Acid Base outside solution.pdf

8/14/2019 Acid Base outside solution.pdf

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ENV 6519 Chapter 4 HW Part 2 (#12-#25) David E. MacNevinJune 14, 2006

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4-122.5*10 -7 mole of NH 3(aq) and 2.5*10 -7 mole NaOH are added to 1 liter of distilled water at25C. What is the pH if =0?

Charge Balance+++ =++ OH H NH Na 4

This condition is satisfied where

pH=7.77

3

4

5

6

7

8

9

10

11

12

13

0 2 4 6 8 10 12 14 16

pH

p C

`


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4-1310 -2 moles NH 4Ac and 10 -2 moles NaOH are added to 1 liter of water at 25C. What isthe pH if =0?

Charge Balance+++ +=++ AcOH H NH Na 4 pH=10.6

0

1

2

3

4

5

6

7

8

9

0 2 4 6 8 10 12 14

pH

p C

` `


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4-14 pH of a Solution Having Two Bases by Graphical MethodWhat is the pH of a 25C solution containing 10 -3 moles NaHCO 3 and 2x10 -3 mole NH 3

per liter if =10 -2?

Write the proton condition, where proton reference levels are HCO 3- and NH 3 [ ] ++ +=++ 23432 COOH H NH CO H

The equilibrium constants need to be adjusted for ionic strength. New equilibriumconstants based on concentration, cK, may be calculated.

For =10 -2, from Fig 3-4 find the activity coefficients using the GntelbergApproximation.1=0.9; 2=0.66

{ }{ } +++ == OH H OH H K OH H w 91.1314

109.09.0

10

+

=

==OH H

ww

c K K

For the carbonate system

{ }{ }{ } [ ]32

3

32

31

32

3

CO H

HCO H

CO H

HCO H K

CO H

HCO H a

++ +== 2.6

3.61

1 109.09.010132

=

==+ OH H

aCO H a

c K K

{ }{ }{ } [ ]

+

+

+

==3

23

3

23

2

3

23

HCO

CO H

HCOCO H

K HCO

CO H a

1.10

3.102

2 1066.09.0109.0

2

3

3

=

==+

CO H

a HCOa

cK

K

For the ammonia system

{ }{ }{ }

[ ][ ]+

+

+

+

+

+==4

3

4

31

4

3

NH

NH H

NH NH H

K NH

NH H a

3.9

3.91

1 1019.0109.0

3

4

=

==+

+

NH H

a NH a

cK

K

Construct the appropriate pC-pH diagram using these modified coefficients, noteCT,CO3 =10 -3M and C T,NH3 =2*10 -3M

To find the solution pH, make the simplifying assumptions, [NH 4+]>>[H 2CO 3]>>[H +] whichreduces the proton condition to

+ += 234 COOH NH Find the ammonium concentration to intersect the sum of the hydroxyl ion and carbonate

ion concentration at

pH = 9.85


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0

1

2

3

4

5

6

7

8

9

0 2 4 6 8 10 12 14

pH

p C

``


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4-15A nitric acid solution, pH=2.7, results from NO x removal from a stack gas. Neglect ionicstrength effects and the temperature is 25C.

a.

How much Na 2CO 3 must be added to neutralize this solution prior to discharge?

(The final pH is 8.3. Assume that no weak acids are present in the scrubberwater.) Hint : What is the predominant carbonate species at pH 8.3?)

b.

What is the buffer intensity of the final solution?

At pH 2.7, almost all of C T,NO3 is present as NO 3-. Therefore, the neutralization reaction isCO 32-+H + HCO 3-.

a. Calculate the quantity of H + to be neutralized,

H+=10 -2.7-10 -8.3 =10 -2.7

Add 10-2.7

moles Na 2CO 3 per liter of discharge

b. Calculate the buffer intensity, , of the final solution using Eq 4-99

( )3,213,103.2 COT COT C C OH H +++= +

[H+]=10 -8.3 [OH -]=10 -5.7 CT,CO3 =10 -2.7

For the carbonate system at pH=8.30=0.01 1=0.98 2= 0.01

Neglect H + as insignificant

( ) 57.27.27.5 105.91001.098.01098.001.0103.2 =++=

=9.5*10 -5M This is the buffering capacity of the solution


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4-16 Neutralization of a Strong Ac id wi th a Strong Base How much soda ash, Na 2CO 3, in moles/liter is required to neutralize a pickle liquorsolution containing 10 -1 mole H 2SO 4/liter? Assume that the H 2SO 4 will react only withthe Na 2CO 3 and that the pH after neutralization is 8.3.

