AKAcidBase2

7/23/2019 AKAcidBase2

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Answer Key to Acid/Base Problem Set

1. This is a strong base problem.

Ba(OH)2 (aq) → Ba2+(aq) + 2 OH‾(aq)

]-[OH -OHM.! Ba(OH)mol"

-OHmol2 #

soln$"

Ba(OH)mol.2

2

2==

pOH % - log[OH‾] % - log(.!) % ".!

pH % "! & pOH % "! & ".! % "2.'

2. This is an aii b*er problem bea*se a ,ea ai an a salt o the ,ea ai are present in appreiable q*antities. rom Table "'./0 the 1 a o H3 % !.4 # "-".

H3(aq) + H2O (l) 3‾(aq) + H/O+(aq)init ./ .! hange -# # #eq*il ./ & # .! + # #

"-/ "#!.4 #-./

#)(#)(.!

H3

]O][H-[3 a1 =

+==

+

5impliies to6

"-/ "#!.4 ./

(.!)(#)

H3

]O][H-[3 a1 ===

+

]O[H M "#/.7 .!

)(./)"#(!.4 # /

"-"-

+

===

[H3] % ./ & (/.7 # "-" ) % ./ M

[3‾] % .! + # % .! + (/.7 # "-" ) % .! M

[3a+] % .! M

8-

"-

"!-

/

, "#2.7 "#/.7"#".

]O[H1 ]-[OH ===

+

[H2O] % 88.8 M

"



3. This is a strong ai problem.

H3O/(aq) + H2O(l) 3O/ ‾(aq) + H/O+(aq)

[H3O/] % M

[3O/ ‾] % [H/O+] % .!8 M

[H2O] % 88.8 M

M"#2.2 .!8

"#".

]O[H

1 ]-[OH

"!-"!-

/

,===

+

4. This is a ,ea ai problem. The 1 a o H3O2 is !.8 # "-!.

H3O2(aq) + H2O(l) 3O2 ‾(aq) + H/O+(aq)

init .!8 hange -# +# +#eq*il. .!8 & # # #

!-

2

/2a "#!.8

#.!8

(#)(#)

][H3O

]O][H-[3O 1 =

−

==

+

5impliies to6

!-

2

/2a "#!.8

.!8

(#)(#)

][H3O

]O][H-[3O 1 ===

+

#2 % !.8 # "-!(.!8) % 2. # "-!

]-[3O ]O[H M"#".! "#2. # 2/

2-!-====

+

[H3O2] % .!8 & # % .!8 & ".! # "-2 % .!! M

M"#7." "#".!

"#".

]O[H

1 ]-[OH

"/-

2-

"!-

/

,===

+

[H2O] % 88.8 M

2



5. This is a h9rol9sis problem. The salt is the salt o a strong ai an strong base0 soneither 3a+ (,ea on:*gate ai) nor 3O/ ‾ (,ea on:*gate base) ,ill h9rol9;e. Boththese ions are spetator ions. Th*s0 the onl9 [H/O+] or [OH‾] is rom the issoiation o ,ater.

[3a+

] % [3O/ ‾] % .!8 M

[H/O+] % [OH‾] % ". # "-7 M

[H2O] % 88.8 M

6. This is also a h9rol9sis problem. The 1 + is a spetator bea*se it is the ,ea on:*gateai o a strong base. 3itrite ion ,ill h9rol9;e6

3O2 ‾(aq) + H2O(l) H3O2(aq) + OH‾(aq)init .!8

hange -# # #eq*il .!8 & # # #

3ee to in 1 b6

""-

!-

"!-

a

, b "#2.2

"#!.8

"#".

1

1 1 ===

""-

2

2 b "#2.2

#!8.

(#)(#)

]-[3O

]-][OH[H3O 1 =

−

==

5impliies to6

""-

2

2 b "#2.2

!8.

(#)(#)

]-[3O

]-][OH[H3O 1 ===

#2 % 2.2 # "-""(.!8) % ". # "-""

]-[OH ][H3O M"#/.2 "#". # 2

'-""-====

[3O2<] % .!8 & # % .!8 & (/.2 # "-') % .!8 M

[1 +] % .!8 M

[H2O] % 88.8 M

/



7. This is a ,ea base problem. The 1 b or H/ 3H2 is o*n on Table "'.8 an has a=al*e o !.! # "-!.

H/ 3H2 (aq) + H2O(l) H/ 3H/+ (aq) + OH<(aq)

init. ./8

hange -# # #eq*il. ./8 & # # #

!-

2/

// b "#!.!

#./8

(#)(#)

3HH

]-][OH 3H[H 1 =

−

==

+

5impliies to6

!-

2/

// b "#!.!

