SEAGER and SLABAUGH
`
`0001
`
`PSG2022
`Catalent Pharma Solutions v. Patheon Softgels
`IPR2018-00422
`
`

`

`Acknowledgment.s appear on pp. 669-670.
`
`Copyright © 1979 Scott. Foresman and Company.
`AU Rights Reserved.
`Printed in the United States of America.
`
`Library of Congress Cataloging in Publication Data
`
`Seager. Spencer L.
`Introductory chemistry
`Includes index .
`I. Chemistry.
`I. Slabaugh, Michael R .. joint author.
`II. Title.
`540
`QD31.2.S412
`ISBN 0-673-15026-7
`
`78- 15936
`
`2345678910-VHJ-85848382818079
`
`0002
`
`

`

`176
`
`9 Acids, Bases,
`and Salts
`
`0003
`
`

`

`OBJECTIVES
`
`9.2 THE BRIZ!NSTED THEORY
`
`177
`
`Show that you understand all key terms given in the margms
`and defined in the text
`2. Write equations that illustrate Br¢nsted acid and base behav1or.
`3. Write equations that illustrate the general characteristiCS of acids
`and bases.
`4. Write equations that illustrate various ways to prepare salts.
`5 Do calculations involving the pH and pOH concepts.
`6. Write equations representing the stepwise dissociation of di(cid:173)
`and triprotic ac1ds.
`7. Solve problems related to the analysis of acids and bases by
`titration.
`8 Explain and wnte equat1ons that Illustrate the hydrolysis of salts.
`9 Explain and write equations that illustrate the action of buffers.
`
`A cids. bases. and salts are among the most common and important
`
`solutes found in solutions. Until late in the nineteenth century,
`these substances were chardcterized by such properties as taste or
`color changes induced in cenain dyes. Acids taste sour: bases, bitter;
`and salts. salty. Litmus. a dye. is red in the presence of acids and blue
`in the presence of bases. These and other observations led to the
`correct conclusions that acids and bases are chemical opposites and
`that salts result when acids and bases react with each other.
`Today we define acids and bases in more precise ways. which we
`will also use to study their characteristics.
`
`9.1 THE ARRHENIUS THEORY
`
`In 1887. Swedish chemist Svante Arrhenius proposed a theory deal(cid:173)
`ing with electrolytic dissociation.
`Arrhenius acid; Acids are substances that dissociate when dissolved in water and
`Arrhenius base produce hydrogen ions, H •. Similarly. bases are substances that
`dissociate and produce hydroxide ions, OH - .
`
`Hydrogen chloride, HCI. and sodium hydroxide, NaOH. are examples
`of an Arrhenius acid and base. They ionize in water as follows:
`.- cr(cid:173)
`NaOH- Na - OH
`
`HCI- H
`
`91
`
`92
`
`Note that the hydrogen ion is a bare proton- the nucleus of a hydrogen
`atom.
`
`9.2 THE BR0NSTED THEORY
`
`Arrhenius did not know that free hydrogen ions cannot exist in water.
`They covalently bond with water molecules to form hydronium ions.
`H,o·.
`
`0004
`
`

`

`178
`
`ACIDS. BASES. AND SALTS
`
`H ' + :~-H- [H~-HJ
`
`new bond
`
`Or more simply.
`
`H ' +H,O- H,O
`
`93
`
`9.4
`
`coordinate covalent
`(dative) bond
`
`Covalent bonds in which both electrons are furnished by a single
`atom or molecule (by water in hydronium ions) are called coordinate
`covaJent or dative bonds.
`
`Johannes Br~nsted in Denmark (and Thomas Lowry in England)
`proposed an acid-base theory in 1923 that took into account this be(cid:173)
`havior of hydrogen ions.
`
`Brtnsted acid ;
`Br-nsted base
`
`An acid is any hydrogen containing substance that is capable of
`donating a proton (hydrogen ion) to another substance. A base is a
`substance capable of accepting a proton.
`
`In conformity with this theory. the acidic behavior of HCI in water
`is written:
`
`HCI .,. H,O ---:::: H,O .,. Cl
`
`95
`
`The HCI behaves as a Br0nsted acid by donating a proton to a water
`molecule. The water molecule, by accepting the proton. behaves as
`a base.
`The double arrow in Equation 9.5 indicates that the reaction is
`reversible, with the equilibrium lying far to the right. In actual water
`solutions approximately 92 percent of the dissolved HCI is in the
`ionic form at equilibrium. Remember from Section 8.6 that both the
`forward and reverse reactions are taking place in reactions at equililr
`rium. This means hydronium ions can donate protons to chloride ions
`and form HCI and H 70 molecules. Thus, HJO · behaves as a Br0 nsted
`acid. and Cl as a Br0nsted base.
`
`Identify all Br~nsted acids and bases in the following
`
`Example 9.1.
`reactions.
`a) HN03 + H20 ~ H30 + No3-
`b) H2SO. + 2 H!O ~ 2H30 + SOl
`c) NH 3 _.. H 10 ;:._ NH 1 ' + OH
`Solution. a) The nitric acid. HNO:~. donates a proton to water in the
`forward reaction. Therefore. HN0.1 is a Brp nsted acid and H!O is a
`Br¢nsted base. In the reverse reaction, H30 ' donates a proton to the
`is a
`• Therefore, H.10' is a Br~nsted acid and N03
`nitrate ion, N0.1
`Br~nsted base.
`b) Similarly, H 2S0 1, sulfuric acid. is a Br~n sted acid and H!O is a
`Br~nsted base (forward reaction). Also. H30 • is a Br~nsted acid , and
`• is a Br~nsted base (reverse reaction).
`the sulfate ion. SO 1
`2
`c) In this react ion the water now donates a proton instead of accepting
`one. Therefore . ammonia. NH.1• is a Br~nsted base. while water is a
`Br~nsted acid (forward reaction ). The ammonium ion. N H ,·. is an
`acid, and the hyd roxide ion. OH , is a base (reverse reaction). Note
`that an Arrhenius base was the source of the OH . while according to
`Br~nsted the OH is a base.
`
`0005
`
`

