. Thus, IfAJb2. P.et.~ utI,; LOW!£/N <:>isdue to the number of solute particles in solution.

Boiling Point Elevation

I. When the temperature of a solution containing a nonvolatile solute is raised to the of the pure solvent, the resulting vapor pressure is still less than atmospheric pressure and the solution will not boil.

2. The solution must be heated to a higher temperature to supply the additional kinetic energy needed to raise the vapor pressure to atmospheric pressure.

3. The temperature difference between a solution's boiling point and a pure solvent's boiling point is called f.Iol LIN 6 Po"..1 T I:::l.l.::-V An 0 Ai maLA L BOILI,J II Pol p. T" 4. It can be expressed as: k/b :: l-LL:>fAnv;J CO,\! S'/'f\N T (07'1'\11)

at t:: blSSoCltlno,J r:A C-TVR.. (~ OF PMnc.u;:-s)

***** Determine the new boiling point ofa 2.75m solution ofNaOH in water. II 16 z: ~,df t

s: (.SI2.)(,;l)(~7~) z: d.3/C,"C

Freezing Point Depression

I. In a solution, the solute particles interfere with the attractive forces among the solvent particles.

2. This prevents the solvent from entering the solid state at its normal freezing point.

3. The freezing pint of a solution is always lower than that of a pure solvent.

4. A solution's FttEEtl,J6 Po/,..\T DEfJ{l£Sf6.er/, is the difference between its freezing point and the freezing point of its pure solvent.

5. It can be expressed as: ~ If :: .l,\dt' 1<'+

<:At ': DISSOCI4 no~ i=At..YOe.. (.f:t OF P..4.~Tl(LbS")

W\ z: MOLAUN M= .sOLVTlu,'-f It!,.. " = h-\OLAL ~~/JoJ(' i'ollJT 1>l?PflksSlo,J evNSmAi'T (0 /~~1r. ) ***** Determine the freezing point of a O.029m aqueous solution of sodium chloride (NaC\).

NaCI(s) - Na" (aq) + CI- (aq)

( • 0 2-q) ( 01) ( ~I n,) = Mr- :: M dt/

***** A lab technician determines the boiling point elevation of an aqueous solution ofa nonvolatile, nonelectrolyte to be 1.12oC. Determine the solution's . ciTb :: MeA ~ kb 1.1a :: ( I) (. S 12..) !vi I, I 2. M - -•s I 2.. ::

Osmosis and

1. 05 t\10S l S is the of solvent particles across a semi permeable membrane from an area of higher solvent concentration to an area oflower solvent concentration.

HETEROGENEOUS MIXTURES

1. Not all mixtures are solutions.

2. Heterogeneous mixtures contain substances that exist in distinct phases.

Suspensions

1. A SuSPeJ.SIO,J is a mixture containing particles that settle out if left undisturbed.

2. Suspensions can also be separated using a filter.

1. Particles in a are larger than atoms, while those in solution are atomic-scale in size.

2. A CO L.LO ID is a heterogeneous mixture of intermediate size particles. These are between the size of solution particles and suspension particles.

3. are classified according to the phases of their dispersed particles and dispersing mediums.

Brownian Motion

1. particles make jerky, random movements. This erratic movement of colloid particles is called B.'2..0WN14tJ MOll 0 ""

2. results from collisions of particles of the medium with the dispersed particles. These collisions prevent the colloid particles from settling out of the mixture. The Tyndall Effect

I. Whereas concentrated colloids are often cloudy or opaque, dilute colloids sometimes appear as clear as solutions.

2. Dilute colloids appear to be homogeneous because their dispersed particles are so small that they cannot be seen by the unaided eye.

