EXPT 7 Heterogeneous Equilibria Objectives

Understand and explain the conditions necessary for precipitation.

Enumerate and explain the conditions for dissolution of precipitates.

Understand and explain the principle of fractional precipitation.

Introduction Chemical equilibriums that involve more than one phase are said to be heterogeneous. A common example of this is the precipitation reaction, which is a formation of an insoluble ionic solid during a chemical reaction between two solutions. Such reaction occurs when certain pairs of oppositely charged ions attract each other strongly that they form precipitate. With the use of centrifugation for the experimental results and the comparison of the solubility product constant, Ksp, and reaction quotient, Qsp, for the expected results, the concept of heterogeneous equilibrium will be easily understood. Centrifugation is a process of increasing the effective gravitational force on a test tube so as to more rapidly and completely cause the precipitate to gather on the bottom of the test tube. Experimental Part A. Precipitation Label test tubes A to D. A. 5 drops 0.1 M CaCl2 + 5 drops 1 M NH4OH + 15 drops

distilled H2O

B. 5 drops 0.1 M CaCl2 + 5 drops 1 M NaOH + 15 drops distilled H2O

C. 5 drops 0.1 M CaCl2 + 5 drops 3 M NaOAc + 15 drops distilled H2O

D. 5 drops 0.1 M CaCl2 + 5 drops 3 M NH4OH + 15 drops 3 M NH4Cl

CENTRIFUGE EACH AND OBSERVE. Calculate the Ion Product

of Ca(OH)2 in each mixture.

Part B. Dissolution of Precipitates

1. Place 10 drops each of Pb(NO3)2 and AgNO3 in two

separate test tubes. To each one, add HCl dropwise.

Shake the test tube after each drop. Continue adding

acid until last drop does not result in precipitation.

Centrifuge and observe. Decant supernatant liquid and

was precipitate with 5 drops of water. Discard the wash

water. In each of test tubes, add water and shake the

tubes. Heat each one in a water bath. Observe. Cool to

room temperature. Observe and compare to previous

observation. Place 10 drops each of Pb(NO3)2 and AgNO3

ions in two separate test tubes. To each one, add HCl

dropwise. Shake after each drop. Continue adding acid

until last drop does not result in precipitation. Centrifuge

and observe. Discard supernatant liquid and was

precipitate with 5 drops of water. Discard the wash

water. Into each test tubes, add 2 mL of concentrated

NH4OH gradually. Observe.

2. Place Pb(NO3) and Ba(NO3)2 in two separate test tubes.

Add K3CrO4. Add HNO3. Neutralize by adding NaOH or

until precipitation occurs. Do not add excess NaOH.

Centrifuge and decant supernatant liquid. To each of the

precipitates, add 15 drops of 6 M NaOH.

3. Place Ba(NO3)2 in each of two separate test tubes. Add

Na2CO3 to the first until precipitation is complete. To the

second, add Na2SO3. Centrifuge. Observe result. Discard

supernate. Add concentrated HCl to precipitate.

4. Place Cu(NO3)2 in each of two separate test tubes. Add

HOAc dropwise until acidic to litmus, then add

thocetamide solution dropwise until precipitation is

complete. Centrifuge then decant supernatant liquid.

Add 2 mL of distilled water to each test tube. Stir,

centrifuge and discard supernate. Add 5 drops of HCl to

precipitate and HNO3 to the other. Heat both mixtures in

water bath. Observe.

Part C. Fractional Precipitation

1. Mix K2CrO4, Na2S, KI and H2O in a test tube. Then add Pb(NO3)2 dropwise.

2. Then add Pb(NO3)2 dropwise. Centrifuge after the addition of each drop. Record result. Continue adding until no precipitate is formed.

RESULTS

Part A. Precipitates Given that the Ksp for Ca(OH)2 is 6.5 x 10

-6. We calculate for the Ion Product Constant and prove the existence of precipitate. Ca

2+ + OH

- Ca(OH)2

M1V1 = M2V2 (0.1M)(5 drops) = M2(25 drops) M2 = 0.02 M Ca

2+

Mixtures of:

A No Precipitation.

B White precipitate formed at the bottom, with few suspended and stuck on the walls of the test tube.

C White precipitate formed, more than B.

