Equilibium Constants |
Key Concepts
Equilibrium Constants (K):
- are derived from experimental data
- are temperature dependent
- provide a measure of the equilibrium position
- if K is large (> 102) products of the reaction are favoured
- if K is small, reactants are favoured.
|
- remain the same value for a constant temperature even if the equilibrium concentrations of the reactants or products are altered
- do NOT change in the presence of a catalyst (a catalyst changes the time taken to reach equilibrium but does not alter the equilibrium position or equilibrium constant)
- For the reaction:
aA + bB
 cC + dD
where A, B, C and D are the chemical species and
a, b, c and d are the stoichiometric coefficients,
Some Special Special Cases
Ion-product of Water (Kw)
- Ion-product of water refers to self-ionization of water:
H2O H+ + OH-
- Since H2O is a pure substance its concentration is a constant at a given temperature and pressure so its concentration is incorporated into the overall equilibrium constant:
Kw = [H+][OH-]
- At 25o (298K), Kw = 10-14 M2
- Since Kw is small, the reactant, H2O, is favoured,
so very few H+ and OH- ions are present at equilibrium.
- Calculating [H+] and [OH-] at equilibrium:
at equilibrium [H+] = [OH-] (from the balanced chemical equation)
Kw = [H+]2 = 10-14
[H+] = [OH-] = 10-7 mol L-1
Solubility Product (Ksp)
- refers to an ionic compound dissolving to form ions
- for the reaction:
aA(s) bB(aq) + cC(aq)
- Since the concentration of the solid is a constant, it is effectively incorporated into the equilibrium constant:
Ksp = [B]b[C]c
- Calculating how much solid will dissolve in water
For example, calculate how much silver bromide will dissolve in 1 L of water
(Ksp at 25oC is 5.0 x 10-13)
AgBr(s) Ag+(aq) + Br-(aq)
Ksp = [Ag+][Br-] = 5.0 x 10-13
At equilibrium [Ag+] = [Br-] (from the balanced chemical equation)
5.0 x 10-5 = [Ag+]2
[Ag+] = [Br-] = 7 x 10-7 mol L-1
The solubility of AgBr at 25oC is 7 x 10-7 M
- Deciding whether a precipitate will form
For example, will a precipitate form if 25.0 mL of 1.4 x 10-9 M NaI and 35.0 mL of 7.9 x 10-7 M AgNO3 are mixed? (Ksp for AgI at 25oC is 8.5 x 10-17)
Calculate the concentration of I- after mixing:
M1 = [I-] before mixing = 1.4 x 10-9 M (assuming full dissociaiton of NaI)
V1 = initial volume = 25.0 mL = 25.0 x 10-3 L
V2 = final volume after mixing = 25.0 + 35.0 = 60.0 mL = 60.0 x 10-3 L (assuming volumes are additive)
M2 = [I-] after mixing
M2 = [I-] after mixing = M1 x V1 ÷ V2 = (1.4 x 10-9 x 25.0 x 10-3) ÷ (60.0 x 10-3) = 5.8 x 10-10 M
Calculate the concentration of Ag+ after mixing:
M1 = [Ag+] before mixing = 7.9 x 10-7 M (assuming full dissociaiton of AgNO3)
V1 = initial volume = 35.0 mL = 35.0 x 10-3 L
V2 = final volume after mixing = 25.0 + 35.0 = 60.0 mL = 60.0 x 10-3 L (assuming volumes are additive)
M2 = [Ag+] after mixing
M2 = [Ag+] after mixing = M1 x V1 ÷ V2 = (7.9 x 10-7 x 35.0 x 10-3) ÷ (60.0 x 10-3) = 4.6 x 10-7
Calculate the ion product:
K = [Ag+][I-] = 5.8 x 10-10 x 4.6 x 10-7 = 2.7 x 10-16
Decide whether a precipitate forms:
If ion product > Ksp a precipitate will form
If ion product < Ksp a precipitate will not form
In this case, 2.7 x 10-16 > Ksp (8.6 x 10-17) so a precipitate will form
Acid Dissociation Constants (Ka)
For the reaction:
HB H+ + B-
- The stronger an acid is, the more it dissociates producing more H+ and B-,
so a stronger acid has a larger Ka
- Strong acids completely dissociate in solution so their Ka values approach infinity,
eg, HCl(aq)
Calculating [H+]
For example: calculate the [H+] in 0.10 mol L-1 HNO2(aq). (Ka = 5.0 x 10-4 at 25oC)
HNO2 H+ + NO2-
let x = moles of HNO2 that dissociate to form H+ and NO2-
| |
HNO2 |
|
H+ |
+ |
NO2- |
| initial concentrations (M) |
0.10 |
|
0 |
|
0 |
| Equilibrium concentrations |
0.10 - x |
|
x |
|
x |
| Ka = |
[H+][NO2-] |
|
| [HNO2] |
| 5.0 x 10-4 = |
[x][x] |
|
| [0.10 - x] |
Since the acid dissociates only slightly, x will be very small compared to 0.10
so 0.10 - x ~ 0.10
5.0 x 10-4 = x2 ÷ 0.10
x = [H+] = 7.1 x 10-3 mol L-1
|
|
|
Bookmark this site!