3. Dispersed colloidal particles are large enough to scatter light, a phenomenon known as the Tyndall Effect.

4. A solution will never scatter light.

ACIDS & BASES

AN INTRODUCTION

Properties of and Bases

1. Acidic solutions taste sour. A tastes bitter and feels slippery.

2. Acids react with some metals to form hydrogen.

in (s) + J.. H CJ (1M ) -~ In CJfJ (~) of lI."J.fJJ b

3. Metal and hydrogen carbonates react with aqueous solutions of acids to produce .

4. Aqueous solutions of acids turn blue litmus paper pink. Aqueous solutions of bases turn red litmus paper blue.

5. Both and base solutions will conduct electricity.

Ions in Solution

1. All aqueous solutions contain hydrogen ions (W) and ions (OH-). The relative amounts of the two ions determine whether an aqueous solution is acidic, basic or neutral.

2. An acidic solution contains more hydrogen ions than hydroxide ions. A basic solution contains more hydroxide ions than hydrogen ions.

MiD 64Mi 3. Water self-ionizes as follows:

This simplifies to:

l/tJ 0 (1) 4. Water is neutral since equal numbers of H" and OH- ions are present.

The Arrhenius Model of Acids and Bases

I. The Arrhenius model states that an acid is a substance that contains hydrogen and ionizes to produce hydrogen ions in an aqueous solution:

STRENGTHS OF ACIDS AND BASES

Strengths of Acids

I. The strength of an acid depends upon how completely it ionizes. -t -- 2. A .sTJlOAi6 J)CI D will ionize completely. ,~t!kt;)f1-/;.0(1) -7 1-130 rfl}J -f CI (o.r

Other strong acids are:

,f(~10l/) HrJ03) I+J., HZ.S04, H£rj I-1C10...2 3. An acid that ionizes only artially in dilute aqueous solutions is called a tJl.::-r:JiC 4ClJ;> H(LI-h ()z. Cc

Strengths of Bases

1. S17lJJiJ6 B/.ISl:;:....s dissociate entirely into metal ions and hydroxide ions in aqueous

NOL014s~~tio~ AI"" t (~)1 OJ.l-raj) Other strong bases are: I~OI-J, IQbO~, CsOl-l

2. W8t ~ e4.sES dissociate partially in dilute aqueous solutions. C l~J",U+2- .,..Hz.0 (~) ~ (2L-/-3 tJf.b T (~) f 0/1 -rNj))

WHAT IS pH? pH and pOH

1. The acidity of a solution is based on the concentration of H+ ions in the solution. Since this number often small and expressed in scientific notation, chemists adopted an easier way based on common .

2. This scale, the f ~ .seAL~ , is based on the following: \-\~ - f lccj. [\4+J~ .nJ M 0 LI-}..e IT y ***** Determine the pH of solutions having the following H+ :

a. [W] = 1.00 x 10·2M

F4:7 -1o

b. [W] = 3.00 x 10·6M

c. [W] = 8.20 x 1O-6M

3. Sometimes chemists use a pOH scale to express the (basicity) of a solution>

4. The relationship between pH and pOH is:

***** Determine the pH and pOH of aqueous solutions having the following ion concentrations:

a. [OH-] = 1.00 x 1O-6M

pOl4 <: -loj (,.C() ''''O-~)~

PII;: lef· 00- c. 00 ~ b. [OH-] = 6.50 x 10-4M

f (J /-1 t: - J~ (~So, ¥. ( 0 - C{) z: 3.19

fll-:: IV. 00 ~ ,3. /9 .: /0, f-j

c. [W] = 3.60 x 1O-9M

pl/::-AJ (3,(;'OyrO-1)::. ~~

;; O}/.: I If·00 - If. r.tt{ -= s..5 f..:,

d. [W] = 0.025M

~.i'!J (.OOl,s): I. c 0

/l/,OO - I.•(PO = /;), V 0 5. One can calculate the [H+] and [OH-] concentrations ofa solution if the pH is known.