D No precipitation

Test Tube A NH4OH NH4

+ + OH- NH4

+ + OH- NH3 + H2O M1V1=M2V2 (1 M NH4OH)(5 drops) = M2(25 drops) M2 = 0.2 M NH4OH To get OH-, which is represented as X, Kb for NH3= [NH4

+][OH-] / [NH3] 1.76 x 10-5 = (x)(x) / 0.2 M [OH-] = 1.88 x 10-3 M Ion Product Constant = [Ca2+][OH-] Ca2+ + 2OH- Ca(OH-)2 (0.02 M) (1.88 x 10

-3 M)

2 = 7.07 x 10-8

IF Ksp > Qsp, then no precipitation will occur. Correct observation. Test tube B Na+ + OH- NaOH M1V1=M2V2 (1 M NaOH)(5 drops) = M2(25 drops) M2 = 0.2 M NaOH Ion Product Constant = [Ca2+][OH-] Ca2+ + 2OH- Ca(OH-)2 (0.02 M)(0.2 M)2 = 8 x 10-4 If Ksp < Qsp, then precipitation will occur. Correct observation. Test tube C NaOAc Na+ + OAc M1V1 = M2V2 (3 M NaOAc)(5 drops)=M2(25 drops) M2 = 0.6 M NaOAc OAc

- + H20 HOAc + OH

-

Since Kw = KaKb ; where kb = 1.0 x 10

-14 / 1.8 x 10

-5

5.56 x 10-10 = (x)(x) / 0.06 [OH-] = 1.83 x 10

-5

Ion Product Constant = [Ca2+][OH-] Ca2+ + 2OH- Ca(OH-)2 (0.02 M)(1.83 x 10-5)2 = 6.69 x 10-12 If Ksp > Qsp, then precipitation will occur. Erroneous observation.

Test tube D NH4OH NH4

+ + OH- NH3 + H2O NH4

+ + OH- M1V1=M2V2 (3 M)(5 drops) = M2(25 drops) M2 = 0.6 M NH4OH NH4Cl NH4

+ + Cl-

M1V1=M2V2 (4 M)(15 drops) = M2(25 drops) M2 = 2.4 M NH4

+ Kb of NH3 = (x)(2.4+x) / (0.6) 1.8 x 10-5 = 2.4x + x2 / 0.6 [OH-] = 4.05 x 10-13 Ion Product Constant = [Ca

2+][OH

-]

Ca2+

+ 2OH- Ca(OH

-)2

(0.02 M)(4.05 x 10-13) = 8.1 x 10-15 If Ksp > Qsp, then precipitation will not occur. Correct observation. Part B. Dissolution of Precipitates In the first experiment where 0.1 M HCl was added dropwise to 10 drops of 0.1 M Pb(NO3)2 and 10 drops of 0.1 M AgNO3, it was observed that no precipitate formed in Pb(NO3)2 and it remained clear, while precipitate settled at the bottom of the test tube in the AgNO3 solution. When the solutions were added with water and heated in a water bath, the precipitate solidified in the AgNO3 solution and no precipitate still was seen in the Pb(NO3)2 solution. When it was cooled to room temperature, it stayed the same and no change happened to the solutions. The reverse happened when NH4OH was added. Pb(NO3) yielded precipitate while the precipitate of AgNO3 was not evident as before. The reactions involved for the first part were: Pb(NO3)2 (aq) + HCl (aq) PbCl2 (s) + HNO3 (aq) AgNO3 (aq) + HCl (aq) AgCl (s) + HNO3 (aq) The reactions involved when NH4OH was added were: PbCl2 (s) + 2NH3 (aq) + 2H2O (l) Pb(OH-)2 (s) + 2NH4Cl (aq) AgCl2 (s) + 2NH3 (aq) + 2H2O *Ag(NH3)2]

+ (aq) + 2Cl

- (aq) +

2H2O In Part B, which concerns the dissolution of precipitates, we have to consider the solubility product constant of the precipitates in each solution. For Pb(NO3)2 added with HCl, the equation is: Pb(NO3)2 (aq) + HCl (aq) PbCl2 (s) + HNO3 (aq) With PbCl2 = Ksp = 1.6 x 10

