1. CHAPTER 2 ELECTROLYTE SOLUTION
2-1 Strong and Weak Electrolyte Solution
2-2 Theory of Acid-base
2-3 Acidity and Calculation of Solution
2-4 Equilibrium Between Dissolution
and Precipitation
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2. 2-1 Strong and Weak Electrolyte Solution
Theory of Strong Electrolyte Solution
Ion-ion Interaction Theory
Figure 2-1 Ion atmosphere
of strong electrolyte solution
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3. Ion Activity and Activity Coefficient
Ion concentration, which can play a real action in solution is ionic effective concentration, is called ion activity.
actual concentration of ion (c) multiply a correction factor - activity coefficient ( f ).
a = f ·c (2-1)
Generally,
a＜c, 0＜f＜1
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4. Activity coefficient are influenced by ion concentration the electric-charge number of ion has nothing to do with the nature of ion.
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5. c is the amount-of-substance concentration of the ion i;
Ionic Strength ( I )
Where, I is ionic strength;
c is the amount-of-substance concentration of the ion i;
z i is the charge number of the ion i.
Note that the activity is for an ion; the ion strength is for a solution.
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6. Table 2-1 Ion activity coefficient and ion strength of solution
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7. (1) calculate the ion strength of the solution:
Example 2-1
25ml 0.02mol /L HCl mixed with 25ml 0.18mol /L KCl, calculate activity of H+ ?
Solution:
(1) calculate the ion strength of the solution:
I =( 0.01× × × ×12)/2 =0.1
(2) look up the activity coefficient of ion has one charge:
when I = 0.1, Z =1, f = 0.78
(3) calculate the activity of H+ (aH+)
cH+ =0.02/2 =0.01mol /L, f = 0.78
So, aH+ = f ·c = 0.78×0.01 = (mol /L)
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8. 2-1.2 Ionization Equilibrium of Weak Electrolyte Solution
The law of Chemical Equilibrium
(Equilibrium Constant)
a A + b B → c C + d D
[C]c[D]d
K = (2-3)
[A]a[B]b
Equilibrium
constant
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9. Ionization Constant (Ki )
HAc + H2O H3O+ + Ac -
or simply
HAc H+ + Ac -
The corresponding equilibrium-constant expression is
[H+][Ac -]
K i = (2-4)
[HAc]
Acid-ionization
constant
Equilibrium c
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10. Ka stands for “acid constant” Kb stands for “base constant”
Remember:
Ka stands for “acid constant”
Kb stands for “base constant”
Point of Emphasis:
Do not confuse a weak acid with a dilute acid.
A weak acid has a small Ka , and a dilute acid has a
low concentration.
Notice:
1. Only a change in temperature will change the value of the equilibrium constant for a particular reaction.
2. The smaller the value for Ka or Kb , the weaker the acid or base.
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11. Degree of Ionization (α)
1. Definition:
Number of ionized molecules
α= ×100%
total number of solute molecules
= ×100%
ionized molecules + non- ionized molecules
concentration of ionized weak electrolyte
= ×100%
initial concentration of weak electrolyte
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12. 2. The factor of influencing degree of ionization
① the nature of solute:
18℃，0.1mol/L, αHAc= 1.33%, αH2S= 0.07%,
αHCN= 0.007%
② the initial concentration of solute:
(the more dilute the solution, the greater the
degree of ionization).
③ temperature:
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13. Dilution Law HA H+ + A - c α Initial c c 0 0 Equilibrium c – cα cα cα
concentration
[H+][A -] cα·cα cα2
KHA = = =
[HA] c- cα α
For Ka is very small, α is very small, 1-α≈ 1
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14. KHA = cα2 or Physical meaning:
Note that above dilute law only is for some given conditions : (1) the weak electrolyte must be monoprotic
(2) α≤ 5%
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15. The Common Ion Effect and Salt Effect
The ionization of a weak electrolyte is markedly decreased by the adding to the solution an ionic compound containing one of the ion of the weak electrolyte, this effect is called the common ion effect.
For example,
HAc
Ac - + H+
NaAc → Ac – + Na+
shift the equilibrium from right to left, decreasing the [H+].
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16. 2. Salt effect The ionization of a weak electrolyte is increased by
adding to the solution an soluble strong electrolyte which not contains the common ion with the weak electrolyte. This effect is called salt effect.
For example:
0.1mol/L [H Ac] α= 1.33%, adding NaCl,
[NaCl]= 0.1mol/L, α= 1. 68%
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17. Example: There is a solution of c(HAc) =0
Example: There is a solution of c(HAc) =0.1mol/L, if we add NaAc, when c(NaAc)=0.1mol/L. Calculate the α of HAc.
