titration+problems+ap

8/2/2019 Titration+Problems+AP

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Titration Problems

1973A sample of 40.0 milliliters of a 0.100 molar HC2H3O2 solution istitrated with a 0.150 molar NaOH solution. Ka for acetic acid =

1.8x10-5

(a) What volume of NaOH is used in the titration in

order to reach the equivalence point?(b) What is the molar concentration of C2H3O2- at the

equivalence point?

(c) What is the pH of the solution at the equivalence

point?

Answer:

(a) MaVa=MbVb

(0.100M)(40.0 mL) = (0.150M)(Vb)

Vb = 26.7 mL

(b) acetate ion is a weak base with

Kb=Kw/Ka = 1.0 10-14/1.8 10-5 = 5.6 10-10

[ CH3

COO

]o=

4 . 00 mmol

( 40 . 0 mL + 26 . 7 mL )

= 0 . 0600 M

[CH3COO-]eq = 0.600M -X

[OH-] = [CH3COOH] = X

5 . 6 10 10

=X

2

( 0 . 0600 X ); X = 9 . 66 10

5M

0.0600M-9.66 10-5M=0.0599M [CH3COO-]eq

(c)[ H

+

] =

Kw

[ OH

]=

1 . 0 10 14

9 . 66 105 = 1 . 04 10

10

M

pH = -log [H+] = -log(1.04 10-10) = 9.98

1978 AA 0.682 gram sample of an unknown weak monoprotic organic

acid, HA was dissolved in sufficient water to make 50 milliliters

of solution and was titrated with a 0.135 molar NaOH solution.

After the addition of 10.6 milliliters of base, a pH of 5.65 was

recorded. The equivalence point (end point) was reached after theaddition of 27.4 milliliters of the 0.135 molar NaOH.

(a) Calculate the number of moles of acid in theoriginal sample.

(b) Calculate the molecular weight of the acid HA.

(c) Calculate the number of moles of unreacted HAremaining in solution when the pH was 5.65.

(d) Calculate the [H3O+] at pH = 5.65

(e) Calculate the value of the ionization constant, Ka,

of the acid HA.

Answer:

(a) at equivalence point, moles HA = moles NaOH

= MbVb = (0.0274 L)(0.135 M) = 3.70x10-3 mol HA

(b)

molec . wt . =mass HA

mol HA

=0 . 682 g

3 . 70 103

mol

= 184 g/mol

(c) HA + OH---> A- + H2O

initial: 0.00370 mol

added: (0.0106L)(0.135M) = 0.00143 moleremaining: (0.00370 - 0.00143) = 0.00227 mol

(d) pH = -log[H3O+]; [H3O+] = 10-pH = 10-5.65

= 2.2x10-6M

(e)

Ka=

[ H 3 O+

][ A

]

[ HA ]

=( 2 . 2 10

6) ( 0 . 00143 / v )

( 0 . 00227 / v )

= 1.4x10-6

1979 BA solution of hydrochloric acid has a density of 1.15 grams per

milliliter and is 30.0% by weight HCl.

(a) What is the molarity of this solution of HCl?

(b) What volume of this solution should be taken in

order to prepare 5.0 liters of 0.20 molar

hydrochloric acid by dilution with water?

(c) In order to obtain a precise concentration, the 0.20

molar hydrochloric acid is standardized againspure HgO (molecular weight = 216.59) by titrating

the OH- produced according to the following

quantitative reaction.

HgO(s) + 4 I- + H2O --> HgI42- + 2 OH-

In a typical experiment 0.7147 grams of HgO

required 31.67 milliliters of the hydrochloric acidsolution for titration. Based on these data what is

the molarity of the HCl solution expressed to four

significant figures.

