Thermodynamics Worksheet (Show Answer on This Sheet, and Show Work on Separate Paper)

CHEM 1412 Solubility and Precipitation Rexn Dr. Ya-Ping HuangName: ______

Predict if precipitate will form for the solutions formed by mixing the reagent from the row with the reagent from the column.

50.0mL 0.0010M Fe(NO3)3 / 25.0mL 0.050M Cu(NO3)2 / 50.0mL 0.0010M AgNO3
25.0mL 0.010M HCl / (1). Fe(NO3)3 +3 Hcl 
FeCl3 + 3HNO3
all products soluble, no ppt / (2). Cu(NO3)2 + 2 Hcl 
CuCl2 + 2 HNO3
all products soluble, no ppt / (3). AgNO3+ Hcl 
AgCl(s) + HNO3
2. AgCl ppt, Ksp=1.8 x 10-10
4.Qsp = [Ag+][Cl-]= 2.21 x 10-6
5. ppt visible
25.0mL 0.10M NaOH / (4). Fe(NO3)3 + 3NaOH 
Fe(OH)3 (s) + 3 NaNO3
2. Fe(OH)3 ppt Ksp = 6.3 x 10-38 4. Qsp =[Fe3+][OH-]3 = 2.4 x 10-8
5. ppt visible / (5). Cu(NO3)2 + 2 NaOH 
Cu(OH)2 (s) + 2 NaNO3
2. Cu(OH)2 ppt. Ksp = 1.6 x 10-9 4. Qsp =[Cu2+][OH-]2 = 6.25 x 10-5 5. ppt visible / (6). 2AgNO3 + K2CrO4
Ag2CrO4(s) +2 KNO3
2. Ag2CrO4 ppt Ksp = 9.0 x 10-12
4. Qsp =[Ag+]2[CrO4-2] = 1.48 x 10-8 5. ppt visible
Calculate the solubility in mol/L and g/L for each of the following compounds in each of the given solutions listed on left column unless indicated otherwise
AgCl / Mg(OH)2 / CaF2 / Ag3PO4 / Fe4[Fe(CN)6]3
MM / 58.3 / 78 / 419 / 859.3
pure water / 7. Ksp = S2
S= 1.34 x 10-5 M
= 1.93 x 10-3 g/L / 8. Ksp = S(2S)2=4S3
S=1.55 x 10-4 M / 9. Ksp = S(2S)2=4S3
S = 2.14 x 10-4M / 10 Ksp = (3S)3S=27S4
S =4.68 x 10-6M
= 1.96 x 10-3 g/L / 11 Ksp = (4S)4(3S)3 = 6912 S7 = 3.0 x 10-41
S= 4.6 x 10-7M
0.01M HCl / 12. Common-ion(Cl-)
effect
Ksp =S(S+0.01)
=0.01S
S= 1.8 x 10-8 M / 13 WB + SA, 100 % neutralization:
Mg(OH)2+ 2H+
Mg2++ 2H2O
½ (.01MH+)=.005M / 14. in 0.5 M KF
Common-ion (F-)
effect
S= 1.56 x 10-10 M
= 1.2 x 10-8 g/L / 15. in 0.1 M AgNO3
Common-ion (Ag+)
effect
Ksp = (3S+0.1)3 S
= 1x 10-3 S
S= 1.3 x 10-17 M

Thermodynamics Worksheet (Show answer on this sheet, and show work on separate paper)Name:______

Dr. Ya-Ping Huang Synonym:______

(1) Reactions: decomposition of a fertilizer 2NH4NO3(s)  2N2(g) + 4H2O(g) + O2(g)
H
-236 kJ/ mol rexn / qp for 500 g NH4NO3(s) / -737.5 kJ / E
-253.35 kJ / S
1.0406kJ/mol.K / Transition Temp
Does not exist
25 C / 300 C
G
-546.1 kJ / K
4.53 x 1095 / Spontaneous?
Yes / G
-832.47 kJ / K
7.08 x 1075 / Spontaneous?
Yes
(2) Reactions: ionization of water H2O(ℓ)  H+ (aq) + OH-(aq)
H
55.8kJ/mol rexn / qp for 500 g NH4NO3(s) / 1550 kJ / E
55.8kJ/mol rexn / S
-0.0806 kJ/mol.K / Transition Temp
Does not exist
25 C / 90 C
G
79.82 kJ / K
1.04 x 10-14 / Spontaneous?
No / G
85.06 kJ / K
5.78 x 10-13 / pH for pure water
6.12
(3) Habor process to produce NH3(g) N2(g) + 3H2(g)  2NH3(g)
H
-92.38kJ
/mol rexn / qp for 500 g NH4NO3(s) / -1356 kJ / E
-87.42kJ
/mol rexn / S
-0.1983 kJ/mol rxn / Transition Temp
465.9 K = 192.7°C
25 C / 300 C
G, kJ/mol rexn
a. -32.97(HS
b. -32.38 (Gf) / K