-5

For Ag(NO3)2 added with HCl, the equation is: Ag(NO3)2 (aq) + HCl (aq) AgCl (s) + HNO3 (aq) With AgCl = Ksp = 1.6 x 10-10 Since, PbCl2 has a higher ksp, it is then more soluble, making the non-evidence of precipitate as compared to AgCl justifiable. When heated, the precipitate in Ag(NO3)2 looked more solid. However, if we incorporate the Le Chateliers Principle, the subjection of heat to the reaction should have created an exothermic process that shifted backward to compensate for the heat. The precipitates should have lessened in amount and should have dissolved. When cooled, it should have done the opposite. When added with NH4OH, PbCl2 from the latter set-up would yield precipitate due to the presence of Pb(OH)2 (Ksp = 1.2 x 10-15). PbCl2 (s) + 2NH3 (aq) + 2H2O Pb(OH)2 (s) + 2NH4Cl (aq) AgCl from the latter set-up; however, would yield [Ag(NH3)2]

+, an ion which is readily soluble with a Ksp of 1.7 x 107. Because of this, no precipitate would form and the liquid will turn clear. In the second experiment, when 3 drops of 0.1 M K2CrO4 and 3 drops of 1 M HNO3 was added to 10 drops of 0.1 M Pb(NO3)2 and Ba(NO3)2, it was observed that Pb(NO3)2 was cloudy yellow. The liquid turned yellow with a yellow orange precipitate. Supposedly, it would have no precipitate. Ba(NO3)2 was initially yellow and clear. It remained the same with no precipitate. The reactions involved were: Pb(NO3)2 (aq) + K2CrO4 (aq) PbCrO4 (s) + 2KNO3 (aq) Ba(NO3)2 (aq) + K2CrO4 (aq) BaCrO4 (s) + 2KNO3 (aq) PbCrO4 (s) + HNO3 (aq) Pb(NO3)2 (aq) + H2CrO4 (aq) BaCrO4 (s) + HNO3 (aq) Ba(NO3)2 (aq) + H2CrO4 (aq) When 6 M NaOH was added, Pb(NO3)2 yielded an orange powdery precipitate while Ba(NO3)2 yielded a white precipitate. The reactions involved were: Pb(NO3)2 (aq) + 2NaOH (aq) Pb(OH)2 (s) + 2NaNO3 (aq) Ba(NO3)2 (aq) + NaOH (aq) Ba(OH)2 (s) + NaNO3 (aq) In the 3rd set of Part B, BaCO3 and BaSO4 are the insoluble salts of the reactions or the precipitates.

The reactions involved were: Ba(NO3)2(aq) + Na2CO3(aq) BaCO3(s) + 2NaNO3 (aq) Ba(NO3)2(aq) + Na2SO4(aq) 2NaNO3(aq) + BaSO4(s) After adding HCl, BaCO3 dissolved and BaSO4 stayed undissolved. Addition of HCl Observation

BaCO3

+ HCl completely dissolved

BaSO4

+ HCl white precipitate is still evident

BaCO3 dissolved because the H

+ ion of HCl combined with CO3

2- forming carbonic acid. This causes the equilibrium to shift to the right to replace the supply of CO3

2- that was consumed. On the other hand, BaSO4 did not dissolve, even though H+ from HCl reacted with SO4

2- to formed HSO4

-. This happened

because HSO4- ion will dissociate since it does not exist as a

molecular compound. There would be no shift in equilibrium because the SO4

2- ion is not consumed, making BaSO4

undissolved. In the 4th set, Cu(NO3)2 was added to thiocetamide to form the precipitate, CuS. Cu(NO3)2 (aq) + CH3CSNH2(aq) CuS(s) CuS(s) Cu2+(aq) + S2-(aq) Observations when HCl and HNO3 were added to the mixture. Reagents mixed Observation

CuS+ HCl precipitate dissolved

CuS+ HNO3 precipitate is still evident

The experimental results were different from the expected results. CuS should have dissolved in HNO3 and did not dissolved when HCl was added. This was expected to happen because of the capability of HNO3 to remove S

2- ion to elementary

sulfur by oxidation favors the forward reaction, so CuS dissolves. The reaction between CuS and HNO3 can be written as follows: 3CuS(s) + 8H

++ 2NO3

-(aq) 3Cu 2+(aq) + 4H2O(l)+ 2NO(g) +

3S(s) Part C. Fractional Precipitation Observations The solution is yellow in color. When Pb(NO3)2 was added, orange and yellow precipitates were formed. Reactions involved: Pb(NO3)2(aq) + Na2S(aq) PbS(s) + 2NaNO3(aq) Pb(NO3)2(aq) + K2CrO4(aq) PbCrO4(s) + 2KNO3(aq) Pb(NO3)2(aq) + 2KI(aq) PbI2 (s) + 2KNO3(aq)