`

`9.3 THE SELF-IONIZATION OF WATER
`
`179
`
`9.3 THE SELF-IONIZATION OF WATER
`
`In Examples 9.l(a) and (b), water behaved as a Br~nsted base. In
`9.1(c), it was a Br~nsted acid. What happens when only pure water
`is present?
`
`self-Ionization Pure water that behaves as both an acid and a base is self-ionized.
`The equation representing this self-ionization is:
`H,O + H,O ~ H,O· + OH-
`
`9.6
`
`:?-H +:1-H ;=...__ [~]· +[:0--H]
`
`H
`
`H
`
`H
`
`..
`
`The transfer of a proton from one water molecule (the acid) to another
`(the base) causes one H30~ and one OH
`to form. Therefore, in pure
`water the concentrations of H3Q + and oH- must be equal. These
`cdncentrations are 10- 7 moles per liter (M). Thus, the equilibrium
`position is far to the left. as indicated by the arrows in Equation 9.6.
`
`neutral solution T he term neutral describes any solution in which the concentrations
`of H30 and OH are equal.
`Thus, pure water is neutral because each liter of pure water contains
`10- 7 moles of H30 .. and 10 7 moles of oH -, at ionization equilibrium.
`In any system that contains water, the product of the molar con(cid:173)
`centrations of H3o· and OH
`is a constant .
`. Put more simply. using pure water as an·example:
`
`acidic solution;
`basic (alkaline) solution
`
`moles per liter of H,O· x moles per hter of OH
`
`a constant
`
`or
`
`1 X 10 7
`
`X
`
`1 X 10 7
`
`- 1 X 1Q• II
`
`It is a chemical convention to express concentration in moles per
`liter by using brackets [].Thus, this statement can be written:
`
`9.7
`
`Equation 9.7 is true not only in pure water. but in any solution in
`which water is the solvent.
`An acidic solution is one in which the concentration of HaO+ is
`higher than the concentration of OH . In a basic or alkaline solution,
`·
`the concentration of oH- is higher than that of H30+.
`
`But the product of the molar concentrations of H30 • and OH will
`be I x 10 u in either case.
`
`Example 9.2. Classify each of the following solutions as acidic, basic,
`or neutral. Calculate the molar concentration of the ion whose con(cid:173)
`centration is not given.
`a) (H 30 • ] = I X to-•
`b) (OH· ] = I X 10-e
`c) (OH- ) = I X 10~
`Solution. a) Since [ H30 • ] = I X 10· •,
`
`0006
`
`

`

`180
`
`ACIDS. BASES AND SALTS
`
`[OH]-~-~~1 10''
`[HP 1
`1 X 10 I
`X
`Thus.
`[H,O·] 1 x 10 •
`(OH ) == 1 X 10 10
`The solution is acidic because the [ H30 ' ) is higher than the [OH ];
`I x 10 1 is larger than I X JQ - 10•
`b) Similarl y.[O H - ] = I x 10 u: therefore
`
`and
`
`1 X 10"
`1 X 10 11
`[H,O'] • (OHT = ,...-x-;-ot
`
`1 X 10 •
`
`This solution is also acidic, since I x JQ-~ is greater than I x JQ- 9•
`c) [oH -] = I x JQ- 6 ; therefore
`1 X 10" 1 X 10 11
`lOHT - "i)('i'(fi - 1 X 10 1
`
`[H,O·]
`
`This solution is basic because the OH concentration (I x JQ- 0) is
`higher than the H30 + concentration ( I x 10-8).
`
`9.4 PROPERTIES OF ACIDS
`
`Acids and bases are used so often in most laboratories that stock solu(cid:173)
`tions are kept readily available at each work space. The common solu(cid:173)
`tions and their concentrations are given in Table 9.1 .
`Acids have different properties which make some more practical
`than others for specific uses. However, all acids have certain properties
`in common. We have already discussed two of these-all acids taste
`sour and react with water to produee H3Q+ ions. In addition, acids
`react characteristicall y with solid oxides. hydroxides, car bonates, and
`bicarbonates.
`
`21(::( ...
`
`Cu O
`copper oxide
`
`- CuCI,+ H,O
`
`Ca(OH).
`2HCI+
`- CaCI,+ 2H,O
`calcium hydroxide
`
`Na, CO,
`2HCI""
`sodium carbonate
`
`-< 2NaCI + CO, + H,O
`
`HCI +
`
`NaHCO,
`sodium b•carbonate
`
`- NaCI + CO,+ H,O
`
`98
`
`9.9
`
`910
`
`9 11
`
`The above reactions are written using full equations.
`
`full equation
`
`In full equations the complete formula is given for all reactants
`and products.
`
`However. some of the substances in these reactions will form ions
`in solution. Ion formation is emphasized by writing full ionic equations.
`
`full Ionic equation
`
`In full ionic equations all substances that dissociate are written
`as ions.
`
`Equatio n 9.8 is written in full ionic form in Equation 9. 12. Since water
`solutions really contain HJO+ and not H •. 2H20 must be added to
`both sides:
`
`0007
`
`