The neutralization reaction isH++CO 32- HCO 3-

H+=2*10 -1M and will be practically 100% neutralized in being reduced to 10 -8.3 M,therefore,

2*10 -1M Na 2CO 3 Supply this much soda ash


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4-17 pH Buffer for Breakpoint Chlor inationAn experiment is to be conducted on breakpoint chlorination (the oxidation of ammonia

by chlorine) in which it is desired to maintain the pH constant within 0.5 units of theinitial pH of 8. Make the assumption that all of the ammonia is in the NH 4+ form (a

reasonable assumption since the pK a of NH 4+

is 9.3 and the pH of interest is pH 8). The breakpoint reaction between Cl 2 and NH 3 proceeds as follows:

( )++ +++ Cl H N NH Cl g 6823 242

The maximum amount of ammonia that will be used in the experiments is 12.5 mg as N/liter (0.89*10 -3 moles/liter). Select an appropriate acid-conjugate base pair anddetermine the concentration of it that will control the pH to within 0.5 units of 8.0 duringthe reaction. Neglect ionic strength effects; the temperature=25C. Determine the bufferintensity of this solution. ( Note : There is a detailed discussion of the breakpointchlorination reaction in Chapter 7.)

a. Select the dihydrogen phosphate (H 2PO 4-), monohydrogen phosphate ion (HPO 42-)pair , as it has the closest pK a to 8.0 of species in Table 4-1, excluding hypochlorite.

Now, to determine the quantity of the acid-base pair to add to provide the desired control.The worst case condition is when 8.9*10 -4M of NH 4+ are added, producing 3.56*10 -3MH+.So, find C T, that will absorb that many equivalents of H +, given the knowledge of thespecies of phosphate at pH 7.5 and 8.0.

38,25.7,2 1056.3

== = T pH T pH C C

( ) 31056.367.086.0 = T C Solve for C T

C T=1.87*10-2

M Add this much phosphate salt (as HPO 42-

) to buffer pH, adjustto pH=8.0 to obtain proper starting distribution of species.

b.Calculate Use Eq 4-99 ( )3,213,103.2 COT COT C C OH H +++= + At pH=8.0 0=0 1=0.137 2=0.863 3=0

Neglect all terms as insignificant, except the last one( )21087.1863.0137.03.2 = =5.09*10 -3 M

=5.09*10 -3 M This is the buffering capacity of the solution


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4-19A sample of natural water contains 1*10 -3 M CO 32- and 3*10 -3 M HCO 3-.

a.

As the pH is lowered during the alkalinity titration, at which pH is the CO 2 in

solution in equilibrium with atmospheric CO 2? b.

What is the pH of the total alkalinity equivalence point, pH CO2 ? (Neglect dilutioneffects.) Give your answer to the nearest 0.1 pH unit.

c.

If 50 percent of the CO 2 formed during the titration escapes, what is the new pH of the total alkalinity equivalence point?

a.

This would be the pH where [H 2CO 3]=10 -5M.

[ ]

[ ] [ ]

53

2

3.103.63.6,32 101041010101

13

+

+

=++==

H H

C CO H COT o

[H+]=1.12*10 -9MpH=8.9

b.

pH CO2 =0.5*(pK a1+pC T,CO3 ) pC T,CO3 =-log(1*10 -3M+3*10 -3)=2.40 pH CO2 =0.5*(6.3+2.4)=4.35

pH CO2 =4.35

c.

Recalculate pC T,CO3 =-log(0.5*4*10 -3)=2.70 pH

CO2=0.5*(6.3+2.7)=4.50

pH CO2 =4.50


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4-20A sample of natural water that has been equilibrated with CaCO 3(s) is isolated from itssurroundings. Indicate whether the addition of small quantities of the following willincrease, decrease, or have no effect on the total alkalinity or total acidity and state very

briefly why. Neglect ionic strength effects.

a.

HCl b.

FeCl 3 c.

Na 2SO 4 d.

CO 2 e.

Na 2CO 3

Species Total Alkalinity Total Acidity CommentHCl Decrease Increase H + is a strong acid

FeCl 3 Decrease Increase Ferric iron is a strong acid Na 2SO 4 No effect No effect Sulfate a very weak baseCO 2 No effect Increase CO 2 is the TAlk endpoint, therefore

cannot include in the expression Na 2CO 3 Increase No effect CO 32- is the TAcid endpoint, therefore

cannot include in the expression

For example, when carbonic acid (CO 2(aq) ) is added to water it has no net affect on totalalkalinity, because it produces a hydrogen ion and bicarbonate ion. Each cancels theeffect of the other.