./8

(#)(#)

3HH

]-][OH 3H[H 1 ===

+

#2 % !.! # "-!(./8) % ".8 # "-!

]-[OH ] 3H[H M."2 "#".8 # //

!-====

+

[H/ 3H2] % ./8 & # % ./8 & ."2 % ./! M

M"#>." ."2

"#"

]-[OH

1 ]O[H "/-

"!-

,/ ===

+

[H2O] % 88.8 M

8. This is an alaline b*er problem.

H/ 3H2 (aq) + H2O(l) H/ 3H/+ (aq) + OH<(aq)

init. .!8 ./ hange -# # #eq*il. .!8 & # ./ + # #

!-

2/

// b "#!.!

#.!8

#)(#)(./

3HH

]-][OH 3H[H 1 =

−

+==

+

5impliies to6

!-

2/

// b "#!.!

.!8

(./)(#)

3HH

]-][OH 3H[H 1 ===

+

!



]-[OH M"#'.' ./

)(.!8)"#(!.! # !-

!-

===

[H/ 3H/+] % ./ + # % ./ - '.' # "-! % ./ M

[H/ 3H2] % .!8 & # % .!8 - '.' # "-! % .!8 M

M"#".8 "#'.'

"#"

]-[OH

1 ]O[H

""-

!-

"!-

,/ ===

+

a. HBr is a strong ai0 th*s it issoiates ompletel96

HBr(aq) + H2O(l) → Br?(aq) + H/O+(aq)

[H3 O+¿=

0.01 0 mol HBr

1 L soln

x1 mol H3O

+

1mol HBr

= 0.010 M

@sing the 1 , o ,ater0 in the [OH?]6

1 , % [H/O+][OH?] % ". # "-"!

[ H3O+

[O H-] =

Kw

¿ ¿=

1.0 x 10-14

0.010=1.0 x 10

-12 M

pH % - log[H/O+] % - log(.") % 2.

pOH % - log[OH?] % - log(". # "-"2) % "2.

or

pH + pOH % "!0 so

pOH % "! & pH % "! & 2. % "2.

b. a(OH)2 is a strong base0 th*s it issoiates ompletel96

a(OH)2(aq) → a2+(aq) + 2 OH?(aq)

O H-] =

0.30 mol Ca(OH ¿2

1 L soln x

2 mol OH-

1 mol Ca(OH ¿2

= 0.60 M O H-

¿

8



[OH-

[H3 O+ ¿=

Kw

¿ ¿ =1.0 x 10

-14

0.60 = 1.7 x 10

-14M

pH % - log[H/O+] % - log(".7 # "-"!) % "/.7>

pOH % - log[OH?] % - log(.') % .22

or

pH + pOH % "!0 so

pOH % "! & pH % "! & "/.7> % .22

c. Ben;oi ai0 'H8OOH0 is a ,ea ai ,ith a 1 a % '.2 # "-8. Arite the

eq*ilibri*m reation an etermine the e#pressions or the eq*ilibri*m amo*nts0 starting,ith an initial amo*nt o .2 M 'H8OOH. Ae are ass*ming that initiall90 there isntan9 aetate or h9roni*m ion present. Cn # amo*nt o ben;oi ai has to be lost to orman # amo*nt o both ben;oate ion an h9roni*m ion. Th*s0 at eq*ilibri*m0 an amo*nt o .2 & # o ben;oi ai is present0 an # amo*nt o both ben;oate ion an h9roni*m ionare present. This an be on=enientl9 s*mmari;e in the table sho,n belo,6

'H8OOH(aq) + H2O (l) 'H8OO?(aq) + H/O+(aq)

initial .2 M hange -# +# +#

eq*il. .2 & # # #

The eq*ilibri*m onstant e#pression or the abo=e eq*ilibri*m is6

[C6 HCOO-][ H3 O

+

C6 H COOH]

¿¿

Ka = ¿ ¿¿

5*bstit*ting the eq*ilibri*m amo*nts into the abo=e e#pression gi=es6

(x!(x!

0.20 - x = 6.2 x 10

−

The abo=e e#pression ma9 be simpliie b9 maing an ass*mption that the D#E inenominator e#pression0 .2 & # is negligibl9 small. F the magnit*e o the onstant ismore than " times smaller than the magnit*e o the no,n initial onentration0 thisass*mption is =ali. Fn this problem0 the magnit*e o the onstant ("-8) is " times

'



smaller than the ertaint9 o the no,n onentration0 .2 M ("-2 or seon eimal plae)0 so the # in .2 & # an be neglete. One a=eat is that a ."G error in the res*lt isaeptable ,hen this ass*mption is mae.

