Soluble Salts in Water
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Transcript of Soluble Salts in Water
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Soluble Salts in Water
• KI and K2CrO4:– Potassium iodide and
potassium chromate are water-soluble.
• AgI and PbCrO4 – Silver iodide and lead
chromate and are sparingly soluble: Ksp(AgI)=8.5 X 10-17
Ksp(PbCrO4)=1.8 X 10-14
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Common Ion Effect
• Silver iodideAgI(s) Ag+ + I-
Ksp=[Ag+][I-] = 8.5 X 10-17
[Ag+] = [I-] = solubility =
• What if the solution were 0.1M in Ag+ ions?• What if the solution were 0.1M in I- ions?
8.5x10−17
=9.2x10−9
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Common Ion Effect
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Common Ion Effect
• Lead chromatePbCrO4(s) Pb2+ + CrO4
2-
Ksp=[Pb2+][CrO42-] =1.8 X 10-14
Let x = solubility of [Pb2+] = [CrO42-]
• What if [Pb2+] = 0.1M?
• What if [CrO42-] = 0.1M?
• What if [Pb2+]=[CrO42-] = 0.1M?
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Common Ion Effect
• Calcium fluorideCaF2(s) Ca2+(aq) + 2F-(aq)
Ksp=[Ca2+][2F-]2 = 1.7 X 10-10
Solubility = x = [Ca2+] = 5.6 X 10-4
Ksp=[Ca2+] [2F-]2 = (x)(2x)2 = 4x3= 1.7 X 10-10
Ksp= 4.25x10−113 =3.5x10−4
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Selective Precipitation
• A solution where [Ba2+] = [Ca2+] = 0.1M – Add sulfate ions:
• Ksp(BaSO4) = 10-10; Ksp(CaSO4) = 10-5
• Sparingly soluble BaSO4 and CaSO4 precipitate.
• Which salt precipitates first?
• How much of the first is left in solution when the second begins to precipitate?
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Dissolving Precipitates
• AgCl– Add ammonia
• Forms soluble Ag(NH3)2+ complex
• AgBr– Add lots of ammonia
• Forms soluble Ag(NH3)2+ complex
• AgI– Add cyanide ion
• Forms soluble Ag(CN2)- complex
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Complex Ion Formation
–AgI(s) Ag+(aq) + I-(aq) Ksp= [Ag+][I-] = 10-16
– Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
– AgI(s) + 2NH3(aq) Ag(NH3)2+(aq) + I-(aq)
KspK f =[Ag(NH3)2
+][I −][NH3]
2 ≅(1012)(10−16) =10−4
K f =[Ag(NH3)2
+][NH3]
2[Ag+]≅1012
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Complex ion formation demos
• AgCl ppt
• AgBr ppt
• AgI ppt
• Dissolve in ammonia and cyanide
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Electrolytes
• Conductivity in water is due to hydrated ions moving about.– Water and aqueous sugar
solutions
– Acetic acid and vinegar
– Aqueous salt solutions
• Experiments– Electrolytes
– Nonelectrolytes
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Autoionization of Water
• H2O(liq) + H2O(liq) H3O+ + OH-
K =[H3O
+][OH−][H2O]2
KW =K[H2O]2 =[H3O+][OH−]=1.0x10−14
[H3O+]=[OH−]= KW = 1.0x10−14 =1.0x10−7M
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Autoionization of Water
• H2O(liq) + H2O(liq) H3O+ + OH-
K =[H3O
+][OH−][H2O]2
KW =K[H2O]2 =[H3O+][OH−]=1.0x10−14
[H3O+]=[OH−]=xmolL−1
KW =[H3O+][OH−]=(x)(x)=x2 =1.0x10−14
x= KW = 1.0x10−14 =1.0x10−7M
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Acid-Base Theories
• Arrhenius HCl H+ + Cl-
• Could not explain alkalinity of aqueous ammonia
• Brönsted NH3 + H2O NH4
+ + OH- HCl + H2O H3O+ + Cl-
• Lewis NH3 + H+ NH4
+
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pH ScaleMeasures acidity of Aqueous
Solutions• pH = -log[H3O+]
by definition
• pKw = pH + pOHbecause…. Kw=[H3O+][OH-]
• For neutral solutions…[H3O+] = [OH-] = 10-7M and
pH = 7
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pH of Aqueous Solutions
• If [H3O+] = 5.25 x 10-3M– Then pH = -log 5.25 x 10-3 = 2.28
• If pH = 5.25– Then [H3O+] = 10-5.25 = 5.6 x 10-6M
• And [OH-] = Kw/[H3O+] = 1.78 x 10-9M
• So pOH = 8.75• And pH = 14 - 8.75 = 5.25
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pH of Aqueous Solutions
• When pH is low……• [H3O+] > [OH-]
• When pH is neutral…..• [H3O+] = [OH-]
• When pH is high…..• [H3O+] < [OH-]
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Dissociation of HA in Water
HA + H2O H3O+ + A-
• For an acid of the general form HA
KHA =[H3O
+][A−][HA]
KHA
[HA][A−]
=[H3O+]
pKHA +log[A−][HA]
=pH
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IndicatorsHInd + H2O H3O+ + Ind-
• Indicator equation
KHInd =[H3O
+][Ind−][HInd]
KHInd
[HInd][Ind−]
=[H3O+]
pKHInd+log[Ind−][HInd]
=pH
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Oxides and Hydrides
• Metallic oxidesNa2O + H2O 2NaOHCaO + H2O Ca(OH)2
• Nonmetallic oxidesSO2 + H2O H2SO3
CO2 + H2O H2CO3
• Metal HydridesNaH + H2O NaOH + H2
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For Strong Acids
• Aqueous hydrochloric acid solutionsHCl + H2O H3O+ +
Cl-
• 0.1M HCl = 0.1M H3O+ – Presumption is complete dissociation– pH = 1
• Add 10 mL to 990 mL of H2O– pH change is… huge!