+ +++323222 HCO H O H CO H O H CO

It can similarly be shown, that soda ash when added to water has no net effect on totalacidity.


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4-21 Alkalin ity Titration, and pH of Buffered Water Receiving StrongBase WasteA natural water has the following partial analysis:

pH=8.3 [Ca 2+]=5*10 -4M

[HCO 3-]=3*10

-3M [Mg

2+]=1*10

-4M

[CO 2(aq) ]=3*10 -5M [SO 42-]=1*10 -4M

a.

What volume of 0.02N H 2SO 4 is required to titrate a 100-ml sample to the totalalkalinity endpoint? What is the total alkalinity in eq/liter and in mg/liter asCaCO 3?

b.

A waste containing 10 -2 moles NaOH/liter is to be discharged to this water. The pH cannot be raised above 9.5. What is the maximum number of liters of wastethat can be added to each liter of the natural water?

a. Total alkalinity consists of titrating all bases. In the given water, HCO 3-, is the only

significant base present for total alkalinity. Sulfate is far too weak a base and would notneutralize before the total alkalinity endpoint.

Alk = 3*10 -3eq/liter = 150 mg/l as CaCO 3

The volume of 0.02eq/liter H 2SO 4 required per liter to provide 3*10 -3 eq is

mlleq

V l

eq100

102.010*3 3 =

Add V=15 ml of 0.02 N H 2SO 4 to titrate to the total alkalinity endpoint b. At pH=8.3, nearly all the carbonates are in the form of HCO 3-. So

CT,CO3 =3*10 -3M. At pH 9.5, [HCO 3-]=1*C T,CO3 =0.863*3*10

-3=2.59*10 -3M

The amount of bicarbonate neutralized by base is then,3*10 -3-2.59*10 -3=4.1*10 -4M

This is the number of equivalents of base that may be added per liter of naturalwater. Use this information to find the volume of waste.

naturallwastel

eqwastel

naturalleq

041.001.01

101.4 4 =

V= 41 ml of waste per liter of natural water


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4-22 Acidity and AlkalinityA partial water analysis is given as follows:CO 2 = 44 mg/liter [Cl -]=1*10 -3M[HCO 3-]=2*10 -3M [SO 42-]=1*10 -4M

a.

What is the solution pH and CO 32- concentration?

b.

What is the caustic, carbonate, and total alkalinity (in eq/liter and in mg/liter asCaCO 3)?c.

What is the mineral, CO 2, and total acidity (in eq/liter)?d.

What is the pH if 5*10 -4 mole OH- (as NaOH) are added per liter of the abovesample?

a. M mg

mmoll

mgCO 32 10144

44 ==

[ ] 333323, 103102101 =+=+= HCOCOC COT

[ ]

[ ] [ ]

333.63,

32 101103

101

1

1

1101 =

+

+

=

+

=

+

==

H

C

H K

M CO COT a

31

0 = pH=6.6

[ ] [ ]

[ ] [ ]

343,

2211

221

3,223 1031033.1

1

++

+ =

++==

COT aaa

aa

COT C

H

K K H K

H

K K

C CO

[CO 32-]= 3.99*10 -7M pH=6.6

b. From Tb 4-11Caustic Alkalinity

[ ] [ ] ( ) ( )67.0233.0103101010

2 36.66.614

013, +=+

+

+ COT w C H

H K

Caustic Alkalinity = negative therefore = 0 , reasonable since pH < 10.3Carbonate Alkalinity

( ) [ ] [ ] ( )6.6

6.6

1443

023, 101010

33.01033.1103

++ +=+ H H

K C wCOT

Carbonate Alkalinity = negative therefore = 0 , reasonable since pH < 8.3

Total Alkalinity

( ) [ ] [ ] ( )6.6

6.6

1443

213, 101010

1033.1267.01032

++ ++=++ H H

K C wCOT

Neglect pH termsTotal Alkalinity = 0.002 eq/liter = 100 mg/l as CaCO 3


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c. From Tb 4-11Mineral Acidity

[ ] [ ] ( ) ( )43

6.6

146.6

213, 1033.1267.01031010

102

+

+ +=+ COT w C H K

H

Mineral acidity = negative therefore = 0 , reasonable since pH > 4.5

CO 2 Acidity

( ) [ ] [ ] ( ) 6.614

6.643203, 10

10101033.133.0103

++ +=+

H K

H C wCOT

CO 2 acidity=9.9*10-4

eq/liter

Total Acidity

( ) [ ] [ ] ( ) 6.614

6.63013, 10

101033.0267.01032

++ ++=++

H K

H C wCOT

Total acidity=4*10 -3 eq/liter

d. Adding 5*10 -4 moles of OH -, the hydroxyl ions react preferentially with CO 2Thus, after neutralization of the added OH -