Answer:

(a)

1 . 15 g

1 mL

1000 mL

1 L

30 . 0 gHCl

100 g

1 mol

35 . 5 g= 9 . 5 M

(b) MfVf = MiVi

(0.20M)(5.0L) = (9.5M)(V)V = 0.11 L

(c)

0 . 7147 g

216 . 59 g mol

= 0 . 003300 mol HgO

mol OH- prod. = 2 (mol HgO) = 0.006600 mol

mol HCl req. = mol OH- prod. = 0.006600 mol

M HCl=

0 . 006600 mol

0 . 03167 L

= 0 . 2084 M


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1982 DA solution of barium hydroxide is titrated with 0.1-M sulfuric

acid and the electrical conductivity of the solution is measured as

the titration proceeds. The data obtained are plotted on the graph

below.

10 2 0 30 4 0 706050 80

t iv i ty ,

Millilitres of 0.1-M H2SO4

(a) For the reaction that occurs during the titration

described above, write a balanced net ionicequation.

(b) Explain why the conductivity decreases, passes

through a minimum, and then increases as the

volume of H2SO4 added to the barium hydroxide isincreased.

(c) Calculate the number of moles of barium hydroxideoriginally present in the solution that is titrated.

(d) Explain why the conductivity does not fall to zeroat the equivalence point of this titration.

Answer:

(a) Ba2++2OH-+ 2H+ + SO42--->BaSO4(s) + 2H2O

(b) The initial conductivity is high because of thepresence of Ba2+ and OH- ions. The conductivity

decreases because Ba2+ forms insoluble BaSO4 with

the addition of SO42-. The conductivity alsodecreases because OH- combines with the addition

of H+ ions by forming H2O.

Beyond the equivalence point conductivity

increases as H+ and SO42- ions are added.

(c) # mol Ba(OH)2 = # mol H2SO4

=0.1M 0.04L = 0.004 mol

(d) BaSO4(s)dissociates slightly to form Ba2+ and SO42-,while the water ionizes slightly to form H+ and

OH-.

1983 B

The molecular weight of a monoprotic acid HX was to be

determined. A sample of 15.126 grams of HX was dissolved in

distilled water and the volume brought to exactly 250.00

millilitres in a volumetric flask. Several 50.00 millilitre portionsof this solution were titrated against NaOH solution, requiring an

average of 38.21 millilitres of NaOH.

The NaOH solution was standardized against oxalic acid

dihydrate, H2C2O4.2H2O (molecular weight: 126.066 gram mol-1)

The volume of NaOH solution required to neutralize 1.2596grams of oxalic acid dihydrate was 41.24 millilitres.

(a) Calculate the molarity of the NaOH solution.

(b) Calculate the number of moles of HX in a 50.00millilitre portion used for titration.

(c) Calculate the molecular weight of HX.

(d) Discuss the effect of the calculated molecularweight of HX if the sample of oxalic acid dihydrate

contained a nonacidic impurity.

Answer:

(a) mol H2C2O4.2H2O =

1.2596 g

126.066gmol =

9.9916x10-3 mol

H2C2O4 + 2 NaOH --> Na2C2O4 + 2 H2O

9 . 9916 10 3

mol 2 molNaOH

1 molH2

C2

O4

= 1 . 9983 10 2

mol

MNaOH

=

1 . 9983 10 2

mol

0 . 04124 L

= 0 . 4846 M

(b) mol HX = mol NaOH

0.03821 L x 0.4846 M = 0.01852 mol HX

(c)

0 . 01852 mol

50 . 00 mL

250 . 00 mL = 0 . 09260 mol HX

MW =15 . 126 g

0 . 09260 mol

= 163 . 3g

mol

(d) The calculated molecular weight is smaller thantrue value, because:

measured g H2C2O4 is larger than true value,

calculated mol H2C2O4 is larger than true value,

calculated mol NaOH is larger than true value,

calculated M NaOH is larger than true value

calculated mol HX is larger than true value

therefore,

MW =g HX ( true value )

mol HX ( calculated , and too large )

1984 CDiscuss the roles of indicators in the titration of acids and bases

Explain the basis of their operation and the factors to be


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considered in selecting an appropriate indicator for a particular

titration.

Answer:

An indicator signals the end point of a titration bychanging color.

An indicator is a weak acid or weak base where the

acid form and basic form of the indicators are of

different colors.

