Problems 1-3.docx

STOICHIOMETRY

1. *The chemical reaction produce precipitate, this is evident from the result of our experiment that formed a blue precipitate or Cu(OH)2 from NaOH and CuSO4 reaction. In other hand, the reaction of KI and Pb(NO3)2 also formed a yellow precipitate *The chemical reaction produces a color changes. After KI and Pb(NO3)2 are mixed, the color change from transparant to yellow. *The chemical produce changes in temperature. In the KI and Pb(NO3)2 reaction is happen temperature changes  The mass of the products is equal with the mass of the reaactant 2. Refer to figure 1 we can see that the more mass of sulfur, then the lenght of copper is reduced/dwindle. (sulfur sebagai reaksi pembatas) 3. a. Refer to figure 2 and calculating above, we can see that the highest precipitate is produced from the reaction with the reactant volume composition 3 mL Pb(NO3)2 and 7 mL KI as following reaction: a. Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) 0,3 mmol 0,7 mmol 0,3 mmol 0,6 mmol 0,3 mmol 0,1 mmol 0,3 mmol b. In the calculation above, we can know that the moles reacted ratio of Pb(NO3)2 and KI is 1:2 4. a. The higher ∆T are mixing from NaOH 20 mL and 10 mL H2SO4 solution b. n NaOH= M x V = 0,2 x 20 = 4 mmol n H2SO4 = M x V = 0,2 x 10 = 2 mmol a. the moles reacted ratio of NaOH and H2SO4 is 2:1

THERMOCHEMISTRY

1. Determine to calorimeter constant by drawing a graph showed relationship between the temperature of the mixture of cold water and hot water vs time intervals. Answer : Graph of Temperature vs Time Interval

Temperature (°C)

36 35 34 33 32 1

2

3

4

5

6

7

8

9

10

Time intervals (min)

V1 V2 T1 T3(constant)

= 20 mL = 20 mL = 26°C = 299 K = 33°C = 306 K

𝜌water cwater T2

= 1 g/mL = 4,2 J/gK = 58°C = 331 K



Temperature is absorbed in calorimeter ∆T = T3 – T1 o = 306 – 299 =7K





A descent temperature in hot water ∆T2 = T2 – T3 o = 331 – 306 = 25 K Heat is absorbed in cold water  Mass of cold water (m1) = 𝜌water . V1 = 1 g/mL .20 mL = 20 gram  q1 of cold water = m1. cwater. ∆T1 = 20 . 4,2. 7 = 588 J Heat is absorbed in hot water  Mass of hot water (m2) = 𝜌water . V2 = 1 g/mL .20 mL = 20 gram  q2 of hot water =mhw. cwater. ∆T2 = 20 . 4,2 . 25 = 2100 J Heat is absorbed in calorimeter  q3 = q2 – q1 = (2100 – 588) J = 1512 J Calorimeter constant

A rise temperature in cold water o ∆T1 = T3 – T1 = 306 – 299 =7K



𝑞3

Ccal

= ∆𝑇 =

1512 7

T NaOH – HCl (°C)

= 216 J/K 2. Calculate the heat of neutralization of NaOH-HCl, HCl-NH4OH, NaOH-CH3COOH by drawing graphs showed the relationship between temperature of mixture of acid-base vs time intervals. Answer: A. Determination of neutralization heat of HCl +NaOH Graph of Temperature vs Time Intervalof HCl + NaOH

40 30 20 10 0 1

2

3

4

5

6

7

8

9

10

Time intervals (minutes)

THCl TNaOH VHCl 𝜌solution TConstant

= 26°C = 299 K = 26°C = 299 K = 20 mL = 1,03 g/mL = 30°C = 303 K

MolNaOH

= MNaOH.VNaOH = 1 M . 20 mL = 20 mmol = 0,02mol

Reaction : NaOH(aq) + M 0,02mol B 0,02mol S 0

HCl(aq)  0,02 mol 0,02 mol 0

Vmix

= VHCl + VNaOH = 20 mL + 20 mL = 40 mL

o ∆T3

= Tconstant – Ts = 303 – 299 =4K = 𝜌s . vs = 1,03 . 40 = 41,2 gram

o ms

MHCl MNaOH VNaOH csolution

=1M =1M = 20 mL = 3,96 Jg-1K-1

MolHCl = MHCl .VHCl = 1 M . 20 mL = 20 mmol = 0,02 mol

NaCl(aq) 0,02 mol 0,02mol

+

H2O(l) 0,02 mol 0,02 mol



q4



q5



Heat is produced by reaction q6 = q4 + q5 = 652,608 + 864 = 1516,608 J

= ms . cs . ∆T3 = 41,2 . 3,96 . 4 = 652,608 J = Ccal . ∆T3 = 216 . 4 = 864 J 

∆H

=-

𝑞6

=−

𝑛 1516,608 0,02

= -75830,4 J/mol (exothermic)