[CO 2]=1*10 -3-5*10 -4=5*10 -4

[HCO 3-]=2*10 -3+5*10 -4=2.5*10 -3

Calculate pH from 0

[ ]

167.0103105

101

13

4

3.60 =

=

+=

+

H

[H+]=10 -7

pH=7.0


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4-23 Titration for Species DeterminationFifty ml of a natural water sample is titrated with 0.02 N H 2SO 4. The titrant volumerequired to titrate to pH 8.3 is 6 ml and an additional 8 ml is required to titrate to pH 4.3.

a.

What is the caustic, carbonate, and total alkalinity in meq/liter?

b.

What is the mineral, CO 2, and total acidity in meq/liter?c.

What is the C T,CO3 ?d.

What is the pH CO2 and pH CO3 2- (to the nearest 0.1 pH unit)?e.

What is the [H+], [OH -],[CO 32-],[HCO 3-] and [H 2CO 3*] in the original sample?Calculate using both the approximate method described in Section 4.13.3 and theexact procedure and compare the results.

a. The caustic alkalinity is 0 meq/liter, since V mo>V p implying that HCO 3- was presentin solution.Carbonate alkalinity = V p*N/V= 6 ml * 0.02 eq/l / 50 ml

Carbonate alkalinity = 2.4 meq/l

Total alkalinity = V mo*N/V= 14 ml * 0.02 eq/l / 50 mlTotal alkalinity = 5.6 meq/l

b. Mineral acidity and CO 2 acidity are zero since initial pH>8.3Total acidity = bicarbonate originally present = (V mo-V p)*N/V = (8-6)ml*0.02 eq/l/ 50 ml

Total acidity=0.84 meq/l

c. The total carbonate concentration is the sum of the bicarbonate and carbonate ionsoriginally present in the solution. Or by Eq 4-118, it is the difference between the totalalkalinity and the carbonate alkalinity. The total alkalinity counts CO 32- twice, subtractingthe carbonate alkalinity corrects for this.

5.6-2.4=3.2C T,CO3 =3.2 meq/l

d. pH CO2 =0.5*(pK a1+pC T,CO3 )=0.5*(6.3+2.49)pH CO2 =4.40

pH CO32- =14-0.5*(pK b1+pC T,CO3 )=14-0.5*(3.7+2.49)pH CO32- =10.91

e. Approximate Method (Tb 4-9) pH=10.3+log(2.4/0.84)= 10.76 pOH=3.24

[CO 32-

]=2.4*10-3

M (from a)[HCO 3-]=8.4*10 -4M (from b)[H 2CO 3

*] =0

Exact Method

CT,CO3 =[HCO 3-]+[CO 32-]=3.2*10 -3 M [ ]+

=3

23

2 HCO

CO H K a

+= OH H K w

Charge Balance [H +]=[OH -]+[HCO 3-]+2[CO 3

2-]


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Make substitutions

[ ] [ ] 3,23,114

210

COT COT C C H

H ++= +

+

[ ] [ ] [ ][ ] [ ]

[ ][ ] [ ]

3

2

3.103.63.6

2

3.103.6

2

3.103.63.6

3.6

14

102.3101010

1

1010

2101010

1

10

10

+

+

+

+

+

+

+

+

+++

+++=

H H

H

H H

H H

H

[H+]=2.82*10 -11

pH=10.55 pOH=3.45[CO 3

2-]= 2*C T,CO3 =0.64*3.2*10-3=2.05*10 -3 M

[HCO 3-]= 1*C T,CO3 =0.36*3.2*10

-3=1.15*10 -3 M [H 2CO 3*] = 0*C T,CO3 =2.02*10 -5*3.2*10 -3=6.46*10 -8 M


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4-24 Effect of Ionic Strength on Carbonate Equil ibriumA solution has a carbonate alkalinity of 1 meq/liter and a total alkalinity of 6 meq/liter.The ionic strength of the solution is 10 -2 M and the temperature is 40C. Calculate [H +],[HCO 3-], and [CO 32-] neglecting ionic strength effects and compare the values obtained

when corrections are made for ionic strength effects. Use the constants given in Tables 4-2 and 4-7 for your calculations.