Reagents mixed Observation

Ba(NO3)2 + Na2CO3 white precipitate formed

Ba(NO3)2 + Na2SO4 white precipitate formed more rapidly

In part C, precipitates PbS (black), PbCrO4 (yellow), PbI2 (yellow-orange) are expected to form. Using the solubility constants as basis, PbS (4 x 10-26)is expected to precipitate out first since it has the lowest Ksp, second to precipitate out should be PbCrO4 (3 x 10

-13) and last is PbI2 is (7.9 x 10-7) which

has the highest solubility constant. The experimental results were slightly different with the expected one. CONCLUSION Heterogeneous equilibria involve species in more than one phase. In precipitation reactions, an insoluble solid called precipitate forms and settles out of the solution. The comparison of solubility product constant, Ksp, to the reaction quotient, Q, can be used to determine whether a precipitate will form with a given concentration of ions. If Q < Ksp, forward process is favored. No precipitate occurs. If Q = Ksp, neither backward nor forward process is favored. Solid and solution are in equilibrium. If Q > Ksp, reverse process is favored. Precipitation occurs to form more solid. The factors affecting dissolution of precipitates are temperature and addition of complexing agents. Higher temperature and addition of oxidizing and complexing agent increases the solubility of salts. In fractional precipitation, the difference in Ksp values of the various reagents is used to identify the order in which salts will precipitate. Fractional precipitation is a technique in which two or more ions in solution, each capable of being precipitated by the same reagent. It is important that there is a significant difference in the Ksp values of the substances being separated to identify the order in which salts will precipitate.

EXPT 8 TITRATION CURVES INTRODUCTION Titration is a laboratory method wherein the concentration of a particular solute in a solution is determined. In an acid-base titration, a solution with a known concentration of base is slowly added to an acid or vice versa. By monitoring the progress of the reaction, a titration curve can be produced. A titration curve is the graph of the pH as a function of the volume of the added titrant. The equivalence point of a titration curve is the point at which the quantities of an acid and base that have been brought together are stoichiometrically equivalent. The titration curve may be used to determine the Ka of the weak acid or the Kb of the weak base being titrated.

Experimental A. Preparation of Standard Titration Curves In a 100 mL beaker, 25 mL of 0.1 M HOAc was

measured and transferred. The initial pH was determined with the use of a pH meter. The titrant, 0.1 M NaOH, was added by 1 mL increments. The solution was stirred then the pH was recorded afterwards. This was done after each addition of the said titrant. After an abrupt change in the pH, the volume of the increments added was changed to 0.5 mL. This was done until the pH was at 5-6 past neutral.

The procedure was repeated but with: (b) 0.1 M HCl as initial solution and 0.1 M NaOH as the titrant and (c) 0.1 M NH4OH as the initial solution and 0.1 M HCl as the titrant.

B. Analysis of Unknown The same procedure in Part A was done but with 0.1

M HCl titrant used for the unknown base and 0.1 M NaOH for the unknown acid.

RESULTS Part A I. Titration of 25 ml 0.1 M HOAc, (which is a weak acid) and 0.1 M NaOH (which is a strong base). Before Titration: Initial pH of 0.1 M HOAc

HOAc + H2O H3O

+ + OAc-

Initial 0.1 M -x

0.1-x

- - -

0 +x x

0 +x x

Change

Equilibrium

Ka= [H3O

+ ] [OAc-] = 1.8 x 10 -5 [HOAc] x2/ (0.1-x)= 1.8 x 10 -5 x= 1.3 x 10 -3 (pH will depend on [H3O

+ ] = x) pH= -log (1.3 x 10 -3 ) = 2.89 Using the Henderson-Hasselbach equation, (pH= pKa + log [base]/[acid]), it is possible to predict the change in pH as the titrant is added. + 5 ml NaOH= 30 ml (solution) (before equivalence point) 0.1 M NaOH = 0.1 mol NaOH/ 1L= 0.1 mmol/ml 5 ml NaOH x 0.1 mmol/ml = 0.5 mmol 25 ml HOAc x 0.1 mmol/ml= 2.5 mmol

