NAME _______________________
Chem 163
Problem Set #3
_____
1. An ion that is common with two solutes. Based on Le Chậtelier, ion will cause shift that leads away
from the excess of that common ion
2. a) Calculate the pH of a 0.100 M hypochlorite buffer solution and 0.100 M LiClO
 0.100 
pH  7.54  Log
  7.54
 0.100 
b) 0.100 M buffer and 0.150 M LiClO
 0.150 
pH  7.54  Log
  7.71
 0.100 
4. Explain how the different combinations of SA-SB, WA-SB, WB-SA, titrations (all same concentrations)
are different in terms of:
what is being neutralized is the 1st substance, titrant is the 2nd substance listed
a) initial pH
b) pH at the equivalence point
At the equivalence point, the moles of acid equal the moles of base, regardless of the type of titration.
SA●SB pH < 7
WA●SB pH < 7
WB●SA pH > 7
pH = 7
pH > 7
pH < 7
5. The titration of 20.00 mL of 0.100 M triethylamine with 0.100 M HCl. (CH3CH2)3N, Kb = 5.2*10-4
HCl & base will react in equal amounts (CH3CH2)3N + HCl -- Cl- + (CH3CH2)3NH+
1st determine initial # mole (CH3CH2)3N = (0.1000 mol/L)(0.02000L) = 0.002 mol
a) pH after add 10.00-mL HCl
total vol = 20 + 10 = 30.00 mL
mols HCl added: = (0.1000 mol/L)(0.01000 L) = 0.001 mols HCl also [(CH3CH2)3NH+]
(CH3CH2)3N + H2O --(CH3CH2)3NH+ + OHInitial
0.002
0.00
0.00
Change
-0.001
+0.001
+x
Equilibr
0.001
0.001
x
(0.033)
(0.033)
if using Molarity: Equilibrium values
to use in Kb expression in ( )
Kb = [(CH3CH2)3NH+][OH-]/[(CH3CH2)3N] => 5.2*10-4 = [0.001][x]/[0.001]
x = [OH-] = 5.2*10-4 -Log(5.2*10-4) = 3.28 = pOH
pH = 14 – 3.28 = 10.71
b) pH after add 19.00-mL HCl
total vol = 20 + 19 = 39.00 mL
mols HCl added: = (0.1000 mol/L)(0.01900 L) = 0.0019 mols HCl also [(CH3CH2)3NH+]
(CH3CH2)3N + H2O --(CH3CH2)3NH+ + OHInitial
0.002
0.00
0.00
Change
-0.0019
+0.0019
+x
Equilibr
0.0001
0.0019
x
(0.00256)
(0.0487)
if using Molarity: Equilibrium values
to use in Kb expression in ( )
Kb = [(CH3CH2)3NH+][OH-]/[(CH3CH2)3N] => 5.2*10-4 = [0.0487][x]/[0.00256]
x = [OH-] = 2.73*10-5 -Log(2.73*10-5) = 4.56 = pOH
pH = 14 – 4.56 = 9.44
6. Write the ion-product expressions for:
a) iron III hydroxide
Fe(OH)3 (s) ↔ Fe+3(aq) + 3 OH-(aq)
b) barium phosphate Ba3(PO4)2 (s) ↔ 3 Ba+2(aq) + 2 PO4-3(aq)
SnS(s) + H2O (l) ↔ Sn+2(aq) + HS-(aq) + OH-(aq)
c) tin II sulfide
Ksp = [Fe+3][OH-]3
Ksp = [Ba+2]3[PO4-3]2
Ksp = [Sn+2][HS-][OH-]
7. Determine the molar solubility of Ag2SO4 in 0.22M Na2SO4
Ag2SO4 ↔ 2 Ag+ + SO4-2
Initial
—
Change
—
Equil
—
0
+2x
0.22
+x
2x
0.22+x
@ equilibrium [SO4-2] = 2[Ag+]
make the assumption that 0.22 + x  0.22 look up Ksp = 1.5*10-5
Ksp = [2Ag+]2[SO4-2] = (2x)2(0.22)
1.5*10-5 = (4x2)(0.22)
1.5 *10-5
x
 4.1*10-3 molar solubility of Ag2SO4
4(0.22)
8. Identify, circle, the more soluble compound in water for each pair listed.
a) strontium sulfate or barium chromate SrSO4
Ksp: 3.2*10-7
>
2.1*10-10
b) calcium carbonate or copper II carbonate CaCO3
Ksp:
3.3*10-9
>
3*10-12
c) barium iodate or silver chromate
Ksp: 1.5*10-9 > 2.6*10-12
BaI2