When =0 and T=40CK w=2.95*10 -14=10 -13.53

K a1=10 -6.30

K a2=10 -10.22

For =10 -2, from Fig 3-4 find the activity coefficients using the GntelbergApproximation.

1=0.9; 2=0.66

{ }{ } +++ == OH H OH H K OH H w 44.13

14

109.09.0

1095.2

+

===

OH H

ww

c K K

For the carbonate system

{ }{ }{ } [ ]32

3

32

31

32

3

CO H

HCO H

CO H

HCO H K

CO H

HCO H a

++ +== 21.6

3.61

1 109.09.010132

=

==+ OH H

aCO H a

c K K

{ }{ }{ } [ ] [ ][ ]+

+

+==3

23

3

232

3

23

HCOCO H

HCOCO H K

HCO

CO H a

04.1022.102

2 1066.09.0109.0

23

3

=

==+

CO H

a HCOa

c K K

Calculate the analytical carbon concentration (use Eq 4-118)

CT,CO3 =Total Alkalinity-Carbonate Alkalinity=6*10 -3-1*10 -3=5*10 -3eq/liter

From Tb 4-11Total Alkalinity=6*10 -3M In general

( ) [ ] [ ] [ ][ ] [ ]

[ ][ ] [ ]

[ ] [ ]+

+

++

+

++

++ +

+++

++=++ H

H K

H

K K

H

K H

K K

H

K K

H

K H

K

H H K C w

aaa

aa

aaa

a

wCOT

2211

221

2211

1

3213,

12

11052

When =0


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[ ]

[ ] [ ]

[ ]

[ ] [ ][ ] [ ]

++

+

+

+

+

+

+

+

+++

++= H

H

H H

H

H H

H TotAlk

53.13

2

22.1030.630.6

2

22.1030.6

2

22.1030.630.6

30.6

3 10101010

1

1010

2101010

1

10

105

[H+

]=2.82*10-10

pH=9.55 When =0

[HCO 3-]=1*C T,CO3 = [ ]

[ ] [ ]

3

255.9

22.1030.6

55.9

30.6

55.9

30.6

105

10

10101010

1

1010

++

[HCO 3-]=4.1*10 -3M When =0

[CO 32-]=2*C T,CO3 = [ ]

[ ] [ ]

3

255.9

22.1030.6

55.9

30.6

255.9

22.1030.6

105

101010

10101

10

1010

++

[CO 32-]=9.2*10 -4M When =0

When =10-2

[ ]

[ ] [ ]

[ ]

[ ] [ ]

[ ] [ ]+

+

+

+

+

+

+

+

+

+++

++= H

H

H H

H

H H

H TotAlk

44.13

2

04.1021.621.6

2

04.1021.6

2

04.1021.621.6

21.6

3 10101010

1

1010

2101010

1

10

105

[H+]=3.98 *10 -10 pH=9.40 When =10-2

[HCO 3-]=1*C T,CO3 = [ ]

[ ] [ ]

3

240.9

04.1021.6

40.9

21.6

40.9

21.6

105

10

10101010

1

1010

++

[HCO 3-]=4.1*10 -3M When =10

-2

[CO 32-]=2*C T,CO3 = [ ][ ] [ ]

3

240.9

04.1021.6

40.9

21.6

240.9

04.1021.6

105

10

10101010

110

1010

++

[CO 32-]=9.3*10 -4M When =10

-2

Increased ionic strength caused pH to decrease slightly, while HCO 3- remainedunchanged, and CO 32- showed a small increase in concentration.


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4-25 pH of a Poorly-Buffered Lake Receiving Acid RainIn some regions carbonate-bearing minerals are lacking in the earth and the lakes in theseregions have a low alkalinity. Acid rains, resulting from conversion of industrial SO 2 emissions to H 2SO 4 , can cause significant pH depressions in such lakes and may result in

fish kills. What volume of acid rain, pH =4.0, is required to lower the pH of a lake to 6.7?The lake has a volume of 20*10 6 ft3, a pH =7.0, and an alkalinity of 30 mg/liter asCaCO 3.

Some alkalinity is present in the water. Assume it is all present as bicarbonate alkalinity,since pH


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19 f 19

The lake has volume 2*10 7ft3, convert to liters.

lgal

l ft

gal ft 83

37 10655.578.348.7

102 =

Over the entire lake that number is(5.655*10 8)*(8.53*10 -5)=4.82*10 4 equivalents

The rain has a concentration 1*10 -4 N

=

l ft

l N

eq27.28

1082.4101

1082.4 384

4

V=1.71*10 7ft 3

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