1. 4,47x105
2. 4,7x105

/ Spontaneous?
Yes / G kJ/mol rexn
21.28 / K
0.0115 / Spontaneous?
No

(1). 2NH4NO3(s)  2N2(g) + 4H2O(g) + O2(g)always spontaneous

H = 2HfN2(g) + 4HfH2O(g) + HfO2(g) - 2HfNH4NO3(s) = 2(0) +4(-241.8) + 0 –2(-365.6) = -236 kJ/mol rexn

q = [500g/(2x80)]mol x (-236) = -737.5 kJ (There are 2 moles NH4NO3for every mole reaction as written)

H = E +nRT,

E = H -nRT = -236 kJ – (7)(8.314E-3)(298.15) = -737.5 kJ –17.35 = -253.35 kJ

S = 2S.N2(g) + 4S.H2O(g) + S O2(g) – 2S.NH4NO3(s) = 2(191.5) +4188.7) + 205 –2(151.1) = 1040.6 J/mol.K =

1.0406kJ/mol.K

Ttran = H/S = -236 kJ/1.0406kJ/mol.K = -227 K, does not existent,(Transition temperature does not exist if T< 0 K.)

(In this reaction, reaction is always spontaneous, favored by lower enthalpy, H < 0 and increased entropy, S> 0)

25 C :G = H -T S = -236kJ – 298(1.0406) = -546.1 kJ/mol rxn

(or G = 2GfN2(g) + 4GfH2O(g) + GfO2(g) - 2GfNH4NO3(s) = 2(0) +4(-228.6) + 0 –2(-184) = -546.4 kJ/mol rexn)

G = -5.709logK logK = G /( -5.709) = -546.1/-5.709 = +95.66, K = 10exp(+95.66) = 4.53 x 1095

300 C :G = H -T S = -236 – (300+273)(1.0406) = -832.47 kJ

G = -2.303RTlogK logK = G /-2.303(8.314E-3)(300+273.15)= -832.47/-10.975 = +75.85,

K = 10exp(+75.85) = 7.08 x 1075

(2) H2O(ℓ)  H+ (aq) + OH-(aq)always nonspontaneous

H = HfH+ (aq) + HfOH-(aq) - HfH2O(l) = 0 + (-230.0) –(-285.8) = 55.8kJ/mol rexn

q = [500g/(18)]mol x 55.8 = 1550 kJ

90 C: G = H -T S =55.8kJ/mol - 363(-0.0806) = 85.06 kJ

logK = G /-2.303(8.314E-3)(90+273.15)= 85.06/-2.303(8.314E-3)(363) = -12.24, K = 10exp(-12.24) = 5.78 x 10-13

K = [H+][OH-] = [H+]2, [H+] = = 7.5 x 10-7, pH = 6.12

Problem Set 3b:

(2)qH2O = (4.184 J/goC)(100 mL)(1.02 g/mL)(26.65 – 23.35oC) = 1408 J

qCal = (24.0 J/oC)(26.65 – 23.35oC) = 79.2 J

(Total heat evolved = qH2O + qCal = 1487 J)

qtotal= qH2O+ qCal+ qrexn = 0

qrexn = -1487 J = nH

To find n, you have to find moles of each reagent and find which reactant is the limiting reagent.

mol rexn based on CuSO4 =

mol rexn based on NaOH = (0.600M)(0.0500L) x

So NaOH is limiting reagent, and n = 0.0150 mole reaction

qrexn = -1487 J = nH = (0.0150 mole reaction) H

H = -1487 J/(0.0150 mole reaction) = - 99133 J/mole rexn = -99.13 kJ/mole rexn

Complete thermochemical equation includes balanced chemical equation and H

CuSO4(aq) + 2NaOH(aq)  Cu(OH)2(s) + Na2SO4(aq)H = -99.13 kJ/mol rxn

4. The thermite reaction, used for welding iron, is as following:

8 Al(s) + 3Fe3O4(s)  9 Fe(s) + 4Al2O3(s)

A. To calculate H, the best method is using Hf:

8 Al(s) + 3Fe3O4(s)  9 Fe(s) + 4Al2O3(s)

H = 9Hf, Fe(s) + 4 Hf,Al2O3(s) – 3 Hf,Fe3O4(s) –8 Hf, Al(s) = 4(-1676) – 3(-1118) = -3350 kJ/mol rxn

1. To find the energy released, given both reactants, it’s a limiting reagent type of problem. The approach is to calculate

the energy released by the reaction of each reactant. The smaller value is the answer.