`

`94 PROPERTIES OF ACIDS
`
`181
`
`Table 9.1. Common laboratory acids and bases
`
`Name
`
`Formula
`
`Label
`concentrat1on
`
`Molartty
`
`AcidS
`Acetic acid
`Acetic acid
`Hydrochloric ac1d
`HydrochloriC ac1d
`N1tric acod
`N1tric acid
`Sulfuric acid
`Sulfunc ac1d
`
`Bases
`Aqueous ammon1a*
`Aqueous ammorua
`Sodium hydroxide
`
`HC,H,O,
`HC,H,O,
`HCI
`HCJ
`HNO,
`HNO,
`H,SO,
`H,SO,
`
`g lacial
`d il '
`con.t
`d ll
`con
`d il
`con
`d il
`
`NH•
`NH,
`NaOH
`
`con
`none usually goven
`none usually goven
`
`18
`6
`12
`6
`16
`6
`18
`3
`
`15
`6
`6
`
`·Do lute
`t Concentrated
`t Often erroneously called ammonoum hydroxode and goven the formula NH,OH
`
`2H,O' + 2CI .,. CuO .... Cu'" + 2CI- + 3H,O
`
`912
`
`2HCI (-..2H,O)
`
`CuCI,
`
`We can simplify this equation somewhat by following the common
`practice of substituting H + for H30 + (Equation 9.13).
`
`2H' +2CI + Cu0 .... Cut++ 2CI + H,O
`
`9,13
`
`We often do this, but you should remember that water solutions really
`contain H3Q · and not H •. These forms of the equation make it ap(cid:173)
`parent that the chloride ions do not participate in the reaction.
`
`Ions which do not undergo any chemical changes in solution reac(cid:173)
`tions are called spectator ions. The net ionic equation of a solution
`reaction drops all spectator ions and represents only those substances
`actually undergoing chemical changes.
`
`914
`
`The general nature of the reaction is emphasized in the net ionic form,
`since the H· could come from any acid.
`Remember that correctly written fuU equations must have their
`atoms balanced (Section 6.6). Full ionic and net ionic equations must
`also have their atoms balanced, but in addition the total charges on
`each side of the equation must balance. In Equation 9.14, for example,
`the 2H ' ions provide two positive charges on the left, which are
`balanced by the two positive charges of Cut • on the right.
`
`Example 9.3. Write Equations 9.9 and 9.10 in full ionic and net ionic
`forms.
`
`Solution. The ionizable substances in Equation 9.9 a.re HCl and
`CaC(,.
`
`Full ionic:
`
`2H ' + 2CI - + Ca(OH),-+ Ca• • + 2Ct + 2H,O
`
`The spectator ion is Ct
`
`spectator lon;
`net Ionic equation
`
`0008
`
`

`

`182
`
`ACIDS, BASES. AND SALTS
`
`Net 10nic:
`
`2H' + Ca(OH).- Ca•· + 2H,O
`
`In Equation 9.1 0, HCl and N aCl are ionizable.
`
`Full ionic·
`
`2H' + 2CI- + Na.CO,-. 2Na '+ 2CI- +CO,+ H,O
`
`Agam. Cl- is a spectator ion.
`
`Net ionic:
`
`2H • + Na,CO, -+ 2Na' + CO,+ H,O
`
`Another property of acids is their ability to react with (and dissolve)
`certain metals to yield hydrogen gas. Metals differ widely in their
`tendencies to undergo such reactions. Some will react with very low
`concentrations of H30 +, such as that found in water. Others will react
`only when the concentration of H 30 + is quite high. Some wiU not
`react with H30 • at all. These tendencies are represented by the ac(cid:173)
`tivity series in Table 9.2 and by the typical reactions in Equations
`9. 15 and 9.16.
`Full equat1on:
`Net ionic equation:
`
`915
`
`Full equation:
`Net ionic equation:
`
`Zn + 2HCI -+ ZnCI, + H1
`Zn + 2H' .... znu + H,
`2K + 2H,O -+ 2KOH + H,
`2K + 2H10 ..... 2K ' + 20W -r H,
`
`91 6
`
`9.5 PROPERTIES OF BASES
`
`Solutions containing bases feel soapy or slippery, and change the color
`of litmus from red to blue. Equation 9.9 illustrates their most charac(cid:173)
`teristic chemical property-they react readily with acids. When a base
`is reacted completely with an acid, the characteristics of both the acid
`and base disappear. Often the resulting solution is also neutral:
`
`[H,O'] =[OW]
`
`Table 9.2. Activity series of the metals
`
`Metal
`
`Symbol
`
`Comments
`
`.,
`0
`!!!
`,g
`>-
`0 c:
`'0 c:
`~
`0>
`c:
`
`Ql
`
`iii .,
`!!!
`0
`£
`
`Potassium
`Sodium
`Calcium
`
`Magnes1um
`Aluminum
`Z1nc
`Chromium
`Iron
`Nickel
`T10
`Lead
`Copper
`Mercury
`Silver
`Platinum
`Gold
`
`K
`Na
`Ca
`
`AI
`Zn
`Cr
`
`Ni
`Sn
`Pb
`Cu
`Hg
`Ag
`pt
`Au
`
`} react violently with cold water
`
`reacts slowly with cold water
`
`but quite rapidly in h1gher
`H,O • concentrations
`
`react in moderately high H,O·
`concentrations
`
`Mg l react very slowly with steam,
`Fe l
`
`do not react w1th H,O •
`
`1
`
`0009
`
`