Solution:
HAc
H Ac -
Initial c
Equilibrium c [H+ ] [H+ ] [H+ ]
≈ ≈0.1
[H+][Ac-] [H+]×0.1
Ka= = = [H+] = 1.8×10-5
[H Ac]
α= [H+]/[HAc] = 0.018% ＜＜1.33%
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18. 2-2 Theory of Acid-base Developing brief of acid-base theory
Arrhenius Acids and Bases
An acid is a substance that ionizes in water to produce H+ ions .
A base is a substance that ionizes in water to produce OH- ions.
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19. Limitations: First, they apply only to acid-base reactions in aqueous
solutions.
Second, they do not explain the fact that substances
such as NH3, which do not contain the hydroxide
group, increase the OH - ion concentration when
added to water.
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20. 2-2.2 Bronsted-Lowry Acids and Bases (Acid-base Proton Theory)
1. Definition of acid and base
Acid - is a substance capable of donating a proton.
HCl, NH4+, HSO4-, H2O
Base - is a substance capable of accepting a proton.
Cl-, NH3, HSO4-, OH-
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21. Conjugate acid-base pair
H+ + B A
Conjugate acid-base pair
HCl
H+ + Cl-
H2CO3
H+ + HCO3-
HCO3-
H+ + CO32-
NH4+
H+ + NH3
H3O+
H+ + H2O
H2O
H+ + OH-
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22. Conclusion: Acid or base may be a molecule, atom, or ion.
Some molecules or ions are capable of
donating a proton, and also accepting a proton, which named ampholyte.
There are no concepts of salt in acid-base
proton theory.
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23. 2. Essence of Acid and Base Reaction
Essence: proton transfer reaction.
For example:
HCl(g) + NH3(g)
NH4+ + Cl- H+
A1 + B2
A2 + B1
conjugate
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24. 3. Relative Strength of Acids and Bases
① Judgment by acid-base reaction:
A1 + B2 → A2 + B1
acid: A1 ＞A2
base: B1 ＜ B2
For example: HCl, H2O, NH3
HCl + OH- →Cl- + H2O
H2O + NH2- → NH3 + OH-
Obviously, acid: HCl ＞ H2O ＞ NH3
conjugate base: Cl- ＜ OH- ＜ NH2-
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25. ② Compare by K a or Kb [H3O+ ][Ac-] K a = ────── HAc + H2O H3O+ + Ac-
The smaller the value for K a， the weaker the acid； The greater the value for K a， the stronger the acid.
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26. Ac- + H2O HAc + OH- [HAc][OH-] K b = ──────
The smaller the value for Kb， the weaker the base；
The greater the value for Kb， the stronger the base.
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27. The relationship between Ka and Kb: Ka × Kb = K w
③ The relationship between acid-base strength
and solvent
H2O strong acid
HAc weak acid
H2SO base substance
HNO3
H2O weak acid
NH strong acid
HAc
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28. 4. The Leveling Effect and Differentiating Effect
The leveling effect: The inability of a solvent to
differentiate among the relative strengths of all acids
stronger than the solvent’s conjugate acid is known
as the leveling effect.
Because the solvent is said to level the strengths of
these acids, making them seen identical.
leveling solvent:
Strong acid such as HClO4, HCl, HNO3, H2SO4 will
appear to be of equal strength in aqueous solution.
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29. For strong acid or base
Strong acids--hydrochloric acid (HCl), nitric acid (HNO3), perchloric acid (HClO4), and sulfuric acid (H2SO4), for example- are all strong electrolytes. They may be assumed to be completely ionized in water.
HCl(a q) + H2O(1) → H3O+(aq) + Cl - (aq)
HNO3(a q) + H2O(1) → H3O+(aq) + NO3-(aq)
HClO4(a q) + H2O(1) → H3O+(aq) + ClO4 - (aq)
H2SO4(a q) + H2O(1) → H3O+(aq) + HSO4 - (aq)
we can not determine their strength, because H3O+ is the strongest acid that can exist in aqueous solution.
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30. The differentiating effect:
The ability of a solvent to differentiate among the
relative strengths of all acids stronger than the solvent’s
conjugate acid is known as the differentiating effect.
If we use a more weakly basic solvent like acetic acid,
acetic acid can function as a base by accepting a proton.
Since acetic acid is a much weaker base than water, it is
not as easily protonated. Thus there are appreciable
differences in the extent.
