Equilibrium Flashcards
Equilibrium
two opposing changes occurring at the same time and same rate
Macroscopic level
no apparent changes (temp, color, concentrations, etc.)
molecular level
constant change
saturated solutions
equilibrium systems where dissolving and undissolving happening at the same rate
ionic compounds in saturated solutions
forward
NaCl (s) ———> Na+ (aq) + Cl- (aq)
<———
reverse
the level of ions in solution are constant, but vary depending on compound and its solubility (the concentration compound in a saturated solution)
equilibrium expressions
equations that show how concentrations stay constant in an equilibrium
example of equilibrium expression
NO (g) + O2 (g) ———> 2NO (g)
<———
equilibrium expression:
Kc = product concentrations ^raised to coefficients
—————————————————————-
reactant concentrations ^raised to coefficients
Kc = [NO2]^2
————
[NO]^2 [O2]^1
write the solubility equilibrium expression
Ksp = [concentration for each product]
example:
NaF (s) ———> Na+ (aq) + F- (aq)
Ksp = [Na+] [F-]
What is the molar solubility of a compound given Ksp?
example:
Ca3(PO4)2 (s) ———> 3Ca+2 (aq) + 2PO4 -3 (aq)
<———
use ICE chart
Ca3(PO4)2 [Ca+2] [PO4-3] Initial: x 0 0 Change: -x +3x +2x Equilibrium: 0 3x 2x 3 referring to the coefficients
Ksp = [Ca+2 ]^3[PO4-3]^2
Ksp = (3x)^3 (2x)^2
given Ksp = 2.07E-33
2.07E-33 = (27x^3)(4x^2)
1.14E-7 M
Another problem with molar solubility
What is the molar solubility of BaSO4 in a solution of 2.00M Na2SO4? The Ksp of BaSO4 is 1.1E-10
BaSO4 (s) ———> Ba+2 (aq) + SO4-2 (aq)
<———
SO4-2 in the COMMON between the two compounds
rate reverse = K[Ba+2]^x [SO4-2]^y
2.00M is a stock concentration
2.00 mol of Na2SO4 (s) were dissolved into each 1L of solution
NaSO4 ———> 2Na+ (aq) + SO4-2 (aq)
2.00M Na2SO4 (1 SO4-2)
———— = 2.00M SO4-2
(1 Na2SO4)
BaSO4 [Ba+2] [SO4-2] I x 0 2.00 C -x +x +x E. 0 x 2.00 + x
Ksp = 1.1E-10 = [Ba+2][SO4-2]
1.1E-10 = (x)(2.00 + x)
1.1E-10 = 2.00x + x^2
EASY WAY:
1.1E-10 = x(2.00) BC 2.00 + x is so small
= 5.5 E-11
ionization of a weak acid
HC2H3O2 (aq) + H2O (l) ———> H3O+ (aq) + C2H3O2 -
<———
Ka = [H+]^1[C2H3O2]^1
————————
[HC2H3O2]^1
what does Ka =
Ka = [H+][c.b.]
————
[acid]
example for solving for Ka
What is the concentration of H3O+ in a 1.00M HC2H3O2 solution (Ka = 1.8E-5)
HC2H3O2 + H2O ———> H3O+ + C2H3O2-
<———
Ka = [H+][C2H3O2-]
——————-
[HC2H3O2]
[HC2H3O2] [H+] [C2H3O2-] I 1.00 0 0 C -x +x +x E 1.00 - x x x
1.8E-5 = (x)(x)
——-
1.00-x
1.8E-5 = x^2
1.00 - x is so small you simplify it to 1
4.2E-3M = x
buffer system
a mixture of a weak acid and its conjugate base in approximately equimolar amounts
example:
a mixture of .500M HC2H3O2 in .500M NaC2H3O2 is a buffer
weak acid: acetic acid HC2H3O2 500. M
conjugate base: acetate C2H3O2 500. M
NaC2H3O2 (s) ———> Na+ (aq) + C2H3O2- (aq)
spectator ion
now a buffer system:
HC2H3O2 + H2O ———> H3O+ + C2H3O2 -
<———
HC2H3O2 and C2H3O2- have the SAME CONCENTRATIONS
pH of a buffer system
What is the pH of our .50M HC2H3O2 in .50M C2H3O2- buffer?
HC2H3O2 + H2O ———> H3O+ + C2H3O2-
<———
Ka = [H+][C2H3O2-]
———————
[HC2H3O2]
[HC2H3O2] [H+] [C2H3O2-] I .50M 0 .50M C -x +x +x E .50M - x x .50M + x
Ka = 1.8E-5
1.8E-5 = (x)(.50)
———
(.50)
1.8E-5 = x
pH = -log1.8E-5 = 4.74 = pKa
