1. Chapter 8 Acids and Bases
Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
Chapter 8 Acids and Bases
Dr Nehdi Imededdine
King Saud University
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2. Some Definitions Arrhenius
An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions.
HCl(g) H2O H+(aq) + Cl-(aq)
A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions.
NaOH H2O Na+(aq) + OH-(aq)
an Arrhenius acid
an Arrhenius base
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3. The hydronium ion, H3O+(aq)
An H+ ion is simply a proton with no surrounding valence electron. This small, positively charged particle interacts strongly with the nonbonding electron pairs of water molecules to form hydrated hydrogen ions. The interaction of a proton with one water molecule forms the hydronium ion, H3O+(aq):
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4. Some Definitions Brønsted-Lowry
An acid is a proton donor ( HCl, H2O, …).
A base is a proton acceptor (NH3, H2O, …).
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5. A Brønsted-Lowry acid… ( examp: HCl)
…must have a removable (acidic) proton.
A Brønsted-Lowry base…( examp: NH3)
…must have a pair of nonbonding electrons.
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6. If a substance can be either acid or base it is amphiprotic
Amphiprotic substance
If a substance can be either acid or base it is amphiprotic
HCO3-
HSO4-
H2O
Base
HCO H2O H2CO OH-
Acid
HCO H2O (CO3) H3O+
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7. What Happens When an Acid Dissolves in Water?
Water acts as a Brønsted-Lowry base and abstracts a proton (H+) from the acid.
As a result, the conjugate base of the acid and a hydronium ion are formed.
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8. Conjugate Acids and Bases
The term conjugate comes from the Latin word “conjugare,” meaning “to join together.”
Reactions between acids and bases always yield their conjugate bases and acids.
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9. Acid and Base Strength
Strong acids are completely dissociated in water.
Their conjugate bases are quite weak.
Weak acids only dissociate partially in water.
Their conjugate bases are weak bases.
Figure 8.1 Relative strengths of some conjugate
acid–base pairs
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10. Acid and Base Strength
Substances with negligible acidity do not dissociate in water.
Their conjugate bases are exceedingly strong.
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11. Acid and Base Strength
In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.
HCl (aq) + H2O (l)  H3O+ (aq) + Cl- (aq)
H2O is a much stronger base than Cl-, so the equilibrium lies so far to the right that K is not measured (K >>1).
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12. Acid and Base Strength
In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.
CH3CO2H (aq) + H2O (l)
H3O+ (aq) + CH3CO2- (aq)
Acetate is a stronger base than H2O, so the equilibrium favors the left side (K< 1).
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13. Proton transfer reaction
A proton transfer reactions are governed by the relative abilities of two acids to transfer protons. For example, consider the proton transfer that occurs when an acid HX acid dissolves in water
HX (aq) + H2O (l)
H3O+ (aq) + X- (aq)
In any acid-base reaction, the position of the equilibrium favors transfer of the proton from the stronger acid to the stronger base.
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14. Autoionization of Water
As we have seen, water is amphoteric.
In pure water, a few molecules act as bases and a few act as acids.
This is referred to as autoionization.
H2O (l) + H2O (l)
H3O+ (aq) + OH- (aq)
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15. Ion-Product Constant The equilibrium expression for this process is
Kc = [H3O+] [OH-]
This special equilibrium constant is referred to as the ion-product constant for water, Kw.
At 25C, Kw = 1.0  10-14
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16. pH
pH is defined as the negative base-10 logarithm of the concentration of hydronium ion.
pH = -log [H3O+]
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17. pH In pure water, Kw = [H3O+] [OH-] = 1.0  10-14
Since in pure water [H3O+] = [OH-],
[H3O+] =  = 1.0  10-7
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18. pH Therefore, in pure water, pH = -log (1.0  10-7) = 7.00
An acid has a higher [H3O+] than pure water, so its pH is <7.
A base has a lower [H3O+] than pure water, so its pH is >7.