This simpliies the algebrai e#pression to6

x2

0.20 = 6.2 x 10

-

5ol=e or #6 #2 % '.2 # "-8(.2) % ".2 # "-8

x = √ 1.2 x 10-=3. x 10

-3M=[C6H COO

-]=[H3 O+ ]

3o, that # is no,n0 sol=e or ['H8OOH] % .2 & # % .2 & ./8 % .2. 3otiethat o*r ass*mption that # ,as negligibl9 small relati=e to the ben;oi ai onentration,as =ali.

@se the 1 , o ,ater to in the [OH?]6

+¿

H3 O¿

¿−¿¿

Kw

¿

To in the pH an pOH6 pH % -log[H/O+] % -log(/.8 # "-/) % 2.!8

pOH % "! & pH % "! & 2.72 % "".88

d. Morphine is a ,ea organi base0 as iniate b9 the 1 b % ".' # "-'. Arite theeq*ilibri*m reation an etermine the e#pressions or the eq*ilibri*m amo*nts0 starting,ith an initial amo*nt o .8 M "7H"4 3O/. Ae are ass*ming that initiall90 there isntan9 on:*gate ai o morphine ("7H"4 3O/H+) or an9 h9ro#ie (OH?) present. Toestablish an eq*ilibri*m an # amo*nt o morphine has be lost to rom an # amo*nt o theon:*gate ai an # amo*nt o h9ro#ie. The res*lting eq*ilibri*m amo*nts are thene*e. One again0 this is most easil9 sho,n b9 the table belo,6

"7H"4 3O/(aq) + H2O(l)

"7H"4 3O/H+(aq) + OH?(aq)

initial .8 hange - # +# +#eq*il. .8 & # # #


7



−¿

[C17H1" # O3

K$=[C17 H1"# O3

H+¿ ][OH

¿

¿ =1.6 x 10

−6


(x!(x!

0.00 - x =1.6x 1 0

−6

The abo=e e#pression ma9 be simpliie b9 maing an ass*mption that the D#E inenominator e#pression0 .8 & # is negligibl9 small. F the magnit*e o the onstant ismore than " times smaller than the magnit*e o the no,n initial onentration0 thisass*mption is =ali. Fn this problem0 the magnit*e o the onstant ("-') is " timessmaller than the ertaint9 o the no,n onentration0 .8M ("-/ or thir eimal plae)0so the # in .8 & # an be neglete. One a=eat is that a ."G error in the res*lt isaeptable ,hen this ass*mption is mae.


x2

0.00 =1.6x10

−6

5ol=e or #6 #2 % ".' # "-'(.8) % >. # "->

x = √ %.0 x 10−% % 2.> # "-! % ["7H"4 3O/H+] % [OH?]

3o, that # is no,n0 sol=e or ["7H"4 3O/] % .8 & # % .8 & (2.> # "-!

) % .8 M 3otie that o*r ass*mption that # ,as negligibl9 small relati=e to the ben;oi aionentration ,as =ali.

@se the 1 , o ,ater to in the [H/O+]6

+¿−¿

[H3 O¿=

Kw

[O H¿ =

1.0 x 10−14

2.% x 10−4

=3.6 x 10−11

To in the pH an pOH6 pH % -log[H/O+] % -log(/.' # "-"") % ".!8

pOH % "! & pH % "! & ".!8 % /.88

1!. Fn general0 an ai an a base ,ill orm a salt an ,ater. The cation o the salt al,a9somes rom the base an the anion o the salt al,a9s omes rom the ai (as a memor9ai0 eep the onsonants an =o,els together). 3ot all salts are ne*tral. The ai or base properties o a salt are etermine b9 the relati=e strengths o the ai an base rom ,hih

>



the salt originate. Hl0 HBr0 HF0 H3O/0 HlO! an the irst issoiation o H 25O! arestrong ais. Cll other ais are ,ea. Organi ais are al,a9s ,ea. Cn eas9 ,a9 toreogni;e an organi ai is to loo or arbon in the orm*la. or e#ample0 aeti ai0H/O2"0 is a ,ea organi ai. The aii h9rogen is sho,n in bol ont. Thisorgani ai is oten reerre to0 in general0 as a arbo#9li ai. The arbo#9li ai gro*p

is oten ,ritten in a orm*la as &O2H or OOH. ro*p FC metal h9ro#ies0 $iOH0 3aOH0 1OH0 IbOH0 sOH0 an in ro*p 2C0 a(OH)20 5r(OH)2 an Ba(OH)2 are t9pialstrong bases. Cmmonia0 3H/0 is a ,ea base an organi ompo*ns alle amines are,ea bases. Cmines are eri=ati=es o ammonia. One or more o the h9rogens onammonia ha=e been replae ,ith a arbon gro*p. or e#ample0 meth9l amine is H/ 3H2.One again0 the presene o arbon in the orm*la is a lear iniation the base is ,ea. Cstrong ai pro*es a ,ea on:*gate baseJ a strong base pro*es a ,ea on:*gate ai.C ,ea ai pro*es a strong on:*gate baseJ a ,ea base pro*es a strong on:*gateai. F a ,ea on:*gate ai or ,ea on:*gate base is pro*e0 it means that thatspeies is too ,ea to beha=e as an ai or base0 respeti=el9. On the other han0 i astrong on:*gate ai or strong on:*gate base is pro*e0 then that speies ,ill beha=e as

an ai or base0 respeti=el9. This proess is alle h9rol9sis.