T NaOH – CH3COOH (°C)

B. Determination of neutralization heat of NaOH + CH3COOH Graph of Temperature vs Time Intervalof NaOH + CH3COOH 35 30 25 20 15 10 5 0

1

2

3

4

5

6

7

8

9

10

Time intervals (minutes)

T CH3COOH TNaOH V CH3COOH 𝜌solution TConstant

= 26°C = 299 K = 26°C = 299 K = 20 mL = 1,098 g/mL = 30°C = 303 K

MolNaOH

= MNaOH.VNaOH = 1 M . 20 mL = 20 mmol = 0,02mol

Reaction : NaOH(aq) + M 0,02mol

M CH3COOH MNaOH VNaOH csolution

CH3COOH(aq)  0,02 mol

=1M =1M = 20 mL = 4,02 Jg-1K-1

MolCH3COOH

= MCH3COOH . VCH3COOH = 1 M . 20 mL = 20 mmol = 0,02 mol

CH3COONa(aq)+ -

H2O(l) -

B S

0,02mol 0

0,02 mol 0

o ∆T3

= Tconstant – Ts = 303 – 299 =4K



q4

= ms . cs . ∆T3 = 43,92 . 4,02 . 4 = 706,2336 J



q5 = Ccal . ∆T3 = 216 . 4 = 864 J



Heat is produced by reaction q6 = q4 + q5 = 706,2336 + 864 = 1570,2336 J 

∆H

=− =−

0,02 mol 0,02mol

0,02 mol 0,02 mol

o ms

= 𝜌s . vs = 1,098 . 40 = 43,92 gram

𝑞6 𝑛 1570,2336 0,02

= -78511,68 J/mol (exothermic)

T HCl – NH4OH (°C)

C. Determination of neutralization heat of HCl +NH4OH Graph of Temperature vs Time Intervalof HCl + NH4OH 30 20 10 0 1

2

3

4

5

6

7

8

9

10

Time intervals (minutes)

b. Determination of neutralization heat of HCl + NH4OH THCl = 26°C = 299 K MHCl TNH4OH = 26°C = 299 K MNH4OH VHCl = 20 mL VNH4OH 𝜌solution = 1,015 g/mL csolution TConstant = 28°C = 301 K MolNaOH

= MNaOH.VNaOH = 1 M . 20 mL = 20 mmol = 0,02mol

=1M =1M = 20 mL = 3,96 Jg-1K-1

MolHCl = MHCl .VHCl = 1 M . 20 mL = 20 mmol = 0,02 mol

Reaction : NH4OH(aq) + M 0,02mol B 0,02mol S 0

HCl(aq)  0,02 mol 0,02 mol 0

NH4Cl(aq) 0,02 mol 0,02mol

+

H2O(l) 0,02 mol 0,02 mol

∆T3 = Tconstant – Ts = 301 – 299 =2K o ms = 𝜌s . vs = 1,015 . 40 = 40,6 gram o





q4



Heat is produced by reaction q6 = q4 + q5

∆H

=− =−

= ms . cs . ∆T3 = 40,6 . 3,96 . 2 = 321,552 J



q5

= Ccal . ∆T3 = 216 . 2 = 432 J = 321,552 + 432 = 753,552 J

𝑞6 𝑛 753,552 0,02

= -37677,6 J/mol (exothermic) 2. Calculate the heat of dilution of ethanol in water based on the experimental. Answer : Determination of heat dilution of ethanol in water A. Composition A (2 mL water dan 8 mL ethanol) Twater = 26°C = 299 K Vwater = 2 mL Tethanol = 28°C = 301 K Vethanol = 8 mL -1 -1 cwater = 4,2 Jg K cethanol = 1,92 Jg-1K-1 ∆TConstant = 30°C = 303 K 𝜌ethanol = 0,793 g/mL Mr C2H5OH = (2.12 + 6.1 + 1.16) = 46 ∆Tm