1. H based on 8.0 g Al(s):

1. H based on 20.0 g Fe3O4

The answer is therefore 96.3 kJ released. And Fe3O4(s) is the limiting reagent.

c. Grams of Fe produced is based on the limiting reagent Fe3O4(s)

Fe3O4

CHEM 1412 Assignment ACID-BASE REXN KEYS Dr. Ya-Ping Huang

Calculate the pH value for the following solutions: You need to show more details than the one shown below

1. 10.0mL 0.10M NaOH +25.0mL 0.10M Hcl
neutralization:
H+ +OH- H2O
[H+]=0.0428M pH=1.37 / 2.10.0mL 0.10M NaOH +25.0mL 0.10M HNO2
neutralization:
HNO2+ + OH-  H2O + NO2-
Then pH = pKa + log(CB/CA)
[H+] =6.73x10-4, pH=3.17
3. 25.0mL 0.10M NaOH +25.0mL 0.10M Hcl
neutralization:
H+ +OH- H2O
complete neutralization of SA & SB
[H+]=[OH-] =1.0x10-7M pH=7.0 / 4. 25.0mL 0.10M NaOH +25.0mL 0.10M HNO2
neutralization:
HNO2+ OH-  H2O + NO2-
Then hydrolysis:
NO2- + H2O HNO2+OH-
. [OH-]=1.05x10-6M pOH=5.98 , pH=8.02
5. 50.0mL 0.10M NaOH +25.0mL 0.10M Hcl
neutralization:
H+ +OH- H2O
[OH-] = 0.034 pOH =1.47 pH = 12.53 / 6. 50.0mL 0.10M NaOH +25.0mL 0.10M HNO2
neutralization:
HNO2+ OH-  H2O + NO2-
then [OH-]=.034M pOH=1.47 , pH = 12.53
7. 25.0mL 0.10M NH3 +25.0mL 0.10M Hcl
neutralization:
NH3+ H+ NH4+
Then hydrolysis: NH4+ NH3+ H+
[H+]=5.3x10-6 M, pH = 5.28 / 8. 50.0mL 0.10M NH3 +25.0mL 0.10M Hcl
neutralization:
NH3+ H+ NH4+
Then buffer,
pH = 9.27

Sample Test 3 Problem #12. Titration curves:

When you draw the titration curves, pay attention to the details,

1. the steep change of pH around equivalence point,
2. the flat first segments (slow change of pH) when SA is titrated w/ SB or when SB is titrated w/ SA.
3. the initial relatively large change in pH when WA is titrated w/ SB or when WB is titrated w/ SA.
4. Draw the curve as ideal curve rather than just connect 4 dots (4 points of a-d). Refer to titration handout fro details

a. 25.0 mL of 0.1 M HCl titrated w/ 50.0 mL 0.10 M NaOH

Volume at equivalence point: [acid](Vacid)(# of acidic H) = [base](Vbase)(# of OH)
(0.10M)(25.0x10-3L)(1) = (0.10M)(VNaOH)(1)
VNaOH = 25.0 mL(25.0x10-3L )
Volume at half-neutralization point = ½ (Volume at equivalence point)
= ½(25.0 mL) = 12.5 mL

1. pH at the starting point ( w/ no second reagent))

[H+] = [HCl] = 0.10 M , pH = -log(0.10) = 1.0

2. Volume of the second reagent and pH at the half-neutralization point

Vol = 12.5 mL of NaOH

[H+] = [HCl] = (0.10 M x 25.0 mL)/37.5 mL = 0.0667 M

[OH-] = [NaOH] = (0.10 M x 12.5 mL)/37.5 mL = 0.0333 M

Neutralization of SA(HCl) and SB(NaOH)

R H+ + OH-  H2O

I 0.0667 0.0333

C -0.0333 -0.0333

E 0.0334 0

Leftover SA [H+] will determine pH, pH = -log(0.0334) = 1.48

3. Volume of the second reagent and pH at the equivalent point

Vol = 25.0 mL of NaOH

[H+] = [HCl] = (0.10 M x 25.0 mL)/50.0 mL = 0.050 M

[OH-] = [NaOH] = (0.10 M x 25.0 mL)/50.0 mL = 0.050 M

Complete neutralization of SA(HCl) and SB(NaOH)

R H+ + OH-  H2O

I 0.050 0.050

C -0.050 -0.050

E  0 0

SA (HCl) is completely neutralized by SB(NaOH), solution is neutral, pH = 7.0

4. pH value when 50 mL of second reagents has been added, total volume = 75.0 mL

[H+] = [HCl] = (0.10 M x 25.0 mL)/75 mL = 0.0333 M

[OH-] = [NaOH] = (0.10 M x 50.0 mL)/75 mL = 0.0667 M

Neutralization of SA(HCl) and SB(NaOH)

R H+ + OH-  H2O

I 0.0333 0.0667

C -0.0333 -0.0333

E  0 0.0334

Leftover SB [OH-] will determine pH,

pOH = -log(0.0334) = 1.48 , pH = 14 –1.48 = 12.52

5. For SA and SB, no Ka or Kb information

**. For WA titrated w/ SB or WB titrated w/ SA, at HN (half-neutralization) point, pH = pKa and [H+] = Ka

b. 25.0 mL of 0.1 M NaOH titrated w/ 50.0 mL 0.10 M HCl

T3worksheetF04key110/12/2018