An indicator changes color when the pH of thesolution equals the pKa of the indicator. In selecting an

indicator, the pH at which the indicator changes color

should be equal to (or bracket) the pH of the solution at

the equivalence point.

For example, when a strong acid is titrated with a

strong base, the pH at the equivalence point is 7, so wewould choose an indicator that changes color at a pH =

7. {Many other examples possible.}

1987 BThe percentage by weight of nitric acid, HNO3, in a sample of

concentrated nitric acid is to be determined.

(a) Initially a NaOH solution was standardized bytitration with a sample of potassium hydrogen

phthalate, KHC8H4O4, a monoprotic acid often used

as a primary standard. A sample of pure KHC8H4O4weighing 1.518 grams was dissolved in water and

titrated with the NaOH solution. To reach the

equivalence point, 26.90 millilitres of base was

required. Calculate the molarity of the NaOHsolution. (Molecular weight: KHC8H4O4 = 204.2)

(b) A 10.00 millilitre sample of the concentrated nitric

acid was diluted with water to a total volume of

500.00 millilitres. Then 25.00 millilitres of thediluted acid solution was titrated with the

standardized NaOH solution prepared in part (a).

The equivalence point was reached after 28.35millilitres of the base had been added. Calculate the

molarity of the concentrated nitric acid.

(c) The density of the concentrated nitric acid used in

this experiment was determined to be 1.42 gramsper millilitre. Determine the percentage by weightof HNO3 in the original sample of concentrated

nitric acid.

Answer:

(a)

1 . 518 g 1 mol

204 . 2 g

= 7 . 434 10 3

mol acid

= mol NaOH required to neut.

7 . 434 10 3

mol

0 . 02690 L

= 0 . 2764 M NaOH

(b)

28 . 35 mLNaOH

25 . 00 mLHNO3

0 . 2764 mol

1 L

1 molHNO

3

1 molNaOH

=

= 0.3134M HNO3

MfVf=MiVi;(0.3134M)(500mL) = (M)(10.00mL)M = 15.67M HNO3

(c)

% HNO 3 in conc . sol n=

g L HNO 3

g L sol n

100 %

grams HNO3 in 1 L conc. soln =15.67 molHNO

3

1 L

63 . 02 g

1 mol= 987 . 5

g

L

grams soln in 1 L conc. Soln

1 . 42 g soln

1 mL

1000 mL

1 L= 1420

g

L

[ HA ] =( 0 . 0152 3 L )( 0 . 2211

mol

L)

0 . 0700 0 L= 0 . 04810 M

K =[ OH

][ HA ]

[ A

]

=

Kw

K a

=

1 . 10 14

7 . 7 10 5

= 1 . 3 10 10

[ A

] =7 . 789 10

3mol

0 . 0852 3 L= 0 . 0914 M

1988 D

X

X XX

X

X

X

X

0

2

4

6

8

10

12

0 5 10 15 20 25 30

pH

millilitres of NaOH

A 30.00 millilitre sample of a weak monoprotic acid was titratedwith a standardized solution of NaOH. A pH meter was used to

measure the pH after each increment of NaOH was added, and the

curve above was constructed.

(a) Explain how this curve could be used to determine

the molarity of the acid.


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(b) Explain how this curve could be used to determinethe dissociation constant Ka of the weak

monoprotic acid.

(c) If you were to repeat the titration using a indicator

in the acid to signal the endpoint, which of the

following indicators should you select? Give thereason for your choice.

Methyl red K a = 1x10-5

Cresol red K a = 1x10-8

Alizarin yellow K a = 1x10-11

(d) Sketch the titration curve that would result if the

weak monoprotic acid were replaced by a strong

monoprotic acid, such as HCl of the same molarity.

Identify differences between this titration curve andthe curve shown above.

Answer:

(a) The sharp vertical rise in pH on the pH-volume

curve appears at the equivalence point (about 23mL). Because the acid is monoprotic, the number

of moles of acid equals the number of moles of

NaOH. That number is the product of the exact

volume and the molarity of the NaOH. Themolarity of the acid is the number of moles of the

acid divided by 0.30L, the volume of the acid.