***** Determine the [W] and [OH-] concentrations of a solution that has a pH = 7.40. po~ = ~·~O - 7. C/O LH-t J s: 10 ::

-7 10- (".(;0 ~, S ( Y fY1 z to

NEUTRALIZATION

The Reaction Between Acids and Bases

1. A neutralization reaction is a reaction in which an acid and a base react in aqueous solution to produce a salt and water.

2. A salt is an made up of a cation from a base and an anion from an acid.

Acid - Base

I. The of an acid-base neutralization reaction is the same as that of any other reaction that occurs in solution.

2. Stoichiometry provides the basis for a procedure called TI TIZA no;J ,which is used to determine the concentrations of acidic and basic solutions.

3. Titration is a method for determining the concentration of a solution by reacting a known volume of the solution with a solution of known concentration.

4. The titration process continues until the reaction reaches the stoichiometric point. The stoichiometric point is the point at which the moles of H+ ions from the acid equal the moles of OH- ions from the base.

5. The stoichiometric point is known as the GQIJI ALaJCE 'FtJ/N T of the titration. 12

10 +- ---.Equivelence Point All ecld converted to conjugate bsee pH g

6 Helf-equivelence Point pH = pKa = 4.75

4

2

OTO--~--'~-~-~~--T--~1~2--~-~1r'--~-~20 Tihn\ "'.'"rno (rrL)

6. Chemists often use a chemical to detect the equivalence point rather than a pH meter.

7. Chemical whose colors are affected by acidic and basic solutions are called f\-CtD- I3A$G J:NDICI!fTOR....J .

8. The point at which the indicator used in a titration changes color is called the ~7iD ?olAf r of the titration. The color change of the indicator selected for an acid- base titration should coincide closely with the equivalence point of the titration.

9. Various indicators are needed because the equivalence point is not always when pH = 7.00. The equivalence point can be higher or lower depending on the strengths of the acids and bases involved.

PHASE DIAGRAMS - HEATING CURVES - ENERGY DIAGRAMS :::Q..\~6 ~ I. A diagram looks like: 'Pe~O'(jsr 0 ~~o 'eP c.R.mCi.\L (J\.),I\~ \...-\ t-l~ / 1'\-1'''- FOI,.Jr

..1lY1.&mSI8LE TO LIQUEFY) I1tf €>AS /.).'3.ovi: TJ-IIJ {>OIJJr \ LI 90 1.D

SbLIO

'~ 2. A heating/cooling curve looks like:

Tern p eratur e

gas 1 aporiz ati on B oilin g temperature not change, liquid due to he at of vaporization

Melting, t emper etu e not change, s id due to he at. of fusi en Temperature rises time

3. An energy diagram looks like:

EXOTHERMIC activated complex ------activation Potential...,:._e_n_e_rg_y1 __ . Fof2. AN /;:f,U::oTti E bl'\lC PfUJU3S. )11-1 ~ f!Wove.. J energy reactants eNetl6 'f Flt-.llS' HES (kJ) 1-4-1&H ~ Il-{ArJ n+ 1: REAc..If.lMI &J...i02..G'f ------~--~~

reaction pathway 30 Solubility Curves for Selected Solutes

150 / IV / 140 III / / v f v 130 / V; 120 / V 1/ ./J 110 i NaN / 03/ / \'..1- Y nn"'3 / / V ~ / / Q) ::l <, - ~ HCI UH"C "-/ V -o / 70 / U) \ -: •••• ~ o 60 \ N V U) "'" NH3 ~ / I< CI" E V-- ~ 50 ,/ / ><----- ~ G V-- =<; V L----- ~ Na ~ •....• '-- /' --- 40 Y'> x / V...--1' c>: <. -: ---- .... / ...... 30 :::=-- xcu )3 ---'"!' <, 20 V------>< <,I'-....-...... »: .>:v r-, ----- <; 10 ~ r<-...... ------r--. /S( )2 ---- o 10 20 30 40 50 60 70 80 90 100

Temperature (DC)

© 1989 Prentice Hall, Inc. Chemistry Transparency 30