HOAc + OH- OAc- + H2O

Initial 2.5 mmol -0.5 mmol

2 mmol

0.5 mmol -0.5 mmol

0

0 +0.5 mmol 0.5 mmol

- - -

Change

Equilibrium

pH= pKa + log ( [base]/[acid]) -log (1.8 x 10 -5) + log (2 mmol/30 ml)/ (0.5 mmol/30) pH= 4.14 + 25 ml NaOH= 50 ml (solution) (Equivalence point) 25 ml NaOH x 0.1 mmol/ml= 2.5 mmol

HOAc + OH

- OAc- + H2O

Initial 2.5 mmol -2.5 mmol

0

2.5 mmol -2.5 mmol

0

0 +2.5 mmol 2.5 mmol

- - -

Change

Equilibrium

OAc- (weak base) will undergo hydrolysis

OAc- + H2O HOAc + OH

-

Assume [OH] is small and [NaOAc]= 2.5 mmol/50 ml= 0.05 M Kb= [OH] [HOAc] / [OAc]= Kw/Ka= 1 x10 -14 / 1.8 x 10 -5 =5.6 x10 -10 Kw/Ka= 1 x10 -14 / 1.8 x 10 -5 = x2/ 0.05 x= 5.27 x 10 -6 pOH= -log (5.27 x 10 -6) pOH=5.28 pH= 14-5.28= 8.72 As 0.1 M NaOH is added in increments of 1 ml, the pH of the solution shows little change because of the buffer solution being set up. Due to its buffer capacity, which is the amount of acid or base (in this case 0.1 M NaOH) the buffer can neutralize before the pH will change to an appreciable degree. However, the change in pH becomes more pronounced as it nears the equivalence point. The pH of the solution depends on the concentration of HOAc that has not been neutralized. The goal of titration is to reach the equivalence point, which is the point where the added solute (NaOH) reacts completely with the substance present (HOAc). In this titration, equivalence point is attained when 25 ml of 0.1 M NaOH is added to 25 ml of 0.1 M HOAc. Thus, the concentration of the acid and base is the same. Since HOAc is a weak acid which reacted to NaOH which is a strong base, the solution becomes slightly basic at the equivalence point, with a pH of 8.72. The pH at the equivalence point is always above 7 when a weak acid-strong base titration happens, because the salt formed has an anion that is a weak base (OAc-), which will undergo hydrolysis(pH upon computation is 8.72). After the equivalence point, the pH is determined by the concentration of OH

- from the excess NaOH.

Theoretical Titration of 0.1 M HOAc with 0.1 M

NaOH ml titrant pH

0 2.89

5 4.14

10 4.56

15 4.92

20 5.34

25 8.72

30 11.96

35 12.22

40 12.36

45 12.46

50 12.52

55 12.57

60 12.61

0

2

4

6

8

10

12

14

0 50 100

pH

ml NaOH

Theoretical Titration:0.1 M HOAc with 0.1 M NaOH

p

0

2

4

6

8

10

12

14

0 20 40

pH

ml NaOH

Experimental Titration Curve of HOAc with NaOH

p

Titration of 0.1 M HOAc with 0.1 M NaOH

ml titrant pH ml titrant pH

0 3.11 20 5.32

1 3.4 21 5.43 2 3.67 22 5.58

3 3.85 23 5.77

4 3.98 24 6.09

5 4.09 25 6.63

6 4.2 26 9.88

7 4.3 26.5 10.22

8 4.37 27 10.82

9 4.45 27.5 10.93

10 4.52 28 11.21

11 4.6 28.5 11.32

12 4.67 29 11.39

13 4.73 29.5 11.46 14 4.81 30 11.52 15 4.88 30.5 11.57 16 4.95 31 11.6 17 5.03 18 5.11 19 5.22

II. Titration of 25 ml 0.1 M HCl (strong acid) with 0.1 M NaOH (strong base) Before Titration: Initial pH of 0.1 M HCl [H+] = [HCl] pH= -log (0.1M) pH= 1 + 5ml= 30 ml (solution)(before equivalence point) Acid mmol= 0.1 mmol/ml x 25 ml= 2.5 mmol Added Base mmol= 0.1 mmol/ml x 5 ml= 0.5 mmol