`

`96 SALTS
`
`183
`
`For this reason, such reactions are often called neutralization reac(cid:173)
`tions (see below).
`More than 20 billion pounds or sodium hydroxide, NaOH, is pro(cid:173)
`duced and used in the United States each year. This useful crystalline
`solid is quite soluble in water and ionizes to form basic solutions.
`NaOH- Na · + OH
`
`9.17
`
`The neutralization reaction between NaOH and HCI is
`HCI + NaOH .... NaCI + H,O
`
`Ful l equat1on
`
`Full 1omc equation
`
`H" .,-CI + Na· + OH - Na ' + CI + H, O
`
`Net ionic equatiOn·
`
`H" + OH - H, O
`
`The full equation (9 . 18) illustrates the following statement:
`
`918
`
`919
`
`920
`
`neutralization reaction During a neutralization reaction, an acid and base combine to form
`a salt and water.
`
`salt;
`cation;
`anion
`
`(We say more about salts in the next section.) The net ionic form of the
`equation (9.20) emphasizes the general nature of neutralization reac(cid:173)
`tions- H • ions (from any source) react with OH-
`ions (from any
`source) to form water.
`Bases also react with fats and oils, and convert them into smaller,
`soluble molecules (see Section 22.6). For this reason, most household
`cleaning products contain basic substances. Lye (NaOH) is the active
`ingredient in numerous drain cleaners. Also, many liquid household
`cleaners contain ammonia.
`
`9.6 SALTS
`
`Salts are solid crystalline substances at room temperature.
`
`Salts contain the cation (positive ion) of a base and the anion
`(negative ion) of an acid.
`
`Thus, ordinary table salt, NaCI, contains the cation of NaOH and the
`anion of HCI (look again at Equation 9.18). Similarly, CuSO. is a salt
`containing the cation of Cu(OH)z and the anion of ~so •. We must
`be careful to think of the term salt in a general way, and not as repre(cid:173)
`senting only NaCI.
`Some acids and bases are not stable enough to be isolated even
`though their salts are. For example, carbonic acid, ~C03, cannot be
`isolated in the pure state. When it forms in water, it promptly de(cid:173)
`composes:
`
`H,CO, -;::: H,O + CO,
`In spite or this characteristic, salts of carbonic acid, such as N~C03
`and NaHC03, are quite stable.
`It is not necessary to identify the parent acid and base in order to
`write correct salt formulas or names. We can simply remember that
`the cation of a salt can be any positive ion except H ?, and it will usu(cid:173)
`ally be a simple metal ion or NH. +.The salt anion can be any negative
`
`9.21
`
`0010
`
`