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31. In acetic acid solvent, their relative strength increase as follows:
HNO3 ＜H2SO4 ＜HCl＜ HClO4
HCl(aq) + CH3COOH(l) CH3COOH2+(aq) + Cl - (aq) HNO3(aq) + CH3COOH (l) CH3COOH2+ (aq) + NO3-(aq) HClO4(aq)+ CH3COOH (l) CH3COOH2+ (aq) + ClO4 - (aq) H2SO4(aq)+ CH3COOH (l) CH3COOH2+ (aq) + HSO4 - (aq)
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32. 2-3 Acidity and Calculation of Solution
Autoionization of Water
H2O(l) + H2O(l)
H3O+ (aq) + OH-(aq)
Water is capable of acting as a proton donor and proton
acceptor toward itself. The process by which this occurs
is called autoionization of water.
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33. Kw = [H3O+][OH-]
Where Kw is the equilibrium constant for water (unitless) , is called ion product of water or autoionization equilibrium constant.
At 25℃, Kw = [H3O+][OH -] = 1.0 ×10 -14
[H+] ＞ [OH-] in acid solutions
[H+] ＜ [OH-] in basic solutions
[H+] = [OH-] in neutral solutions
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34. 2-3.2 Acidity of solution (The pH Function)
pH = - log aH+ = -log [H3O+]
pOH = - log aOH- = -log[OH-]
For, [H+][OH-] = Kw= 1×10-14
So, pH + pOH =pKw= 14.00
In neutral solutions, pH＝7＝pOH,
In acid solutions, pH ＜7 ＜pOH
In basic solutions pH ＞7 ＞pOH
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35. 2-3.3 Calculation of Acidity of Solution
● For Strong Acids and Bases
pH = - log[H+] = -log[acid]
pOH = - log[OH-] = -log[base]
Example:
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36. ● Monoprotic Weak Acids and Bases
H Ac
H+ + Ac –
initial c
equilibrium c- [H+] [H+] [Ac -] = [H+]
[H+] [Ac -] [H+]2
Ka= =
[H Ac] c-[H+]
[H+]2 + Ka[H+] – Kac = 0
Solve this equation:
[H+] = - Ka /2 + √ Ka2 /4 +Ka c
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37. Note that above equation is limited for some given condition:
When: c/Ka ≥103,
or α≤5%, c - [H+] ≈ c
thus,
Similarly, for weak base , there is an equation:
Note that above equation is limited for some
given condition:
Example 4-5: P41
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38. 2-4 Equilibrium between Dissolution and Precipitation
2-4.1 Solubility Product Constant (Ksp)
dissolution
AgCl(s)
Ag+ + Cl-
precipitation
Ksp = [Ag+][Cl-]
solubility product
constant
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39. Ksp = [M g2+][OH-]2 Ksp = [A g+]2[CrO42-] Ksp = [Fe3+][OH-]3
● Mg(OH) 2 (s)
Mg2+ + 2OH-
Ksp = [M g2+][OH-]2
2Ag+ + CrO42-
● Ag2CrO4 (s)
Ksp = [A g+]2[CrO42-]
● Fe(OH) 3 (s)
Fe3+ + 3OH-
Ksp = [Fe3+][OH-]3
mAn+ + nBm-
● AmBn(s)
Ksp = [An+]m [Bm-]n
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40. 2-4.2 Exchange between Solubility and Ksp
● AB (s , ksp)
AB(s)
A + B
In saturated solution: [A]=[B]=s (mol/L)
Ksp = [A] [B]= s2
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41. or ● AB2 (A2B) AB2(s) = s×(2s)2= 4s3 A + 2B
In saturated solution: [A]=s (mol/L)
[B]=2s(mol/L)
Ksp = [A] [B]2
= s×(2s)2= 4s3 or
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42. Solution: Example2-4: s = 2.1× 10- 4 mol /L
The Ksp for CaF2 is 3.9× What is its solubility in water, in grams per liter?
Solution:
s = 2.1× mol /L
2.1× mol /L × 78.1 g /mol = 1.6 × 10-2 g / L
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43. 2-4.3 Formation and dissolution of precipitation
● Rule of Solubility Product
HAc
H+ + Ac-
AgCl(s)
Ag+ + Cl-
Qi (ion product quotient): the product of the ion concentration in solution when the system is under any situation ( at equilibrium or not ).
Qi (AgCl)=cAg+ cCl-
Difference between Qi and Ksp:
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44. The relationship between Qi and Ksp
1. If Qi = Ksp, equilibrium is reached - no precipitate will
form. Saturated solution
2. If Qi > Ksp , a precipitate will form (until Q i decreases
to Ksp). Supersaturated solution
3. If Qi < Ksp, any precipitate in solution will dissolve until
Qi increases to Ksp. Unsaturated solution
The state above is called rule of solubility product.