1! a. 1Br is the salt o 1OH an HBr. 1OH is a strong base. Th*s0 1 + ation is a ,ea on:*gate ai bea*se it originate rom a strong base (1OH). Ft ,ill not beha=e as anai0 hene it oes not h9rol9;e an ,ill solel9 be a spetator ion in sol*tion. 5imilarl90Br? anion is a ,ea on:*gate base bea*se it originate rom a strong ai (HBr). Ft ,illnot beha=e as a baseJ hene it oes not h9rol9;e an ,ill solel9 be a spetator ion insol*tion. 5ine neither the potassi*m ion nor the bromie ion is ontrib*ting to theh9roni*m ion or h9ro#ie ion onentration in sol*tion0 the onl9 so*re or H/O+ or OH? in sol*tion is rom the issoiation o ,ater. Th*s0 the onentration o h9roni*m ion ,ill be "-70 an the res*lting pH o the sol*tion ,ill be 7. This is ,h9 1Br is a ne*tral salt.

1! b. Ib3O2 is the salt o IbOH an H3O2. IbOH is a strong base. Th*s0 Ib+ ation isa ,ea on:*gate ai an ,ill not h9rol9;e. Ft ,ill beha=e as a spetator ion in sol*tionan not aet the pH o the sol*tion. 3itrite ion0 3O2 ?0 is the strong on:*gate base o the,ea ai0 H3O2. Th*s it ,ill beha=e as a base in sol*tion b9 aepting a proton rom,ater6

3O2 ? (aq) + H2O(l) H3O2(aq) + OH?(aq)

5ine h9ro#ie ion is generate in sol*tion0 the sol*tion ,ill be alaline ,ith a pH K 7.

". . 3H! 3O/ is the salt o 3H/ an H3O/. The nitrate ion0 3O/ ?0 is the ,ea on:*gate base o the strong ai0 H3O/0 th*s it ,ill not h9rol9;e. Ft ,ill beha=e as a spetator ionin sol*tion. On the other han0 3H/ is a ,ea base. Th*s0 3H!

+ is the strong on:*gateai o a ,ea base0 3H/. Th*s0 the 3H!

+ ation ,ill *nergo h9rol9sis an beha=e as anai6

3H!+(aq) + H2O(l) 3H/(aq) + H/O+(aq)

5ine h9roni*m ion is being generate0 the sol*tion is aii ,ith a pH L 7.

4



1! d. 3H!'H8O2 is the salt o ammonia0 3H/0 a ,ea base0 an ben;oi ai0'H8O2H0 a ,ea ai. Th*s0 the ammoni*m ation0 3H!

+0 is a strong on:*gate ai an,ill h9rol9;e beha=ing as an ai an the ben;oate ion0 'H8O2 ? is a strong on:*gate base an ,ill h9rol9;e beha=ing as a base6

3H!+(aq) + H2O(l) 3H/(aq) + H/O+(aq)

'H8O2 ?(aq) + H2O(l) 'H8O2H(aq) + OH?(aq)

There are ompeting eq*ilibria present an the pH ,ill be eie b9 ,hihe=er eq*ilibri*m lies *rther to,ars pro*ts. This ,ill be etermine b9 omparing themagnit*e o eah o the eq*ilibri*m onstants. The onstants ma9 be al*late rom theat that the 1 a1 b % 1 , or an9 aion:*gate base pair or baseon:*gate ai pair.

al*late the 1 a or 3H!+ *sing the 1 , or ,ater an 1 b or ammonia6

Ka=Kw

K$

=1.0 x 10

−14

1.% x 10− =.6 x 10

−10

al*late the 1 b or 'H8O2 ? *sing the 1 , o ,ater an the 1 a or 'H8O2H6

K$=Kw

K a

=1.0x10

−14

6. x 10− =1. x 10

−10

omparison o the 1 a an the 1 b sho,s that the 1 a or ammoni*m ion is slightl9 larger than the 1 b or ben;oate ion. Th*s the h9rol9sis o ammoni*m lies *rther to,ars pro*ts than the h9rol9sis o ben;oate ion. More h9roni*m ion is pro*e in sol*tionthat h9ro#ie ion0 so the sol*tion is aii ,ith a pH L 7.