= =

𝑇𝑤𝑎𝑡𝑒𝑟+𝑇𝑒𝑡ℎ𝑎𝑛𝑜𝑙 2 299 𝐾 +301 𝐾 2

= 300 K

Mass of water



q7

= ∆TConstant - ∆Tm = (303 – 300) K =3K

Heat is absorbed in water 

∆T

= 𝜌water .Vwater = 1 g/mL . 2 mL = 2 gram = mwater.cwater. ∆T = 2 . 4,2 . 3 = 25,2 J

Heat is absorbed in ethanol 



Mass of ethanol = 𝜌ethanol .Vethanol

q8

mol C2H5OH =

= 0,793 g/mL . 8 mL

=

= 6,344 gram = methanol. cethanol. ∆T = 6,344 . 1,92 . 3 = 36,54144 J

= 0,1379mol

Heat is absorbed in calorimeter 

= Ckal . ∆T = 216 . 3 = 648 J

q9

𝑔 𝑀𝑟 C2H5OH 6,344 𝑔𝑟𝑎𝑚 46

Heat is produced by solution q10

= q7+ q8+ q9 = 25,2 + 36,54144 + 648 = 709,74144 J

Enthalpy change of the reaction 𝑞10

∆HA

= − 𝑚𝑜𝑙 C2H5OH =−

709,74144 1,379

= - 514,68 J/mol (exothermic) B. Composition B (4 mL water dan 6 mL ethanol) Twater = 26°C = 299 K Vwater Tethanol = 28°C = 301 K Vethanol -1 -1 cwater = 4,2 Jg K cethanol ∆TConstant = 29°C = 302 K 𝜌ethanol Mr C2H5OH = (2.12 + 6.1 + 1.16) = 46 ∆Tm

= =

𝑇𝑤𝑎𝑡𝑒𝑟+𝑇𝑒𝑡ℎ𝑎𝑛𝑜𝑙 2 299 𝐾 +301 𝐾 2

= 300 K Heat is absorbed in water  Mass of water = 𝜌water .Vwater = 1 g/mL . 4 mL = 4 gram 

= mwater.cwater. ∆T = 4 . 4,2 . 2 = 33,6 J Heat is absorbed in ethanol  Mass of ethanol = 𝜌ethanol .Vethanol



∆T

= 4 mL = 6 mL = 1,92 Jg-1K-1 = 0,793 g/mL

= ∆TContaant - ∆Tm = (302 – 300) K =2K

q7

q8

𝑔

molC2H5OH = 𝑀𝑟 C2H5OH 4,758 𝑔𝑟𝑎𝑚

= 0,793 g/mL . 6 mL

=

= 4,758 gram

= 0,1034mol

= methanol. cethanol. ∆T = 4,758. 1,92 . 2

46

= 18,27072 J Heat is absorbed in calorimeter  q9 = Ckal . ∆T = 216 . 2 = 432 J Heat is produced by solution 

q10

= q7+ q8+ q9 = 33,6 + 18,27072 + 432 = 483,87072 J

Enthalpy change of the reaction 𝑞10

∆HB

= − 𝑚𝑜𝑙 C2H5OH =-

483,87072 0,1034

= - 4679,60 J/mol (exothermic) C. Composition C (6 mL water dan 4 mL ethanol) Twater = 26°C = 299 K Vwater Tethanol = 28°C = 301 K Vethanol -1 -1 cwater = 4,2 Jg K cethanol ∆TConstant = 30°C = 303 K 𝜌ethanol Mr C2H5OH = (2.12 + 6.1 + 1.16) = 46

∆Tm

= =

𝑇𝑤𝑎𝑡𝑒𝑟+𝑇𝑒𝑡ℎ𝑎𝑛𝑜𝑙 2 299 𝐾 +301 𝐾 2

= 300 K

q8

= ∆TConstant - ∆Tm = (303 – 300) K =3K

Heat is absorbed in water  Mass of water = 𝜌water .Vwater = 1 g/mL . 6 mL = 6 gram  q7 = mwater.cwater. ∆T = 6 . 4,2 . 3 = 75,6 J Heat is absorbed in ethanol  Mass of ethanol = 𝜌ethanol .Vethanol