`

`184
`
`ACIDS. BASES. AND SALTS
`
`salt hydrate;
`water of hydration
`
`ion except OH-. Most of the polyatomic anions we will use were
`listed in Table 4.6.
`As we have seen, salts in solution are dissociated into ions. The salt
`can be recovered by evaporating away the water solvent.
`When recovery by evaporation is done carefully, some salts retain
`definite numbers of water molecules as a part of the solid crystalline
`structure. Such salts are called hydrates, and the retained water is
`the water of hydration.
`
`Most hydrates lose all or part of the water of hydration when they are
`heated to moderate or high temperatures. A number of useful hydrates
`are given in Table 9.3.
`Many salts occur in nature, and some are used as industrial raw
`materials. Examples are sodium chloride, NaCI (a source of Cl~ and
`NaOH), caJcium carbonate or limestone, CaC03 (a source of cement
`and building stone), and calcium phosphate or rock phosphate,
`Ca3(P04)~ (a source of fertilizer). In the laboratory we can prepare
`salts by reacting an appropriate acid with a metal, metal oxide, metal
`hydroxide, metal carbonate, or metal bicarbonate. These reactions,
`given earlier as examples of acid properties (Equations 9.8, 9.9, 9.10,
`9 .I I, and 9 .15), are given below in a general fonn.
`
`acid +metal ..... salt + H,
`acid + metal oxide -> salt+ H:O
`
`ac1d +metal hydrox1de ..... salt+ H,O
`
`acid - metal carbonate ..... salt .._ H:O ... CO,
`ac1d - metal bicarbonate _. salt + H,O - CO,
`
`9.22
`
`9.23
`
`9.24
`
`9.25
`
`9.26
`
`to represent the preparation of
`Example 9.4. Write equations
`Mg(N03)~. using Reactions 9.22 through 9.26. Use full equations to
`emphasize the salt fonnation.
`
`Table 9.3. Some useful and common hydrates
`
`Formula
`
`caso.-H,O
`
`CaS0,·2H,O
`
`MgS0.-7H,O
`
`Na,B.01·1 OH,O
`
`Na, C0,-1 OH,O
`
`Na,PO, ·12H,O
`
`Na,SO ,· tOH,O
`
`Na,S,O, ·SH,O
`
`Chemical name
`
`Common name
`
`Uses
`
`calcium sulfate
`monohydrate
`calcium sulfate
`dihydrate
`magnesium sullate
`heptahydrate
`sodium tetraborate
`decahydrate
`sodium carbonate
`decahydrate
`sodium phosphate
`dodecahydrate
`sodium sullate
`decahydrate
`sodium thiosullate
`pentahydrate
`
`plaster of paris
`
`gypsum
`
`Epsom salts
`
`plaster, casts.
`molds
`casts. molds.
`wallboard
`cathartiC
`
`borax
`
`laundry
`
`washing soda
`
`water softener
`
`tnsodium
`phosphate. TSP
`Glauber's salt
`
`water softener
`
`cathartic
`
`hypo
`
`photography
`
`0011
`
`

`

`9 7 THE pH CONCEPT
`
`185
`
`Solution. In each case we react nitric acid. HN03 , with magne(cid:173)
`sium metal or the appropriate magnesium compound.
`2HN01 + Mg .... Mg(NO,), + H,
`2HN01 + MgO -+ Mg(NO,), + H,O
`2HNO, + Mg(OH), - Mg(NO,), + 2H,O
`2HNO I + MgCO, - Mg(NO,): + H,O .,. co,
`2HNO,-"- Mg(HCO,J, - Mg(NO,), ~ 2H,O + 2CO,
`
`9.7 THE pH CONCEPT
`
`In Section 9.3 we learned that the concentration of H • in pure water is
`I x 10 7 M. Chemists. technologists, and other laboratory personnel
`routinely work with solutions in which the H- concentration may be
`anywhere from I 0 to I 0 11 M. As it is inconvenient to work with num(cid:173)
`bers that extend over such a wide range, chemists long ago adopted
`a shoncut notation known as the pH. Mathematically, the pH is de(cid:173)
`fined in terms of the logarithm (log) of [H ·] as follows :
`pH= - log [H )
`
`927
`
`Upon rearrangement, this equation becomes
`[H" J = 1 )< 10 ...
`pH pH is the negative of the exponent used to express the hydrogen
`ion concentration in moles per liter.
`
`9 28
`
`Example 9.5. Express the following concentrations in terms of pH.
`a) [W) = I X 10 ~
`c) [W] = I X 10 7
`d) [ H +] = I X I 0 II
`b) [OH ) = I X 10 ''
`Solution. a) pH is the negative of the exponent used to express
`[H •]. Therefore the pH =-(-5) = 5.
`b) Here [OH ] is given and [H • ] must be calculated. We remember
`Equation 9. 7, and get
`
`(H']
`
`1 X 10 11
`1 X 10 11
`[OHT=~ 1 X 10 >
`Therefore the pH = -(-5) = 5.
`c) This pH = -(-7) = 7. [ H"] = I x 10 7 corresponds to pure water
`and neutral solutions. Thus. a pH of 7 represents neutrality.
`d) pH =-(- 11) = II.
`
`The pH value of 5 obtained in Example 9.5(a) is for a solution in
`which the H - concentration is higher than the 0 H concentration(cid:173)
`the solution is acidic. Any pH lower than 7 means an acidic solution.
`Any pH higher than 7 means a basic or alkaline solution. The pH of
`some familiar solutions is given in Table 9.4. The pOH values listed
`in Table 9.4 are based on the following relationships:
`-log [OH J
`pOH
`
`9 29
`
`[OH]= 1 x 10"''"
`
`930
`
`0012
`
`