(b) At the half-equivalence point (where the volume of

the base added is exactly half its volume at the

equivalence point), the concentration [HX] of theweak acid equals the concentration [X-] of its

anion. Thus, in the equilibrium expression [H

+

][X

-

]/[HX] = Ka, [H+] = Ka. Therefore, pH at the half-equivalence point equals pKa.

(c) Cresol red is the best indicator because its pKa(about 8) appears midway in the steep equivalence

region. This insures that at the equivalence point

the maximum color change for the minimal changein the volume of NaOH added is observed.

(d)

X

X X X

X

X

X

X

0

2

4

6

8

10

12

0 5 10 15 20 25 30

pH

millilitres of NaOH

Longer equivalence

region for strong acid

Same volume of NaOH

added for equivalence point

Acid portion at lower

pH for strong acid

1989 AIn an experiment to determine the molecular weight and the

ionization constant for ascorbic acid (vitamin C), a studen

dissolved 1.3717 grams of the acid in water to make 50.00millilitres of solution. The entire solution was titrated with a

0.2211 molar NaOH solution. The pH was monitored throughoutthe titration. The equivalence point was reached when 35.23

millilitres of the base has been added. Under the conditions of this

experiment, ascorbic acid acts as a monoprotic acid that can be

represented as HA.

(a) From the information above, calculate the

molecular weight of ascorbic acid.

(b) When 20.00 millilitres of NaOH had been addedduring the titration, the pH of the solution was

4.23. Calculate the acid ionization constant for

ascorbic acid.

(c) Calculate the equilibrium constant for the reaction

of the ascorbate ion, A-, with water.

(d) Calculate the pH of the solution at the equivalence

point of the titration.

Answer:

(a) (0.2211M)(0.03523L) = 7.789x10-3 mol

1.3717g/7.789x10-3 mol = 176.1g/mol

(b) at pH 4.23, [H+] = 8.0x10-8M

[ A

] =( 0 . 02000 L )( 0 . 2211 mol _L

1)

0 . 07000 L

= 0 . 06317 M

[ HA ] =( 0 . 01523 L )( 0 . 2211 mol _L

1)

0 . 07000 L

= 0 . 04810 M

K =[ H

+][ A

]

[ HA ]

=( 5 . 9 10

5) ( 0 . 06317 )

( 0 . 04810 )

= 7 . 7 10 5

(c) A- + H2O HA + OH-


5/8

K =[ OH

] [ HA ]

[ A

]

=K

w

Ka

=1 . 10

14

7 . 7 10 5

= 1 . 3 10 10

(d) at equiv. pt.

[ A

] =7 . 789 10

3mol

0 . 08523 L

= 0 . 0914 M

[OH-]2 = (1.3x10-10)(9.14x10-2) = 1.2x10-11

[OH-] = 3.4x10-6M

pOH=-log(3.4x10-6)=5.47; pH=(14-5.47)=8.53

1994 DA chemical reaction occurs when 100. milliliters of 0.200-molar

HCl is added dropwise to 100. milliliters of 0.100-molar Na3P04solution.

(a) Write the two net ionic equations for the formation

of the major products.

(b) Identify the species that acts as both a Bronsted

acid and as a Bronsted base in the equation in (a),Draw the Lewis electron-dot diagram for this

species.

(c) Sketch a graph using the axes provided, showing

the shape of the titration curve that results when100. milliliters of the HCl solution is added slowly

from a buret to the Na3PO4 solution. Account for

the shape of the curve.

p H

m L H C l0

(d) Write the equation for the reaction that occurs if afew additional milliliters of the HCl solution are

added to the solution resulting from the titration in

(c).

Answer:(a) PO43-+H+-->HPO42-; HPO42-+H+--> H2PO4-

(b) HPO42-

O : P : O : H

: O :

: O :

. .

. .

. .. .

. .

. .

. . . .

:

(c)

p H

m L H C l0

P O + H H P O

H P O + H H P O 4

3

2

4 4

2

4

2

+

+

(d) H+ + H2PO4---> H3PO4

1998 D (Required) [repeated in lab procedures section]

An approximately 0.1-molar solution of NaOH is to bestandardized by titration. Assume that the following

materials are available.