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19. pH These are the pH values for several common substances.
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20. Other “p” Scales
The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions).
Some similar examples are
pOH: -log [OH-]
pKw: -log Kw
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21. -log [H3O+] + -log [OH-] = -log Kw = 14.00
Watch This!
Because
[H3O+] [OH-] = Kw = 1.0  10-14,
we know that
-log [H3O+] + -log [OH-] = -log Kw = 14.00
or, in other words,
pH + pOH = pKw = 14.00
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22. How Do We Measure pH? For less accurate measurements, one can use
Litmus paper
“Red” paper turns blue above ~pH = 8
“Blue” paper turns red below ~pH = 5
Or an indicator.
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23. How Do We Measure pH?
For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.
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24. Strong Acids
You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.
These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.
For the monoprotic strong acids,
[H3O+] = [acid].
Molarity of strong acid solution
pH = -log [acid]
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25. Strong Bases
Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).
Again, these substances dissociate completely in aqueous solution.
For the monoprotic strong bases,
[OH-] = [base].
pOH = -log [base]
Molarity of strong base solution
pH = pKw - pOH
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26. Dissociation Constants
For a generalized acid dissociation,
the equilibrium expression would be
This equilibrium constant is called the acid-dissociation constant, Ka.
HA (aq) + H2O (l)
A- (aq) + H3O+ (aq)
[H3O+] [A-]
[HA]
Ka =
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27. Dissociation Constants
The greater the value of Ka, the stronger is the acid.
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28. Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is Calculate Ka for formic acid at this temperature.
We know that
[H3O+] [COO-]
[HCOOH]
Ka =
HCOOH (aq) + H2O (l)
HCOO- (aq) + H3O+ (aq)
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29. Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is Calculate Ka for formic acid at this temperature.
To calculate Ka, we need the equilibrium concentrations of all three things.
We can find [H3O+], which is the same as [HCOO-], from the pH.
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30. Calculating Ka from the pH
pH = -log [H3O+]
2.38 = -log [H3O+]
-2.38 = log [H3O+]
= 10log [H3O+] = [H3O+]
4.2  10-3 = [H3O+] = [HCOO-]
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31. Calculating Ka from pH HCOOH (aq) + H2O (l) HCOO- (aq) + H3O+ (aq)
Now we can set up a table…
0.10 – 4.2 X 10-3 M ≈ 0.10 M
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32. Calculating Ka from pH [4.2  10-3] [4.2  10-3] Ka = [0.10]
= 1.8  10-4
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33. Calculating Percent Ionization
[H3O+]eq
[HA]initial
Percent Ionization =  100
In this example
[H3O+]eq = 4.2  10-3 M
[HCOOH]initial = 0.10 M
4.2  10-3
0.10
Percent Ionization =  100
= 4.2%
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34. Calculating Percent Ionization
If the percent Ionization < 5%
We can make the simplifying approximation
that x, the amount of acid that dissociates, is small compared with the initial concentration
of acid; so [HA] eq ≈ [HA]initial – x ≈ [HA]initial
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35. CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO - (aq)
Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic acid, CH3COOH, at 25C.
CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO - (aq)
Ka for acetic acid at 25C is 1.8  10-5.
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36. Calculating pH from Ka The equilibrium constant expression is
[H3O+] [CH3COO-]
[CH3COOH]
Ka =
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37. Calculating pH from Ka We next set up a table…
[CH3COOH], M
[H3O+], M
[CH3COO-], M
Initially
0.30
Change -x +x
At Equilibrium
x  0.30 x
We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.
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38. Calculating pH from Ka (x)2 (0.30) 1.8  10-5 =
Now,
(x)2
(0.30)
1.8  10-5 =
(1.8  10-5) (0.30) = x2
5.4  10-6 = x2
2.3  10-3 = x
Our approximation (2.3  10-3/0.3)*100 = 0.76 % < 5 % is appropriate
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39. Calculating pH from Ka pH = -log [H3O+] pH = -log (2.3  10-3)
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40. Polyprotic Acids… …have more than one acidic proton
If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.