`

`186
`
`ACIDS BASES. AND SALTS
`
`Table 9.4. Relationships between (H 1 (OH ), pH, and pOH
`
`[OH] pH pOH
`
`[H" I
`
`10"
`10 I
`
`10 .. 0
`10 ,
`1
`
`Examples
`(soltds are dissolved m H,O)
`
`HCI ( 1 mole/Iller)
`
`gastnc JUICe
`
`lemon Ju•ce
`
`v1negar carbonated dr•nk
`asp1r1n
`orange JUICe
`apple JUice
`black coffee
`normal unne (average value)
`m1lk. hqu1d d1shwashmg detergent
`saliva pure water
`blood
`soap (not synthetiC detergent)
`bakmg soda
`phosphate~onlammg detergent
`
`m•lk of magnesia
`powdered household cleanser
`
`phosphate-free detergent
`household ammonia
`hqu1d household cleaner
`
`NaOH (0 1 mole/ Iller)
`NaOH ( 1 mole/Iller)
`
`14
`13
`
`12
`
`11
`
`10
`
`9
`8
`
`7
`
`6
`
`5
`
`4
`
`3
`
`2
`1
`0
`
`~
`"0
`0
`
`"' Cl c
`u;
`"' II>
`u
`E
`
`10'
`
`10 " 2
`
`10 I
`
`10 II 3
`
`10 I
`
`" 4
`10 1
`
`10'
`10 •
`
`10" 5
`10 • 6
`
`Neutral
`
`10 l
`
`10 l
`
`7
`
`i!:'
`
`s
`iii
`~
`"'
`Cl
`c
`u;
`"' ~
`0
`E
`
`10.
`
`10 • 8
`
`10 •
`
`10' 9
`
`10 10
`
`10 I 10
`
`10 II
`
`10 • 11
`
`tO "
`10 1
`''
`10 "
`
`10 l 12
`13
`10
`14
`10"
`
`pOH pOH is the negati ve of the exponent used to express OH con(cid:173)
`centration in moles per li ter.
`
`When we combine Equations 9.7, 9.28, and 9.30. we get the following
`useful relations hip:
`
`pH +pOH
`
`14
`
`931
`
`ion concentrations tha t
`Example 9.6. Determine the H · a nd OH
`correspond to the following pH or pOH values.
`
`a) pH = 9
`
`b) pH = 3
`
`c) pOH = 8
`
`Solution. In each case. the relationships pH - pOH = 14. [ H ] =
`1 x 10- 1' 11 , and [OH -] = 1 x t0-"0 11 can be used .
`a) Since p H = 9. pOH = 14- pH = 14 - 9 = 5.
`[H· 1 = 1 x 1o· •" - 1 x 10 •
`(OH I = 1 X 10 "'"1
`1 X 10 '
`b) Since pH = 3. pOH = 14 - pH = 14 - 3 = II.
`
`(H ] = 1 x 10 "11
`• 1 < 10 '
`(OH I = 1 )( 10 ..... "' 1 X 10 I
`
`0013
`
`

`

`9 7 THE pH CONCEPT
`
`187
`
`c) SincepOH = 8,pH = 14 -pOH = 14 -8=6.
`[H 1 == 1 X 10 PH - 1 )( 10 1
`[QH ] = 1 X 10-•"" = 1 X 10 <
`
`It is apparent from Table 9.4 that not all solutions have pH values
`that are neat whole numbers. For example, the pH of vinegar is about
`3.3. How do we deal with such numbers? The H • concentration of
`vinegar could be written I x 10 -3·3 , but we do not like to work with
`negative exponents that are not whole numbers. When exact values
`are not needed. we can express pH , [H • ] . etc., as a range. Thus, the
`H. concentration of vinegar is between I X I 0 - 3 and I X I o-• moles
`per liter. Similarly. a solution with [H ] = 2 x 10-s has a pH between
`4 and 5.
`When more-exact values are needed, we can use T able 9.5 , which
`contains the negative logarithms of numbers such as 2 X 10-s. (Recall
`that pH = -log [H'].) In this case, 2 x 10-s is the [H ' ] or [OH-]
`concentration, which makes 2 the coefficient, I o-s the power of I 0, and
`(reading across and down the appropriate columns) 4.70 the negative
`logarithm which represents the pH or pOH. Thus, from a given con·
`centration we can easily obtain the pH or pOH value, and from either
`of those values we can obtain the concentration. In the above example
`of vinegar, we find the pH value of 3.30 in the table, which gives us the
`H · concentration of 5 X I 0 4
`, a much easier figure to work with than
`3
`I x 10 3
`• Following are mo re exam ples.
`·
`
`Table 9.5. Relationships between concentration and pH or pOH values
`
`Powers of 10
`
`10 •
`
`10-7
`
`Coeff•c•ent
`
`1(1'
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`
`000
`-030
`-048
`-060
`- 070
`- 0.78
`-085
`-Q90
`-Q95
`- 100
`
`10 I
`
`100
`070
`052
`040
`030
`022
`015
`010
`005
`0.00
`
`Coelftctent
`
`10 •
`
`10.
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`
`8.00
`7 70
`7.52
`7 40
`730
`7 22
`7 15
`7.10
`705
`700
`
`900
`8 70
`8 52
`840
`830
`822
`815
`8 10
`805
`800
`
`10 '
`200
`1 70
`1 52
`140
`130
`1 22
`115
`110
`105
`100
`
`10 '"
`
`1000
`9 70
`952
`940
`930
`9 22
`915
`9.10
`905
`900
`
`10 '
`300
`2 70
`252
`240
`2.30
`222
`215
`210
`205
`200
`
`10 I
`
`400
`3.70
`3.52
`340
`330
`322
`315
`310
`305
`300
`
`10 J
`
`500
`4.70
`4.52
`4.40
`4.30
`4.22
`4.15
`410
`4 05
`4.00
`
`6.00
`5 70
`5.52
`5.40
`5.30
`522
`5.15
`510
`505
`5.00
`
`10 II
`
`10 It
`
`10 IJ
`
`10 II
`
`1100
`10 70
`1052
`1040
`1030
`1022
`10 15
`10.10
`1005
`1000
`
`12.00
`1170
`1152
`1140
`11.30
`1122
`11 15
`11.10
`11 .05
`11.00
`
`13.00
`12 70
`1252
`1240
`1230
`1222
`1215
`12.10
`12,05
`12.00
`
`1400
`13 70
`1352
`1340
`13 30
`13 22
`13.15
`1310
`1305
`1300
`
`7.00
`6.70
`6.52
`6.40
`630
`6.22
`6.15
`610
`6.05
`6.00
`to-••
`
`15.00
`14.70
`14.52
`14 40
`14.30
`14 22
`14.15
`14 10
`14.05
`14.00
`
`0014
`
`