Clean, dry 50 mL buret

250 mL Erlenmeyer flask

Wash bottle filled with distilled water

Analytical balance

Phenolphthalein indicator solution

Potassium hydrogen phthalate, KHP, a pure solid

monoprotic acid (to be used as the primarystandard)

(a) Briefly describe the steps you would take, using the

materials listed above, to standardize the NaOH

solution.

(b) Describe (i.e., set up) the calculations necessary todetermine the concentration of the NaOH solution.

(c) After the NaOH solution has been standardized, iis used to titrate a weak monoprotic acid, HX. The

equivalence point is reached when 25.0 mL of

NaOH solution has been added. In the spaceprovided at the right, sketch the titration curve

showing the pH changes that occur as the volume

of NaOH solution added increases from 0 to 35.0mL. Clearly label the equivalence point on the

curve.


6/8

(d) Describe how the value of the acid-dissociation

constant, Ka, for the weak acid HX could bedetermined from the titration curve in part (c).

(e) The graph below shows the results obtained by

titrating a different weak acid, H2Y, with the

standardized NaOH solution. Identify the negative

ion that is present in the highest concentration atthe point in the titration represented by the letterA

on the curve.

Answer

(a) exactly mass a sample of KHP in the Erlenmeyerflask and add distilled water to dissolve the solid.

add a few drops of phenolphthalein to the flask.

rinse the buret with the NaOH solution and fill.

record starting volume of base in buret.

with mixing, titrate the KHP with the NaOHsolution until it just turns slightly pink.

record end volume of buret.

repeat to check your results.

(b)

mass of KHP

molar mass KHP = moles of KHP

since KHP is monoprotic, this is the number of

moles of NaOH

moles of NaOH

L of titrant = molarity of NaOH

(c)

equivalence point

(d) from the titration curve, at the 12.5 mL volumepoint, the acid is half-neutralized and the pH = pKaKa = 10pKa

(e) Y2- (could it be OH- ?)

1999 A Required

NH3(aq) + H2O(l) NH4+(aq) + OH(aq)

In aqueous solution, ammonia reacts as represented

above. In 0.0180MNH3(aq) at 25C, the hydroxide ion

concentration, [OH] is 5.60x104M. In answering the

following, assume that temperature is constant at 25Cand that volumes are additive.

(a) Write the equilibrium-constant expression for the

reaction represented above.

(b) Determine the pH of 0.0180MNH3(aq).

(c) Determine the value of the base ionization

constant, Kb, of NH3(aq).

(d) Determine the percent ionization of NH3 in

0.0180MNH3(aq).

(e) In an experiment, a 20.0 mL sample of 0.0180M

NH3(aq) was placed in a flask and titrated to the

equivalence point and beyond using 0.0120 M

HCl(aq).


7/8

(i) Determine the volume of 0.0120 MHCl(aq)

that was added to reach the equivalence

point.

(ii) Determine the pH of the solution in the flask

after a total of 15.0 mL of 0.0120 MHCl(aq)

was added.

(iii) Determine the pH of the solution in the flask

after a total of 40.0 mL of 0.0120 MHCl(aq)

was added.

Answer

(a) Kb =

(b) pOH = -log(5.60x104) = 3.252

pH = 14 pOH = 10.748

(c) Kb = = 1.80x105

(d) x 100% = 3.11%

(e) (i) NAVA = NBVB

(0.0120 N)(VA) = (0.0180 N)(20.0 mL)

VA = 30.0 mL

(ii) at 15.0 mL it is half-titrated and [NH4+] =

[NH3], then the Kb = [OH] = 1.80x105.

pOH = -log(1.80x105) = 4.745

pH = 14 4.75 = 9.255

(iii) at 40.0 mL, there is an excess of 10.0 mL of

HCl past equivalence point,

(10.0 mL)(0.0120M) = 0.120 mmol H+ ions

= 0.00200M=[H+]

pH = -log(0.00200) = 2.70


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