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41. Bases react with water to produce hydroxide ion.
Weak Bases
Bases react with water to produce hydroxide ion.
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42. Weak Bases The equilibrium constant expression for this reaction is
[HB] [OH-]
[B-]
Kb =
where Kb is the base-dissociation constant.
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43. Kb can be used to find [OH-] and, through it, pH.
Weak Bases
Kb can be used to find [OH-] and, through it, pH.
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44. pH of Basic Solutions What is the pH of a 0.15 M solution of NH3?
NH3 (aq) + H2O (l)
NH4+ (aq) + OH- (aq)
[NH4+] [OH-]
[NH3]
Kb =
= 1.8  10-5
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45. pH of Basic Solutions Tabulate the data. [NH3], M [NH4+], M [OH-], M
Initially
0.15
At Equilibrium
x  0.15 x
Approximation :
0.15- x ≈ 0.15
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46. pH of Basic Solutions (x)2 (0.15) 1.8  10-5 =
Our approximation : (x/0.15 ) x 100% = 1.06 % < 5% , is appropriate
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47. pH of Basic Solutions Therefore, [OH-] = 1.6  10-3 M
pOH = -log (1.6  10-3)
pOH = 2.80
pH =
pH = 11.20
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48. Ka and Kb Ka and Kb are related in this way: Ka  Kb = Kw
pKa +pKb = pKw
Therefore, if you know one of them, you can calculate the other.
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49. Reactions of Anions with Water
Anions are bases.
As such, they can react with water in a hydrolysis reaction to form OH- and the conjugate acid:
X- (aq) + H2O (l)
HX (aq) + OH- (aq)
Example : Hypochlorite anion
ClO- (aq) + H2O (l)
HClO (aq) + OH- (aq)
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50. Reactions of Cations with Water
Cations with acidic protons (like NH4+) will lower the pH of a solution.
Most metal cations that are hydrated in solution also lower the pH of the solution.
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51. Reactions of Cations with Water
Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water.
This makes the O-H bond more polar and the water more acidic.
Greater charge and smaller size make a cation more acidic.
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52. Effect of Cations and Anions
Six summary points
An anion that is the conjugate base of a strong acid will not affect the pH (Cl-, NO3-,ClO4- …). These anions are called ‘’spectator’’ anions
An anion that is the conjugate base of a weak acid will increase the pH (CH3COO-, ClO-,...).
A cation that is the conjugate acid of a weak base will decrease the pH (NH4+, CH3NH3+ ...)
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53. Effect of Cations and Anions
Cations of the strong Arrhenius bases will not affect the pH: the alkali metal of the group IA (Na, K, …) and heavier alkaline earth metal of the group IIA Mg, Ca, Ba, …) . These cations are called ‘’spectator’’ cations
Other metal ions will cause a decrease in pH.
When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the Ka and Kb values.
Sample 8-18, 8-19
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54. Actual Yield Theoretical Yield Percent Yield = x 100
One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield).
Actual Yield
Theoretical Yield
Percent Yield = x 100
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55. The Common-Ion Effect Consider a solution of acetic acid:
If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO−(aq)
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56. The Common-Ion Effect
“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”
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57. The Common-Ion Effect
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
Ka for HF is 6.8  10−4.
[H3O+] [F−]
[HF]
Ka =
= 6.8  10-4
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58. The Common-Ion Effect HF(aq) + H2O(l) H3O+(aq) + F−(aq)
Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.
[HF], M
[H3O+], M
[F−], M
Initially
0.20
0.10
Change −x +x
At Equilibrium
0.20 − x  0.20
x  0.10 x
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59. The Common-Ion Effect (0.10) (x) 6.8  10−4 = (0.20)
(0.20) (6.8  10−4)
(0.10)
= x
1.4  10−3 = x
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60. The Common-Ion Effect Therefore, [F−] = x = 1.4  10−3
[H3O+] = x =  10−3 = 0.10 M
So, pH = −log (0.10)
pH = 1.00
Sample 8.20, 8.21
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61. Buffers Buffers are solutions of a weak conjugate acid-base pair.