11. One m*st o a stoihiometr9 problem to etermine the moles o H/O+ present ater eah aition o 3aOH an then al*late the onentration o H /O+ base on the total=ol*me o the sol*tion.

11 a. The initial onentration o H/O+ ,ill be ." M bea*se Hl is a strong ai an

issoiates ompletel9 into H/O+ an l?6

Hl(aq) + H2O(l) → H/O+(aq) + l?(aq)initial ." M M Mhange -." M +." M +." MDeq*il.E M +." M +." M

rom the [H/O+]0 al*late the pH6

"



pH % -log[H/O+] % -log(.") % ".

To etermine the [H/O+] one OH? ha=e been intro*e0 one m*st *se stoihiometr9.irst in the total moles o Hl a=ailable0 al*late the moles o OH? intro*e an

etermine ,hih ion is in e#ess. Ahihe=er ion is in e#ess ,ill etermine the pH.

irst in the total moles o Hl. This ,ill be *se repeatel9 thro*gho*t the problem.

2.0mL HCl soln x1 x 1 0

−3 L soln

1 mL HCl solnx

0.10 mol HCl

1 L soln =0.002 mol HCl

11 b. 3e#t in the moles o h9ro#ie ater the aition o ". m$ o 3aOH6

10.0 mL soln x1 x 1 0

−3 L soln

1 mL solnx

0.10 mol #aOH

1 L soln =0.0010 mol #aOH

The moles o 3aOH ae eq*als the moles o Hl that reats. The moles o Hlremaining is etermine b96

.28 total moles Hl & ." mol Hl reate % ."8 mol Hl remaining

The ne, total =ol*me o the sol*tion is6 ". m$ soln + 28. m$ soln % /8. m$ soln.This =ol*me in liters is6

3.0 mL soln x1 x 10

−3 L soln

1 mL soln= 0.030 L soln

al*late the ne, h9roni*m ion onentration6

+¿

1mol H3

O+¿

1 mol HCl =[ H3O

¿=0.043M

❑

0.001 mol HCl

0.030L soln x ¿

Calculate the new pH: pH % -log[H/O+] % -log(.!/) % "./!

11 c. in the moles o h9ro#ie ater the aition o 28. m$ o 3aOH6

2.0 mL soln x1 x 10

−3 L soln

1 mL solnx

0.10 mol #aOH


""



The moles o 3aOH ae eq*als the moles o Hl that reats. The moles o Hlremaining is etermine b96

.28 total moles Hl & .28 mol Hl reate % mol Hl remaining

The eq*i=alene point o the titration has been reahe. The eq*i=alene point is ,here themoles o ai eq*al the moles o base. The salt orme rom this reation is soi*mhlorie. 3either the soi*m ion nor the hlorie ion *nergoes h9rol9sis an arespetator ions. Cs a res*lt0 the onl9 so*re o h9roni*m ion in sol*tion is rom theissoiation o ,ater. Aater pro*es ". # "-7 M H/O+ in sol*tion. @se thisonentration to al*late the pH6

pH % -log[H/O+] % -log(". # "-7) % 7.

11 d. in the moles o h9ro#ie ater the aition o /. m$ o 3aOH6

30.0 mL soln x1 x 1 0

−3 L soln

1 mL solnx

0.10 mol #aOH


Notice the moles of hydroxide is in excess as the equivalence point hasbeen exceeded.Find the moles of excess hydroxide:

0.0030 mol H! " 0.00#$ mol H! reacted % 0.000$ mol H!in excess

&he total volume of solution is: 30.0 m' ( #$.0 m' % $$.0 m'

Convert this volume into liters:

.0 mL soln x1 x 10

−3L soln

1 mL soln = 0.00 L soln

Find the concentration of hydroxide ion:

−¿ ] = ".0 x 10−3

M

0.000 mol OH

−¿

0.00 L soln = [OH

¿

¿

)olve for pH *rst+ then convert to pH:

"2



pH % "lo,-H ?] % -log(4. # "-/) % 2.!

pH % "! & pOH % "! & 2.! % "".4'

12 a. Fnitiall90 the ,ea ai0 ben;oi ai is issoiating an ,e sol=e the pH as i theq*estion rea6 DAhat is the pH o a ." M ben;oi ai sol*tionNE Th*s0 the problem is,ore similar to roblem 46

Ben;oi ai0 'H8OOH0 is a ,ea ai ,ith a 1 a % '.2 # "-8. Arite the eq*ilibri*mreation an etermine the e#pressions or the eq*ilibri*m amo*nts0 starting ,ith an initialamo*nt o .2 M 'H8OOH. Ae are ass*ming that initiall90 there isnt an9 aetate or h9roni*m ion present. Cn # amo*nt o ben;oi ai has to be lost to orm an # amo*nt o both ben;oate ion an h9roni*m ion. Th*s0 at eq*ilibri*m0 an amo*nt o .2 & # o ben;oi ai is present0 an # amo*nt o both ben;oate ion an h9roni*m ion are present.This an be on=enientl9 s*mmari;e in the table sho,n belo,6


initial .2 M hange -# +# +#eq*il. .2 & # # #


[C6 HCOO-][ H3 O

+

C6 H COOH]

¿¿Ka = ¿ ¿

¿


(x!(x!