`

`188
`
`ACIDS. BASES. AND SALTS
`
`Example 9.7. Use Table 9.5 to do the following.
`a) Express the following molar concentrations as pH values:
`(H ") = 7 X I0 - 3
`(H +) = 3 X IO - ~
`b) Express the following pH or pOH values as molar concentrations:
`
`pH = 6.40
`pH = 12.15
`pOH = 11.52
`Solution. a) r H . ] = 3 X lO !I. We first find the power-of-[ 0 column
`headed by I0 - 9
`. We then come down the lower left column to the
`coefficient value of 3. The number in the body of the table at the inter(cid:173)
`section is 8.52, which corresponds to the pH because we are working
`with [H"].
`
`pH = 8 52
`[ H ·] = 7 X I 0 ". We find the number in the table corresponding to the
`coefficient 7 and I 0 ·'. The number is 2.15.
`
`pH = 2 15
`
`[OH ] = 9 x 10 !. The pOH is the number corresponding to the
`coefficient 9 and 10 !. pOH = 1.05.
`
`pH
`
`14 -pOH
`
`14 - 105 - 12.95
`
`b) pH = 6.40. The values in the body of the table correspond to pH or
`pOH values. Therefore. we locate the value 6.40. This value repre(cid:173)
`sents the coefficient 4 and I 0 7
`• Thus,
`[H" I
`
`4 X 10 ;
`
`pH = 12.15. Similarly, the table value 12. 15 represents the coefficient
`7 and 10 11
`
`(H ]
`
`7 X 10 "
`
`pOH = 11.52. The table value 11 .52 represents the coefficient 3 and
`10 ·~
`
`[OH I 3 X 10 "
`
`9.8 STRENGTH OF ACIDS AND BASES
`
`When salts dissolve in water, they generally dissociate completely.
`This behavior is not characteristic of all acids and bases.
`
`strong acid;
`strong base;
`weak acid;
`weak base
`
`Acids and bases that dissociate almost completely are classified as
`strong acids and bases. Those that dissociate to a much smaller
`extent are called weak or moderately weak, depending on the degree
`of dissociation.
`
`Examples of strong and weak acids are given in Table 9 .6.
`A 0. 1 M solution of hydrochloric acid could be prepared by dissolv(cid:173)
`ing 0.1 mole (3.65 g) of HCI gas in enough water to give 1.0 liter of
`solution. According to Table 9 .6, 92 percent of the gas would dissoci(cid:173)
`ate into H · and Cl . but 8 percent would remain dissolved in the form
`of HCI molecules. In spite of such behavior. we generally carry out
`
`0015
`
`

`

`9.8 STRENGTH OF ACIDS AND BASES
`
`189
`
`Table 9.6. Some common strong and weak aclda
`
`Name
`
`Formula
`
`% Dissociation'
`
`Classification
`
`Hydrochloric acid
`Hydrobromic ac1d
`Nitric acid
`Sulfuric acid
`
`Phosphoric acid
`Sulfurous ac•dt
`
`Acetic acid
`Boric acid
`Carbonic ac•d t
`Nitrous acidt
`
`HCI
`HBr
`HNO,
`H,SO,
`
`H,PO,
`H1SO,
`HC,H,O,
`H,BO,
`H,CO,
`HNO,
`
`• Based on 0.1 M soluloons a1 25 ·c
`t Unslable acod
`
`92
`92
`92
`61
`27
`20
`13
`001
`02
`15
`
`strong
`strong
`strong
`strong
`moderately weak
`moderately weak
`
`weak
`weak
`weak
`weak
`
`calculations involving strong acids (and bases) as if the solute dis(cid:173)
`sociated 100 percent. Thus, the concentration of H+ in a 0.1 M HCl
`solution is usually assumed to be 0.1 M.
`It is important that we remember that the terms weak and strong
`apply to the extent of dissociation, and not to the concentration of an
`acid or base. For example, gastric juice (0.05% HCI) is a dilute (not
`weak) solution of a strong acid.
`Acid behavior is linked to the loss of protons. Thus, acids must con(cid:173)
`tain hydrogen atoms that can be removed to form H •.
`
`Monoprotic acids can lose only one proton per molecule, while
`diprotic and triprotic acids can lose two and three, respectively.
`For example, HCI is monoprotic, ~S04 is diprotic, and H3 P04 is
`triprotic. Di- and triprotic acids dissociate in steps, as shown for
`fizS04 in Equations 9.32 and 9.33.
`H,SO, -:= H • + HSO,(cid:173)
`Hso.- ;::._ w + so,•
`9.33
`The second proton is not as easily removed as the first because it must
`be pulled away from a negatively charged particle, HS04 - . Accord(cid:173)
`ingly, HSO.- is a weaker acid than H2S04 •
`We cannot always determine the number of ionizable hydrogens
`from the molecular formula for an acid. For example, acetic acid,
`HC~H302, is monoprotic even though the molecule contains four
`hydrogen atoms. The dissociation of acetic acid is represented by
`Equation 9.34. Structural formulas are used to emphasize the different
`H atoms in the molecule.
`
`9.32
`
`H 0
`H 0
`I
`I
`I
`II
`H-C- C- 0- H ;:::_w + H- C-c-o-
`1
`I
`H
`H
`
`934
`
`Only the hydrogen bound to the oxygen is ionizable. Those hydrogens
`bound to Care too tightly held to be removed. Table 9.7 contains other
`examples.
`
`monoprotlc acid;
`dlprotic acid;
`trlprotlc acid
`
`0016
`
`