They are particularly resistant to pH changes, even when strong acid or base is added.
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62. Buffers
If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.
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63. Buffers
Similarly, if acid is added, the F− reacts with it to form HF and water.
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64. Buffer Calculations
Consider the equilibrium constant expression for the dissociation of a generic acid, HA:
HA + H2O
H3O+ + A−
[H3O+] [A−]
[HA]
Ka =
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65. Buffer Calculations Rearranging slightly, this becomes
[HA]
Ka =
[H3O+]
Taking the negative log of both side, we get
base
[A−]
[HA]
−log Ka =
−log [H3O+] + −log
pKa pH
acid
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66. Buffer Calculations So pKa = pH − log [base] [acid]
Rearranging, this becomes
pH = pKa + log
[base]
[acid]
This is the Henderson–Hasselbalch equation.
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67. Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  10−4.
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68. Henderson–Hasselbalch Equation
pH = pKa + log
[base]
[acid]
pH = −log (1.4  10−4) + log
(0.10)
(0.12)
pH = (−0.08) pH pH
pH = 3.77
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69. pH Range
The pH range is the range of pH values over which a buffer system works effectively.
It is best to choose an acid with a pKa close to the desired pH.
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70. When Strong Acids or Bases Are Added to a Buffer…
…it is safe to assume that all of the strong acid or base is consumed in the reaction.
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71. Addition of Strong Acid or Base to a Buffer
Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.
Use the Henderson–Hasselbalch equation to determine the new pH of the solution.
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72. Calculating pH Changes in Buffers
A buffer is made by adding mol CH3COOH and mol CH3COONa to enough water to make 1.00 L of solution. The pH of the buffer is Calculate the pH of this solution after mol of NaOH is added.
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73. Calculating pH Changes in Buffers
Before the reaction, since
mol CH3COOH = mol CH3COO−
pH = pKa = −log (1.8  10−5) = 4.74
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74. Calculating pH Changes in Buffers
The mol NaOH will react with mol of the acetic acid:
CH3COOH(aq) + OH−(aq)  CH3COO−(aq) + H2O(l)
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75. Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to calculate the new pH:
pH = log
(0.320)
(0.280) pH
pH = 4.80
pH =
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76. Solubility Products
Consider the equilibrium that exists in a saturated solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42−(aq)
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77. Solubility Products
The equilibrium constant expression for this equilibrium is
Ksp = [Ba2+] [SO42−]
where the equilibrium constant, Ksp, is called the solubility product.
Sample 8.26
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78. Solubility Products Ksp is not the same as solubility.
Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).
Sample 8.27
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79. Will a Precipitate Form?
In a solution,
If Q = Ksp, the system is at equilibrium and the solution is saturated.
If Q < Ksp, more solid can dissolve until Q = Ksp.
If Q > Ksp, the salt will precipitate until Q = Ksp.
Sample 8-31, 8-32
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80. Factors Affecting Solubility
The Common-Ion Effect
If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.
BaSO4(s)
Ba2+(aq) + SO42−(aq)
Sample 8-28
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81. Factors Affecting SolubilitypH
If a substance has a basic anion, it will be more soluble in an acidic solution.
Substances with acidic cations are more soluble in basic solutions.
Sample 8-29
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82. Factors Affecting Solubility
Complex Ions
Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.
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83. Factors Affecting Solubility
Complex Ions
The formation of these complex ions increases the solubility of these salts.
Sample 8-30
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84. Factors Affecting Solubility
Amphoterism
Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.
Examples of such cations are Al3+, Zn2+, and Sn2+.
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85. Selective Precipitation of Ions
One can use differences in solubilities of salts to separate ions in a mixture.
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