0.20 - x = 6.2 x 10

−

The abo=e e#pression ma9 be simpliie b9 maing an ass*mption that the D#E inenominator e#pression0 .2 & # is negligibl9 small. F the magnit*e o the onstant is

more than " times smaller than the magnit*e o the no,n initial onentration0 thisass*mption is =ali. Fn this problem0 the magnit*e o the onstant ("-8) is " timessmaller than the ertaint9 o the no,n onentration0 .2 M ("-2 or seon eimal plae)0 so the # in .2 & # an be neglete. One a=eat is that a ."G error in the res*lt isaeptable ,hen this ass*mption is mae.


"/



x2

0.20 = 6.2 x 10

-

5ol=e or #6 #2 % '.2 # "-8(.2) % ".2 # "-8

x = √ 1.2 x 10-=3. x 10

-3M=[C6 HCOO

-]=[H3 O+]

3o, that # is no,n0 sol=e or ['H8OOH] % .2 & # % .2 & ./8 % .2. 3otiethat o*r ass*mption that # ,as negligibl9 small relati=e to the ben;oi ai onentration,as =ali.

To in the pH6 pH % -log[H/O+] % -log(/.8 # "-/) % 2.!8

12 b. ining the pH *pon the aition o ". m$ o strong base is a more ompliate

problem. One has to o a stoihiometr9 problem irst. The balane hemial reationtaing plaes is6

'H8OOH(aq) + OH? (aq) → 'H8OO?(aq) + H2O(l)

in the total moles o ben;oi ai present an the 'H8OOH(aq)6

2.0 mL soln x1 x 1 0

−3L soln

1 mL soln x

0.20 mol C6 HC O2 H

1 L soln =.0 x 10

−3 mol C6 H CO2 H

in the moles o h9ro#ie ae6

−¿

10.0 mL soln x1 x 1 0

−3 L soln

1 mL soln x

0.10 mol #aOH

1 L solnx

1 mol O H-

1mol #aOH = 1.0 x 10

−3 mol OH

¿

rom the balane reation0 the moles o h9ro#ie ae are eq*al to the moles o ben;oate ion orme an the moles o ben;oi ai that has reate.

−¿ &orm'(

1 mol OH−¿

= 1.0 x 1 0−3

mol C6 H CO2

¿

1 mol C6 HC

O2

−¿

¿

−¿ x ¿

1.0 x 10−3

mol OH¿

"!



1 mol OH−¿

= 1.0 x 1 0−3

mol C6 H CO2 H r'a)*'

−¿ x1 mol C6 H CO2 H

¿1.0 x 10

−3 mol OH

¿

5o ar0 the moles o h9ro#ie ae are less than the total moles o ben;oi ai0 th*ssome o the ben;oi ai remains *nreate. 5ine some ben;oi ai0 a ,ea ai0 is present an some ben;oate ion0 the on:*gate base o the ,ea ai0 is also present0 a b*er sol*tion has been orme.

in the moles o *nreate ben;oi ai6

8. # "-/ mol 'H8O2H & ". # "-/ mol 'H8O2H reate % !. # "-/ mol *nreate 'H8O2H

The total =ol*me o the sol*tion at this point is the s*m o the =ol*mes o ben;oi ai

sol*tion an soi*m h9ro#ie sol*tions6

28. m$ soln + ". m$ soln % /8. m$ soln

on=ert this =ol*me into liters in preparation to al*late the ne, initial onentrations o ben;oi ai an ben;oate ion6

3.0 mL soln x1 x 10

−3 L soln

1 mL soln =0.030 L soln

in the ne, initial onentrations o ben;oi ai an ben;oate ion6

[C6 HC O2 H] =4.0 x 10

−3 mol C6 HC O2 H

0.030 L soln = 0.11 M C6H CO2 H

−¿−¿

1.0 x 10−3

mol C6 HCO2

−¿

0.030 L soln = 0.02" M C6 HC O2

¿

[C6 HC O2¿=¿

@se these ne, initial onentrations to set *p the al*lation or the b*er eq*ilibri*msho,n belo,6


initial ."" M .24 M hange -# +# +#

"8



eq*il. ."" & # .24 + # #


[C6 HCOO-][ H3 O

+

C6 H COOH]¿¿

Ka = ¿ ¿¿


(0.02" + x!(x!