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45. ● Formation of precipitation condition: Qi ＞Ksp
Example 2-5:
Does a precipitate form if L of 3.0 × 10 -3mol/L
Pb(NO3)2 is added to L of 5.0 × mol /L Na2SO4?
possible precipitate form is PbSO4 ( Ksp = 1.6 × 10-8 )
[Pb2+] = 6.0 × 10-4 mol /L
[SO42-] =4.0 × 10-3 mol /L
Qi = [Pb2+][SO42-] = (6.0 × )(4.0 × 10-3)
= 2.4 ×10 - 6
Because Q i> Ksp, PbSO4 will precipitate!
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46. ☆ Selective Precipitation of Ions
Selective precipitation is a technique of separating ions in an aqueous solution by using a reagent that precipitates one or more of the ions, while leaving other ions in solution.
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47. Example: Consider a solution that is 0
Example: Consider a solution that is 0.010mol/L in both Cl- and I-, if we add AgNO3 solution drop by drop, which precipitation will produce first? Ks p(Ag Cl) = 1.8×10 -10, K s p(AgI) = 8.5×10 -17
Solution:
Ag+ + Cl- = AgCl↓
Ag+ + I- = AgI↓
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48. Formed AgI , [Ag+] Ksp 8. 5×10 -17 [Ag+] = ----- = ------------- = 8
Formed AgI , [Ag+] Ksp × [Ag+] = = = 8.5×10 -15(mol/L) [I-]
Formed AgCl , [Ag+] Ksp × [Ag+] = = = 1.8×10 -8(mol/L) [Cl-]
When AgCl precipitate,
Ksp × [I-] = = = 4.7×10 -9(mol/L) [Ag+] ×10 -8
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49. There are several methods to dissolve precipitation.
● Dissolution of precipitation condition: Qi ＜ Ksp (ion product ＜ solubility product)
There are several methods to dissolve precipitation.
1. Forming weak electrolytes by adding some compounds make precipitation dissolve.
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50. For example: Mg(OH)2 not only dissolve in acid, but also dissolve in NH4Cl solution.
Mg(OH)2 + 2HCl = MgCl2 + 2H2O
Mg(OH)2(s)
Mg2+ + 2OH- +
2HCl
2Cl H+
2H2O
Because formed weak electrolyte H2O, [OH-]
decrease, shift the equilibrium from left to right.
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51. AgCl(s) Ag+ + Cl- + 2NH3 [A g(NH3)2 ]+
2. Forming coordination compounds by adding some agents make precipitation dissolve.
For example, AgCl precipitation dissolve in NH3·H2O.
AgCl(s) + 2NH3 = [A g(NH3)2 ]+ + Cl-
AgCl(s) Ag+ + Cl- +
2NH3
[A g(NH3)2 ]+
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52. 3. Producing oxidation-reduction reactions by adding oxidizing agents or reducing agents make precipitation dissolve.
For example :
To the CuS precipitation add the dilution HNO3, CuS might dissolve.
3CuS(s) + 8HNO3(dilution) = 3Cu(NO3)2 + 3S + 2NO + 4H2O
Cu S(s)
S2- + Cu2+ +
HNO3
S↓+ NO↑ + H2O
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53. SUMMARY:
Problems
1. Discuss the effect on solubility of AgCl by adding (a) HCl (b) AgNO3
(c) KNO (d) aqueous ammonia to a
saturated AgCl solution
2. If 50.00mL of 0.1mol·L-1 HA is added to 20.00mL of 0.10 mol·L-1 KOH, and then diluted to 100.0mL, the pH of the resulting solution is What is the Ka of HA?
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54. (a) What precipitations first?
3. A solution contains 0.01mols of KCl and 0.10 moles of K2CrO4 per liter. AgNO3 is added slowly and volume changes are so small that they can be neglected. Ksp(AgCl)=1.77×10-10 and Ksp(Ag2CrO4)=1.12×
(a) What precipitations first?
(b) What is the concentration of the first anion
in solution just as the second anion begins
to precipitate from solution?
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55. 4. Add 50ml of 1. 8mol/L NH3·H2O to 50ml solution which contains 1
4. Add 50ml of 1.8mol/L NH3·H2O to 50ml solution which contains 1.14g MgCl2 . If we prevent Mg(OH)2 precipitation, what grams NH4Cl solid will be added to the mixed solution?
Ksp=1.2 × 10-11, Kb= 1.8 × 10-5,
M(MgCl2)=95, M(NH4Cl)=53.5
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56. 5. A volume of 100 mL of 0.2mol·L-1 MnCl2
is mixed with an equal volume of a NH4Cl
solution containing 0.10mol·L-1 NH3. What
mass of NH4Cl must be added to resulting
solution to prevent the precipitation of
Mn(OH)2? Ksp{Mn(OH)2}=2.06×10-13,
Kb (NH3·H2O)= 1.8 × 10-5.
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