∆T

= 6 mL = 4 mL = 1,92 Jg-1K-1 = 0,793 g/mL

𝑔

molC2H5OH = 𝑀𝑟 C2H5OH 3,172 𝑔𝑟𝑎𝑚

= 0,793 g/mL . 4 mL

=

= 3,172 gram

= 0,0689mol

= methanol. cethanol. ∆T = 3,172. 1,92 . 3

46

= 18,27072 J Heat is absorbed in calorimeter  q9 = Ckal . ∆T = 175,63 . 3 = 526,89 J Heat is produced by solution q10 = q7+ q8+ q9  =75,6 + 18,27072 + 526,89 = 620,76072 J

Enthalpy change of the reaction 𝑞10

∆HC

= − 𝑚𝑜𝑙 C2H5OH =−

620,76072 0,0689

= - 9009,59 J/mol (exothermic)

D. Composition D (8 mL water dan 2 mL ethanol) Twater = 26°C = 299 K Vwater Tethanol = 28°C = 301 K Vethanol -1 -1 cwater = 4,2 Jg K cethanol ∆TConstant = 29°C = 302 K 𝜌ethanol Mr C2H5OH = (2.12 + 6.1 + 1.16) = 46 ∆Tm

= =

𝑇𝑤𝑎𝑡𝑒𝑟+𝑇𝑒𝑡ℎ𝑎𝑛𝑜𝑙 2 299 𝐾 +301 𝐾

= 4 mL = 6 mL = 1,92 Jg-1K-1 = 0,793 g/mL

∆T

= ∆TContaant - ∆Tm = (302 – 300) K

2

= 300 K

=2K

Heat is absorbed in water  Mass of water = 𝜌water .Vwater = 1 g/mL . 8 mL = 8 gram Heat is absorbed in ethanol  Mass of ethanol = 𝜌ethanol .Vethanol



q8



q7

= mwater.cwater. ∆T = 8 . 4,2 . 2 = 67,2 J

𝑔

molC2H5OH = 𝑀𝑟 C2H5OH 1,586 𝑔𝑟𝑎𝑚

= 0,793 g/mL . 2 mL

=

= 1,586 gram

= 0,0345mol

= methanol. cethanol. ∆T = 1,586 . 1,92 . 2

46

= 6.09024 J Heat is absorbed in calorimeter  q9 = Ckal . ∆T = 216 . 2 = 432 J Heat is produced by solution q10 = q7+ q8+ q9 = 60,2 + 6.09024 + 432 = 498,29024 J

Enthalpy change of the reaction ∆HD

𝑞10

= − 𝑚𝑜𝑙 C2H5OH =−

498,29024 0,0345

= - 14443,20 J/mol (exothermic) So, enthalpy change the heat of dilution of ethanol in water based on the experimental is : ∆Hdilution

= = =

∆HA + ∆HB + ∆HC + ∆HD 4 − 514,68 +(− 4679,60) +(− 9009,59 ) +(−14443,20) 4 −28647,07 4

= - 7161,7675 J/mol (exothermic)

ACID-BASE TITRATION

a. Determine the molarity of Oxalc acid as primary standard solution  Mass of oxalic acid = 1.2610 gram  Volume of solution = 100 mL 𝑚

M oxalic acid

= 𝑀𝑟 x 1,2610

= 126,070 x

1000 𝑉

1000 10

= 0,1 M b. Determine molarity NaOH solution  Volume of Oxalic acid = 10 mL  Volume of NaOH (1) = 19,6 mL  Volume of NaOH (2) = 19,6 mL  Average volume of NaOH = 19,6 mL (COONa)2 + 2H2O  2NaOH + (COOH)2 (N.V) oxalic acid = (N.V)NaOH 0,2 . 10 = N . 19,6 NNaOH

2

=

19,6

= M NaOH

= =

0,1 N 𝑀 𝑎 0,1 1

= 0,1 M c. Determine molarity HCl solution  Volume of HCl = 10 mL  Volume of NaOH (1) = 9 mL  Volume of NaOH (2) = 8,9 mL  Volume of average NaOH = 8,95 mL NaOH(aq) + HCl (aq)  NaCl(aq) + H2O(l) VHCl . NHCl 10 . NHCl

