大学无机化学汇总整理 (1)PPT课件
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AnBm (s) nAm+ (aq) + mBn- (aq) K sp (AnBm) ={c(Am+)}n{c(Bn-)}m
6.1.3 Relationship between solubility and the solubility product
The conversion between K sp and solubility Because concentrations in the K sp expression
and solubility product
6.1.1 Solubility
Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent when dynamic equilibrium is established between undissolved solute and the solution at a specific temperature. We usually use the symbol “S” to express it.
K W ={c(H3O+ )}{c (OH- )}
Ka
=Kw , Kb
o:rKa Kb=Kw
pH scale pH + pOH = pK W = 14
20 5 C , pK a +pK b=14
1
percent ionization (a) Buffer solution
α=c0 -ceq 100% c0
K sp = (nS)n (mS)m
nAm+ (aq) + mBn- (aq)
nS
mS
AB: S = K sp
Example ：The solubility of AgCl is found experimentally to be 1.92×10-3 g·L-1 at 25oC. Calculate the value of K sp for AgCl.
Example：The K sp for Ag2CrO4 is 1.1×10-12 at
25oC. Calculate the solubility of Ag2CrO4 in
Chapter 6 Precipitation-Solubility
Equilibria
How to dissolve precipitate?
3
Chapter 6 Precipitation-Solubility Equilibria
§ 6.1 Solubility and solubility product
§ 6.2 Forming and dissolving of precipitates
*§ 6.3 Equilibria between two precipitates
§ 6.1 Solubility and solubility product
6.1.1 Solubility 6.1.2 Solubility product 6.1.3 Relationship between solubility
pH of a buffer solution
pH=pKa(HA )-lgcc((H A-A ))
pH =14-pKb(B)+lgc(cB (B)+ H )
Dissociation / stability constant for complexes
Kf
=1 Kd
2
Inorganic Chemistry – 2016 Fall
Solubility usually is expressed as grams of solute per 100g of water (g/100g) for aqueous solution.
6.1.2 Solubility product
The process which involves the dissolution and precipitation of an insoluble electrolyte occurs when it is added to water at a certain temperature.
Inorganic Chemistry – 2016 Fall
Review for Previous Chapter 5. Acid-Base Equilibrium：
Bronsted acid / base
Lewis acid / base
Ion product constant of water
must in molarity and the unit of solubility is g
solute /100g water, So we need convert the solubility data to molarity(mol·L－1).
AnBm (s)
Equilibmri/umolL-1
Answer：We know Mr(AgCl)= 143.3 S = 1.92×10-3 mol L-1 = 1.34×10-3 mol L-1 143.3
AgCl(s) Ag+ (aq) + Cl- (aq)
Equilibrium / mol.L-1
S
S
K sp (AgCl) = {c(Ag+ )}{c(Cl- )} = S 2 = 1.80×10-10
When the rate of precipitation is equal to the rate of dissolving, a dynamic multiphase equilibriumHale Waihona Puke Baiduis established.
B4 a (s S )O B 2 + (a a + S q 2 4 - ( O )) aq
K sp (BaSO4 ) = [c(Ba2+ )/c ][c(SO24- )/c ] or simply： K sp (BaSO4 ) = {c(Ba2+ )}{c(SO24- )} K sp — solubility product constant
For general precipitation reactions：
6.1.3 Relationship between solubility and the solubility product
The conversion between K sp and solubility Because concentrations in the K sp expression
and solubility product
6.1.1 Solubility
Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent when dynamic equilibrium is established between undissolved solute and the solution at a specific temperature. We usually use the symbol “S” to express it.
K W ={c(H3O+ )}{c (OH- )}
Ka
=Kw , Kb
o:rKa Kb=Kw
pH scale pH + pOH = pK W = 14
20 5 C , pK a +pK b=14
1
percent ionization (a) Buffer solution
α=c0 -ceq 100% c0
K sp = (nS)n (mS)m
nAm+ (aq) + mBn- (aq)
nS
mS
AB: S = K sp
Example ：The solubility of AgCl is found experimentally to be 1.92×10-3 g·L-1 at 25oC. Calculate the value of K sp for AgCl.
Example：The K sp for Ag2CrO4 is 1.1×10-12 at
25oC. Calculate the solubility of Ag2CrO4 in
Chapter 6 Precipitation-Solubility
Equilibria
How to dissolve precipitate?
3
Chapter 6 Precipitation-Solubility Equilibria
§ 6.1 Solubility and solubility product
§ 6.2 Forming and dissolving of precipitates
*§ 6.3 Equilibria between two precipitates
§ 6.1 Solubility and solubility product
6.1.1 Solubility 6.1.2 Solubility product 6.1.3 Relationship between solubility
pH of a buffer solution
pH=pKa(HA )-lgcc((H A-A ))
pH =14-pKb(B)+lgc(cB (B)+ H )
Dissociation / stability constant for complexes
Kf
=1 Kd
2
Inorganic Chemistry – 2016 Fall
Solubility usually is expressed as grams of solute per 100g of water (g/100g) for aqueous solution.
6.1.2 Solubility product
The process which involves the dissolution and precipitation of an insoluble electrolyte occurs when it is added to water at a certain temperature.
Inorganic Chemistry – 2016 Fall
Review for Previous Chapter 5. Acid-Base Equilibrium：
Bronsted acid / base
Lewis acid / base
Ion product constant of water
must in molarity and the unit of solubility is g
solute /100g water, So we need convert the solubility data to molarity(mol·L－1).
AnBm (s)
Equilibmri/umolL-1
Answer：We know Mr(AgCl)= 143.3 S = 1.92×10-3 mol L-1 = 1.34×10-3 mol L-1 143.3
AgCl(s) Ag+ (aq) + Cl- (aq)
Equilibrium / mol.L-1
S
S
K sp (AgCl) = {c(Ag+ )}{c(Cl- )} = S 2 = 1.80×10-10
When the rate of precipitation is equal to the rate of dissolving, a dynamic multiphase equilibriumHale Waihona Puke Baiduis established.
B4 a (s S )O B 2 + (a a + S q 2 4 - ( O )) aq
K sp (BaSO4 ) = [c(Ba2+ )/c ][c(SO24- )/c ] or simply： K sp (BaSO4 ) = {c(Ba2+ )}{c(SO24- )} K sp — solubility product constant
For general precipitation reactions：