`

`190
`
`ACIDS, BASES. AND SALTS
`
`Table 9.7. Eumplu ol monoprotlc, dlprotlc, and trlprotlc acids
`
`Name
`
`Formula
`
`Stluctura/ fotmula
`
`Classdtcalton
`
`Butyric acid
`
`HC,H,O,
`
`Carbon1c ac1d
`
`H,C0.1
`
`Form1c acid
`
`HCHO,
`
`Nunc ac1d
`
`HNO,
`
`Phosphene ac1d
`
`H,PO,
`
`Phosphorous ac1d
`
`H,PO,
`
`H H H 0
`I
`I
`I
`:1
`H-C-G-G-G-Q-H
`I
`I
`I
`H H H
`0
`II
`H-D-G-D-H
`0
`II
`H-C-0-H
`0
`II
`N-0-H
`II
`0
`0
`II
`H -0-P-0-H
`I
`0
`I
`H
`0
`l
`H-0-P-Q- H
`I
`H
`
`monoprot1c
`
`diprotic
`
`moMprollc
`
`monoprottc
`
`lnpro11c
`
`dipro11c
`
`We have focused our attention on the strength of acids, using the
`extent of dissociation as a basis. However, all dissociations are re(cid:173)
`versible to some degree. In the reverse reactions, anions produced by
`the forward reaction behave as bases. What can we say about their
`strengths? Consider Equations 9.32 and 9.33. The arrow lengths indi(cid:173)
`cate that the equilibrium position in 9.32 is to the right. Thus. H 2SO~
`gives up a proton more readily than HSO,- accepts one. Therefore.
`HSO.- behaves weaker as a base than H !SO~ does as an acid. In
`Equation 9.33, so,'- accepts a proton more readily than HSo,- gives
`one up. as indicated by the position of equilibrium to the left. Thus,
`SOl- is stronger as a base than HSOt- is as an acid.
`In general, the anions produced by the dissociation of strong Bre-nsted
`acids are weak Bre-nsted bases. The anions of weak acids are stronger
`bases, with their strengths dependent upon the strength of the parent
`acid.
`Ammonia, N H 3 • is the weak base we most often encounter. in addi(cid:173)
`tion to the anions of strong acids. The dissociation reaction of gaseous
`NH3 in water, given earlier in Example 9. 1. is:
`NH, + H,O ;:._ NH,• ..- OH
`
`9 35
`
`The most common strong bases we will work with are the hydroxides
`of group lA metals (NaOH. KOH. etc.) and the hydroxides of group
`IIA metals (Mg(OH)~. Ca(OH)~. etc.).
`
`0017
`
`

`

`9 9 ANALYSIS OF ACIDS AND BASES
`
`191
`
`9.9 ANALYSIS OF ACIDS AND BASES
`
`The analysis of solutions for total acidity or basicity is a regular activity
`in many laboratories.
`
`acidity The acidity of a solution is its capacity to neutralize a base.
`
`The pH is related to the concentration of H + in solution, whiJe the
`capacity to neutralize a base depends on the total amount of acid
`available. For example, a 0.1 M acetic acid solution has an H+ concen(cid:173)
`tration of about 1.3 X I o-3 (pH= 2.9). However, one liter of the solu(cid:173)
`tion can neutralize 0.1 mole of OH- , not just 1.3 x w-a moles. The
`reason is that the dissociation equilibrium of acetic acid is:
`HC,H,O, ~ H • + C,H10,-
`9.36
`As OH- is added, H'" reacts to form water (see Equation 9.20). The
`removal of H• causes the equilibrium to shift right in accordance with
`Le Chatelier's principle. The continued addition of OH- will even(cid:173)
`tually cause all of the acetic acid molecules to dissociate and react.
`
`titration The procedure often used to analyze acids and bases is called
`titration (Figure 9 .I).
`Suppose we want to determine the total acidity of an unknown acid.
`We first measure out a known volume of the acidic solution by drawing
`
`\
`
`Calibration
`\ - - mark
`
`Pipet
`
`Volume is read before and alter
`the