= =

NHCl

=

NHCl NHCl

= =

M HCl

= =

VNaOH . NNaOH 8,95 . 0,1 . 1 0,895 10

0,0895 0,01 N

𝑀 𝑎 0,01 1

= 0,01 M d. Based on the data d above, draw a titration curve on graph paper by plotting pH and volume of NaOH added

12

10

pH

8 6 4 2 0 1

2

3

4

5

6

7

8

9

10

11

12

Volume NaOH 0,1 M curve 1. Titraion 0,1 M HCl by 0,1 NaOH

12 10

pH

8 6 4 2 0

1

2

3

4

5

6

7

8

9

10

11

12

Volume NaOH 0,1 M curve 2. Titration 0,1 M CH3COOH by 0,1 M NaOH

e. Determine the pH of the mixture in the equivalent point by (i) theoretical, (ii) the titration curve 1) NaOH-HCl Molarity of HCl = 0,1 M Volume of HCl = 10 mL NaOH(aq) + HCl (aq)  NaCl(aq) + H2O(l)  Eqiuivalence point at pH = 7 (theoritical)

12

10

pH

8 6 4 2 0 1

2

3

4

5

6

7

8

9

10

11

12

Volume NaOH 0,1 M curve 3. Titraion 0,1 M HCl by 0,1 NaOH

2) NaOH-CH3COOH

Molarity of CH3COOH Volume of CH3COOH Molarity of NaOH Ka CH3COOH

= 0,1 M = 10 mL = 0,1 M = 1,8 . 10-5

VCH3COOH . NCH3COOH 10 . 0,1 . 1 VNaOH

= = =

n NaOH

= (M .V) NaOH = 0,1 . 10 = 1 mmole

n CH3COOH

m r s

[OH-]

VNaOH . NNaOH VNaOH . 1 . 0,1 10 mL

= (M .V)CH3COOH = 0,1 . 10 = 1 mmole

NaOH(aq) + CH3COOH (aq)  CH3COONa(aq) + H2O(l) 1 mmole 1 mmole 1 mmole 1 mmole 1 mmole 1 mmole 0 0 1 mmole 1 mmole

𝐾𝑤

= √ 𝐾𝑎 [CH3COONa] 10−14

= √1,8 .10−5 [0,1] = 5,5 . 10-6



pOH = - log [OH-] = - log 5,5 . 10-6 = 6 – log 5,5 = 5,26

pH

= 14 – pOH = 14 – 5,26 = 8,74 (theoritical)

12 10

pH

8 6 4 2 0 1

2

3

4

5

6

7

8

9

10

11

12

Volume NaOH 0,1 M curve 4. Titration 0,1 M CH3COOH by 0,1 M NaOH



Eqiuivalence point at pH =

6+10 2

= 8 (practice)

f. From the titration curve of acetic acid-NaOH, determine Ka of acetic acid NaOH(aq) + CH3COOH (aq)  CH3COONa(aq) + H2O(l) m 1 mmole 1 mmole r 1 mmole 1 mmole 1 mmole 1 mmole s 0 0 1 mmole 1 mmole pH =8 pOH = 6 [OH-] = 10-6 𝐾𝑤

[OH-]

= √ 𝐾𝑎 [CH3COONa]

10-6

=√

10-12

= 10

Ka

= 10-3

10−14 𝐾𝑎

−14

𝐾𝑎

[0,1]

[0,1]

g. What is the approximate pH of the mixture of CH3COOH-NaOH in the equivalent point 12 10

pH

8 6 4 2 0 1

2

3

4

5

6

7

8

9

10

11

12

Volume NaOH 0,1 M curve 5. Titration 0,1 M CH3COOH by 0,1 M NaOH



Eqiuivalence point at pH =

6+10 2

= 8 (practice)

h. What is the pH of the equivalence mixture of CH3COOH-NaOH if Ka CH3COOH = 1,8 . 10−5 NaOH(aq) + CH3COOH (aq)  CH3COONa(aq) + H2O(l) m 1 mmole 1 mmole r 1 mmole 1 mmole 1 mmole 1 mmole s 0 0 1 mmole 1 mmole

[OH-]

𝐾𝑤

= √ 𝐾𝑎 [CH3COONa] 10−14

= √1,8 .10−5 [0,1] = 5,5 . 10-6

pOH = - log [OH-] = - log 5,5 . 10-6 = 6 – log 5,5 = 5,26  pH = 14 – pOH = 14 – 5,26 = 8,74 (theoritical)