0.11 - x = 6.2 x 10

−

The abo=e e#pression ma9 be simpliie b9 maing an ass*mption that the D#E in both the

n*merator e#pression0 .24 + # an the enominator e#pression0 ."" & # are negligibl9small. F the magnit*e o the onstant is more than " times smaller than the magnit*eo the no,n initial onentration0 this ass*mption is =ali. Fn this problem0 the magnit*eo the onstant ("-8) is " times smaller than the ertaint9 o the no,n onentrations0One a=eat is that a ."G error in the res*lt is aeptable ,hen this ass*mption is mae.


(0.02"!(x!

0.11= 6.2 x 10

−

Iearrange the algebrai e#pression to sol=e or #6

+¿

x =(6.2x10

−)(0.11)0.02"

= 2.4 x 1 0−4=[H3 O

¿

5ine # % [H/O+]0 the pH an be al*late as6

pH % -log[H/O+] % -log(2.! # "-!) %/.'/

12 c. ining the pH *pon the aition o 28. m$ o strong base is =er9 similar to the

pre=io*s problem. One has to o a stoihiometr9 problem irst. The balane hemialreation taing plaes is6



"'




−3L soln

1 mL soln x

0.20 mol C6 HC O2 H

1 L soln =.0 x 10

−3 mol C6 H CO2 H


−¿2.0 mL soln x

1 x 10−3

L soln

1 mL soln x

0.10 mol #aOH

1 L solnx

1 mol O H-

1mol #aOH = 2. x 10

−3 mol O H

¿

rom the balane reation0 the moles o h9ro#ie ae are eq*al to the moles o ben;oate ion orme an the moles o ben;oi ai that has reate.

−¿ &orm'(

1 mol OH−¿

= 2. x 1 0−3

mol C6 H CO2

¿

1 mol C6 HCO2

−¿

¿

−¿ x ¿2. x 10

−3 mol OH

¿

1 mol OH−¿

= 2. x 1 0−3

mol C6 H CO2 H r'a)*'

−¿ x1 mol C6 H CO2 H

¿2. x 10

−3 mol OH

¿

5o ar0 the moles o h9ro#ie ae are still less than the total moles o ben;oi ai0 th*ssome o the ben;oi ai remains *nreate. 5ine some ben;oi ai0 a ,ea ai0 is present an some ben;oate ion0 the on:*gate base o the ,ea ai0 is also present0 a

b*er sol*tion has been orme.

in the moles o *nreate ben;oi ai6

8. # "-/ mol 'H8O2H & 2.8 # "-/ mol 'H8O2H reate % 2.8 # "-/ mol *nreate 'H8O2H

The total =ol*me o the sol*tion at this point is the s*m o the =ol*mes o ben;oi aisol*tion an soi*m h9ro#ie sol*tions6

28. m$ soln + 28. m$ soln % 8. m$ soln

on=ert this =ol*me into liters in preparation to al*late the ne, initial onentrations o ben;oi ai an ben;oate ion6

0.0 mL soln x1 x 10

−3 L soln

1 mL soln =0.000 L soln

in the ne, initial onentrations o ben;oi ai an ben;oate ion6

"7



[C6 HC O2 H] =2. x 10

−3 mol C6 HC O2 H

0.000 L soln = 0.00 M C6 H CO2H

−¿−¿

2. x 10−3 mol C6 HCO

2

−¿

0.000 L soln = 0.00 M C6 HC O2¿

[C6 HC O2¿=¿

@se these ne, initial onentrations to set *p the al*lation or the b*er eq*ilibri*msho,n belo,6


initial .8 M .8 M hange -# +# +#eq*il. .8 & # .8 + # #


[C6 HCOO-][ H3 O

+

C6 H COOH]

¿¿

Ka = ¿ ¿¿


(0.00 + x!(x!0.00 - x

= 6.2 x 10−

The abo=e e#pression ma9 be simpliie b9 maing an ass*mption that the D#E in both then*merator e#pression0 .8 + # an the enominator e#pression0 .8 & # are negligibl9small. F the magnit*e o the onstant is more than " times smaller than the magnit*eo the no,n initial onentration0 this ass*mption is =ali. Fn this problem0 the magnit*eo the onstant ("-8) is " times smaller than the ertaint9 o the no,n onentrations0One a=eat is that a ."G error in the res*lt is aeptable ,hen this ass*mption is mae.


(0.00!(x!

