Chapter 15. Applying equilibrium. The Common Ion Effect. When the salt with the anion of a weak acid is added to that acid, Lowers the percent dissociation of the acid. Ex: NaF and HF mixed together F- is the common ion HF ↔ H + + F -
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Chapter 15
Applying equilibrium
HF ↔ H+ + F-
Adding F- shifts the equilibrium to the left and therefore fewer H+ ions present
The same principle applies to salts with the cation of a weak base.
The calculations are the same as last chapter.
NH4Cl and NH3 (NH4+ is the common ion)
NH3 + H+ NH4+
Na+ is a spectator and the reaction we are worried about is
HAc H+ + Ac-
Initial 0.50 M 0 0.25 M
-x
x
x
Change
Final
0.50-x
x
0.25+x
HAc H+ + Ac-
Initial 0.50 M 0 0.25 M
Change
-x
x
x
Final
0.50-x
x
0.25+x
x (0.25)
x (0.25+x)
=
1.8 x 10-5 =
(0.50)
(0.50-x)
x = 3.6 x 10-5
HAc H+ + Ac-
0.25 mol
0.50 mol
0.26 mol
0.49 mol
HAc H+ + Ac-
0.25 mol
0.50 mol
0.26 M
0.49 M
HAc H+ + Ac-
0.25 mol
0.50 mol
0.26 M
0.49 M
HAc H+ + Ac-
Initial 0.49 M 0 0.26 M
-x
x
x
Final
0.49-x
x
0.26+x
HAc H+ + Ac-
Initial 0.49 M 0 0.26 M
-x
x
x
Final
0.49-x
x
0.26+x
x (0.26)
x (0.26+x)
=
1.8 x 10-5 =
(0.49)
(0.49-x)
x = 3.4 x 10-5
HC3H5O3 H+ + C3H5O3-
Initially 0.75 mol 00.25 mol
After acid 0.77 mol0.23 mol
Compared to 3.38 before acid was added
HC3H5O3 H+ + C3H5O3-
Initially 0.75 mol 00.25 mol
After acid 0.70 mol0.30 mol
Compared to 3.38 before acid was added
NH4+H+ + NH3
Initially 0.40 mol 00.25 mol
After acid 0.42 mol0.23 mol
Compared to 9.05 before acid was added
NH4+H+ + NH3
Initially 0.40 mol 00.25 mol
After acid 0.35 mol0.30 mol
Compared to 9.05 before acid was added
HAcH+ + Ac-
Initially 5.00 mol 05.00 mol
After acid 5.04 mol4.96 mol
Compared to 4.74 before acid was added
HAcH+ + Ac-
Initially 0.050 mol 0 0.050 mol
After acid 0.090 mol 0 0.010 mol
Compared to 4.74 before acid was added
Titrations
7
pH
mL of Base added
>7
7
pH
mL of Base added
7
pH
mL of acid added
7
<7
pH
mL of acid added
Calculate the pH after 0.0mL, 15.0mL, 25.0 mL, and 30.0mL of 0.10 M NaOH are added to 25.0mL of 0.10M of Nicotinic acid Ka= 1.4 x 10 -5
Titrate 50.0 mL of 0.10 M HF (Ka = 7.2 x 10-4) with 0.10 M NaOH
Solubility Equilibria
Will it all dissolve, and if not, how much?
Ex: Calc. Solubility of CaC2O4 if the value of Ksp= 4.8 x 10 -5 mol/L
CaC2O4 (s) ↔ Ca+2(aq) + C2O4 -2 (aq)
Ksp = [Ca+2] [C2O4-2]
Ksp= [s] [s] = [s]2 = 4.8x10-5 = s2
s= 6.9 x 10 -3 M
2.0 x 10 -4 mol/L. Calc. Ksp value.
CuBr(s) ↔ Cu+1(aq) + Br-1 (aq)
I .0002M0M 0M
C - .0002M+.0002M +.0002M
E 0.0002M .0002M
Ksp = [.0002] [.0002] = 4.0 x 10 -8
(units usually omitted)
The solubility of Calcium Carbonate is 8.7 x 10 -9 mol/L. Calc. Ksp.
Ksp= 4.0 x 10 -11 in a 0.025 M NaF solution.
Chemical analysis gave [Pb2+] = 0.012 M, and [Br-] = 0.024 M in a solution. From a table, you find Ksp for PbBr2 has a value of 4x 10-5. Is the solution saturated, oversaturated or unsaturated?
(Hint: set up ice problem, determine L.R.)
This problem is found on pg. 768-769 in your book. Check your work!
The dissolving of solid AgCl in excess NH3 is represented as…
AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+ (aq) + Cl-(aq)
However, this is a result of several steps
AgCl(s) ↔ Ag+ + Cl- Ksp= 1.6 x 10-10
Ag+ + NH3 ↔ Ag(NH3)+ K1 = 2.1 x 103
Ag(NH3)+ + NH3 ↔ Ag(NH3)2+ K2= 8.2 x 103
So the Equilibrium Expression is written as…. K= [Ag(NH3)2+] [Cl-]
[NH3]2
= Ksp x K1 x K2
= (1.6 x 10-10) ( 2.1 x 103) (8.2 x 103)
= 2.8 x 10-3
3 Factors that affect solubility:
- Common Ions
- pH
- Complexing Agents/Ions
Mixtures of Ions can be separated by selective precipitation (you know this as Quantitative Analysis)
- Steps listed on pg.765 or in your lab