H3O + + OH- 2 H2O

Initial 2.5 mmol 0.5 mmol - - -

Change -0.5 mmol -0.5 mmol

Equilibrium 2mmol 0

(pH will depend on [H3O

+ ])

pH= -log (2mmol/30 ml) pH= 1.18 + 25 ml= 50 ml (solution) (equivalence point) Added base mmol= 0.1 mmol/ml x 25 ml= 2.5 mmol

H3O + + OH- 2 H2O

Initial 2.5 mmol 2.5 mmol - - -

Change -2.5 mmol -2.5 mmol

Equilibrium 0 0

Since no excess acid or base is present, the solution is neutral. pH=7 +30 ml= 55ml (solution) (after equivalence point) Added base mmol= 0.1 mmole/ml x30 ml= 3 mmol

H3O + + OH- 2 H2O

Initial 2.5 mmol 3 mmol - - -

Change -2.5 mmol -2.5 mmol

Equilibrium 0 0.5 mmol

(pH depends on excess [OH]) pOH= -log (0.5 mmol/55ml) pOH= 2.04 pH= 14-2.04= 11.96 In a titration of a strong acid and a strong base, the solution of the strong acid will always have a higher initial pH (initial pH of HCl is 1, whereas initial pH of HOAc is 2.89). Just as with HOAc, the initial pH depends on the concentration of HCl. Before the titration reaches the equivalence point, pH increases slowly, but experiences a steeper rise of pH as it reaches the equivalence point. Theoretically, in a strong acid-strong base titration, the pH at the equivalence point is always 7, because the cation of the strong base (Na

+) and the

anion of the strong acid (Cl-) do not hydrolyze. The pH of the

solution after the equivalence point will be determined by the concentration of excess NaOH.

Theoretical Titration of 0.1 M

HCl with 0.1 M NaOH

ml titrant pH

0 1 5 1.18

10 1.37 15 1.6 20 1.95 25 7 30 11.96 35 12.22 40 12.36 45 12.46 50 12.52 55 12.57 60 12.61

Theoretical Titration of 0.1 M HCl w ith 0.1 M NaOH

0

2

4

6

8

10

12

14

0 20 40 60 80

ml NaOH

pH pH

Titration of 0.1 M HCl with 0.1 M NaOH

ml titrant pH ml titrant pH 0 1.67 20 1.95

1 1.63 21 2.01

2 1.59 22 2.12

3 1.57 23 2.24

4 1.58 24 2.45

5 1.67 25 2.92

6 1.63 26 6.36

7 1.62 27 9.67

8 1.61 27.5 10.38

9 1.62 28 10.77

10 1.64 28.5 10.91

11 1.65 29 11.07

12 1.67 29.5 11.22

13 1.69 30 11.29

14 1.71 30.5 11.33

15 1.74 31 11.4

16 1.77 31.5 11.43

17 1.8 32 11.46

18 1.84 32.5 11.51 19 1.9 33 11.51

20 1.67 33.5 11.52

34 11.56

34.5 11.57

35 11.59

35.5 1.95

III. Titration of 25 ml 0.1 M NH4OH (weak base) with 0.1 M HCl (strong acid). Before Titration: Initial pH of 0.1 M NH4OH

NH4OH NH4+ + OH

-

Initial 0.1 M 0 0 +x x

Change -x +x

Equilibrium 0.1 - x x

Kb= 1.8x 10 -5= [NH4

+] [OH-]/[NH4OH] Assume: 0.1 >>x 1.8x 10 -5 = x 2 X= 1.3 x 10 -3 pOH= -log (1.3 x 10 -3) = 2.89 pH= 14-2.89= 11.11 + 5ml= 30 ml (solution) (before equivalence point) Base mmol before titration= 0.1 mmol/ml x 25 ml= 2.5 mmol Added acid mmol= 0.1mmol/ml x 5 ml= 0.5 mmol

NH4OH + H3O

+- NH4+ + H2O

Initial 2.5 mmol -0.5 mmol

2 mmol

0.5 mmol -0.5 mmol

0

0 +0.5 mmol 0.5 mmol

- - -

Change

Equilibrium

pOH=pKb + log [acid]/[base]

4.75 + log9.25 + log (0.5mmol/30 ml)/ (2mmol/30 ml) = 4.15 pH= 14-4.15= 9.85 +25 ml= 50 ml (solution) (equivalence point) Added acid mmol= 0.1 mmol/ml x 25= 2.5 mmoles