0.00= 6.2 x 10

−

The abo=e e#pression simpliies to # % [H/O+] % '.2 # "-8

5ine # % [H/O+]0 the pH an be al*late as6

">



pH % -log[H/O+] % -log('.2 # "-8) % !.2"

3otie ho, ['H8O2H] % ['H8O2 ?]. This *niq*e point is oten reerre to as the hal eq*i=alene point. Ct the hal eq*i=alene point0 pH % p1 a.

12 d. One again0 one m*st o the stoihiometr9 problem irst to sol=e or the pH *ponaition o 8. m$ o strong base. The balane hemial reation taing plaes is6




−3L soln

1 mL soln x

0.20 mol C6 HC O2 H

1 L soln =.0 x 10

−3 mol C6 H CO2 H


−¿


−3 L soln

1 mL soln x

0.10 mol #aOH

1 L solnx

1 mol O H-

1mol #aOH = .0 x 10

−3 mol OH

¿

The moles o ai eq*al the moles o base. The ben;oi ai has been ne*trali;e an onl9soi*m ben;oate an ,ater are present in sol*tion. The moles o ben;oate ion eq*al themoles o ben;oi ai ne*trali;e6

−¿

1mol C6 HC

O2

−¿

1 mol C6 HC O2 H = .0 x 10−3

mol C6 H CO2

¿

.0 x 10−3

mol C6 H CO2 H x ¿

The total =ol*me o the sol*tion is6 28. m$ + 8. m$ % 78. m$. on=ert this =ol*meto liters in preparation or ining the onentration o ben;oate ion6

7.0 ml soln x1 x 10

−3 L soln

1 mL soln = 0.070 L soln

in the onentration o ben;oate ion in sol*tion6

−¿

.0 x 10−3

mol C6 H CO2

−¿

0.070 soln =0.067 M C6 HC O2

¿

−¿ ] = ¿[C6 HC O2

¿

"4



Ieall at the eq*i=alene point0 onl9 the salt is mo=ing aro*n in aq*eo*s sol*tion. Fn thisase the salt is soi*m ben;oate. The soi*m ion is the ,ea on:*gate ai o a strong base0 an th*s ,ill not *nergo h9rol9sis. Ft ,ill be a spetator ion. On the other han0the ben;oate ion is the strong on:*gate base o a ,ea ai an ,ill *nergo h9rol9sisan beha=e as a base6

'H8O2 ?(aq) + H2O(l) 'H8O2H(aq) + OH?(aq)

To e=al*ate the pH0 ,e nee to o a h9rol9sis eq*ilibri*m problem. irst0 al*late theeq*ilibri*m onstant0 1 b0 or this reation rom the at that the 1 a1 b % 1 , or an9aion:*gate base pair (or in other instanes0 or an9 baseon:*gate ai pair).

K$ =Kw

Ka

=1.0 x 10

−14

6.2 x 10−

= 1.6 x 1 0−10

5et *p a table to etermine the eq*ilibri*m onentrations in sol*tion. Fnitiall90 .'7 M ben;oate is present in sol*tion0 to establish eq*ilibri*m an D#E amo*nt o ben;oate m*st be*se to orm an D#E amo*nt o ben;oi ai an D#E amo*nt o h9ro#ie. ombiningthese q*antities gi=es *s the eq*ilibri*m q*antities6

'H8O2 ?(aq) + H2O(l) 'H8O2H(aq) + OH?(aq) initial .'7 M hange -# + # + # eq*il. .'7 & # # #

5*bstit*te these q*antities into the eq*ilibri*m onstant e#pression6

−¿−¿

K$ =[C6HCO2H][OH

¿

[C6 HC O2¿ =

(x!(x!

0.067 - x = 1.6 x 1 0

−10

The D#E in the enominator term0 .'7 & # an be neglete bea*se the magnit*e o theonstant is "7 times smaller than the last ertain igit o the no,n onentration0 henethe e#pression ,ill simpli9 to6

x2

0.067=1.6x 10

−10

5ol=e or #6 #2 % (".' # "-")(.'7) % "." # "-""

2



−¿O H

¿

x = √ 1.1 x 10−11=3.3 x 10

−6=¿

There are t,o ommon methos or ining the pH. One metho ,o*l be to in the pOHan then on=ert to pH6

pOH % -log[OH?] % -log[/./ # "-'] % 8.!>

pH % "! & pOH % "! & 8.!> % >.82

Clternati=el90 one o*l irst al*late the [H/O+] *sing the 1 , e#pression or ,ater anthen in the pH6

+¿−¿

[H3 O¿=Kw

[O H¿ = 1.0 x 10−14

3.3 x 10−6

= 3.04 x 1 0−"

M

pH % -log[H/O+] % -log(/.! # "-4) % >.82

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