NH4OH + H3O

+ NH4+ + H2O

Initial 2.5 mmol -2.5 mmol

0

2.5 mmol -2.5 mmol

0

0 +2.5 mmol 2.5 mmol

- - -

Change

Equilibrium

Hydrolysis of NH4

+

NH4

+ + H2O NH3 + H3O+

Initial 0.05 M -x

0.05-x

- - -

0 +x x

0 +x x

Change

Equilibrium

Ka= [NH4OH] [H3O

+] /[ NH4

+] =5.6 x 10

-10

5.7 x 10 -10 = [H3O+]2 /(2.5mmol/50 ml)

5.6 x 10 -10 = x2/0.05-x x=5.29 x 10 -6 pH= -log (5.29 x 10 -6) pH= 5.27

+30 ml=55 ml (solution) (after equivalence point) Added acid mmol= 0.1mmol/ml x 30 ml= 3 mmol

NH4OH + H3O

+- NH4

+ + H2O

Initial 2.5 mmol -2.5 mmol

0

3 mmol -0.5 mmol 0.5 mmol

0 +2.5 mmol 2.5 mmol

- - -

Change

Equilibrium

pH= -log[H3O

+]= -log (0.5 mmol/55ml)= 2.04 The pattern of the titration of a weak base-strong acid is quite similar with the titration of a weak acid-strong base, except that the pH of the solution decreases as the titration continues. When titration reaches the equivalence point, the cation of the weak base (NH4

+) will undergo hydrolysis, thus making the pH less than 7 (upon calculation, pH at equivalence point is 5.28). Similarly, the pH after the equivalence point will depend on the excess HCl present.

Theoretical Titration of 0.1 M NH4OH w ith 0.1 M HCl

0

2

4

6

8

10

12

0 20 40 60 80

ml HCl

pH pH

Theoretical Titration of 0.1 M

NH4OH with 0.1 M HC

ml titrant pH

0 11.11 5 9.85

10 9.43 15 9.07 20 8.65 25 5.28 30 2.04 35 1.78 40 1.64 45 1.54 50 1.48 55 1.43 60 1.39

Titration of 0.1 M NH4OH with 0.1 M HC

ml titrant pH ml titrant pH

0 9.9 11 2.66

1 9.64 11.5 2.48

2 9.49 12 2.36

3 9.28 12.5 2.28

4 9.12 13 2.21

5 8.98 13.5 2.16

6 8.78 14 2.11

7 8.56 14.5 2.07

8 8.26 15 2.04

9 7.75 15.5 2.01

10 5.73 16 1.98

10.5 3.08

Part B Titration of Unknown with 0.1 M NaOH

Upon comparison with the theoretical titration curves, the unknown can be classified as a weak acid, due to the less steep change in pH as compared to the theoretical titration curve of a strong acid. Concentration of unknown acid [Concentration of NaOH] (Vol. NaOH)= [C. of unknown acid](Vol.at eq. point) (0.1 M) (25 ml) = x [ (25 ml NaOH) + (27 ml unknown acid)] x= 0.05 M When comparing all the theoretical titration curves with their respective experimental values, there is little difference in terms of the pattern the plot follows. The difference is accounted for by the different initial pH values before titration as compared to the theoretical pH values, which affects the equivalence point during titration. In these titration curves, determining the pH at each step is essential. Before the titration reaches the equivalence point, the pH of the solution depends on the amount of the substance not yet neutralized by the titrant. After reaching the equivalence point, pH levels depend on the excess amount of the titrant in the solution. Determining the pH at equivalence point depends of the substances used in titration. For a strong acid-strong base titration, there is no

need to compute for the pH level is always seven, because no cations or anions undergo hydrolysis, which will have an effect on its pH level. If one of the components is a weak acid or a weak base, the hydrolysis of the cation/anion needs to be taken into account Sources:

Brown, Theodore L., Bursten, Bruce E., Lemay, H. Eugene Jr. (2000). Chemistry: The Central Science. 8

th ed.

Silberberg, Martin S. (2007). Principles of General

Chemistry. McGraw-Hill Intl. ed.

Experimental Titration Curve of NH4OH w ith HCl

0

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6

8

10

12

0 5 10 15 20

ml HCl

pH pH