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For Weak Acids
• Aqueous acetic acid solutionsHOAc + H2O H3O+ + OAc-
ka=1.75 X 10-5 0.1M HOAc << 0.1M H3O+ Incomplete dissociation: pH = 2.87
• Aqueous HF HF + H2O H3O+ + F- ka=7.2 X 10-4 and pH = ?
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For Very Weak Acids
• Aqueous hydrogen cyanide solutions HCN + H2O H3O+ + CN-
• 0.1M HCN <<< 0.1M H3O+ • Ka = 4.0 X 10-10 and pH = 5.2
• 0.1M NaCN pH = ?• 0.1M NaF pH = ?• 0.1M NaOAc pH = ?
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0.10M Solutions of HA: Ka pKa pH
• HCl ˜107 -7 1HOAc ˜10-5 5
3 HCN ˜10-11 11 6 H2O ˜10-14
14 7 NH3 ˜10-
16 16 8
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IndicatorsHInd + H2O H3O+ + Ind-
• Indicator equation
KHInd =[H3O
+][Ind−][HInd]
KHInd
[HInd][Ind−]
=[H3O+]
pKHInd+log[Ind−][HInd]
=pH
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Solvolysis/Hydrolysis
• For a weak acid HA…..– the anion is strong and reacts with the
solvent in a proton-transfer reaction:A- + H2O HA +
OH-
• For a weak base B…..– the cation ion is strong and reacts with the
solvent in a proton-transfer reaction:BH+ + H2O B + H3O+
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Hydrolysis
• Alkaline solutions: NaOAc, NaF, NaCN
• Acidic solutions: NH4Cl
• Neutral solutions from NH4OAc
• NaCN and Na2CO3 solutions?
• AlCl3 and Fe(NO3)3 solutions?
• NH4CN solutions?
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pH of 0.10M Aqueous NaOAc
• NaOAc(s) Na+(aq) + OAc-(aq)
• OAc- + H2O HOAc + OH-
• pOH = 5 and pH =14 - 5 = 9
Kb =Kw
Ka
=[H3O
+][OH−][H3O
+][OAc−][HOAc]
=[HOAc][OH−]
[OAc−]=
10−14
10−5 =10−9
10−9 =(x)(x)(.1−x)
≅x2
.1and x=[OH−] =10−5M
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Buffers
• Solutions of a weak acid and its conjugate base:
HA + H2O H3O+ + A-
• Solutions of a weak base and its conjugate acid:
B + H2O BH+ + OH-
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Buffers
KHA =[H3O
+][A−][HA]
KHA
[HA][A−]
=[H3O+]
pKHA +log[A−][HA]
=pH
• Buffers act across an [A-]/[HA] range of 10/1 to 1/10 … or 2 pH units:
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Buffers
• Aqueous acetic acid solutionsHOAc + H2O H3O+ + OAc-
• For 0.1M HOAc: pH = 2.87• For HOAc/NaOAc buffer: pH = pKa = 4.76• PROBLEM: Add 50mL 0.10M HCl to 950 mL
H2O containing 0.050 mole HOAc and 0.050 mole NaOAc.
– Buffer soaks up the acid – pH change is relatively small.
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Polyprotic Acids
• Carbonic acid (H2CO3) Ka
H2CO3 + H2O H3O+ + HCO3-
10-7 HCO3- + H2O H3O+ +
CO32- 10-11
– Acidity is essentially supplied by Ka(1)
– Blood is buffered by [CO2]/[HCO3-]
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Polyprotic Acids
– Sulfuric acid (H2SO4) pKa H2SO4
+ H2O H3O+ + HSO4- -5 HSO4
- + H2O
H3O+ + SO42- 2
– Phosphoric acid (H3PO4) pKa H3PO4
+ H2O H3O+ + H2PO4- 2.1 H2PO4
- +
H2O H3O+ + HPO42- 7.2 HPO4
- +
H2O H3O+ + PO43- 12.3
• Note:K values are lower by several orders of magnitude due to increasing negative charge.
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Titration curves
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Titration curves
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Titration curves
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Finding Ka for Weak Acids
KHA =[H3O
+][A−][HA]
KHA
[HA][A−]
=[H3O+]
pKHA +log[A−][HA]
=pH
– For a monoprotic acid, at the mid-point, [HA] = [A-]
pH = pKa
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ACID/BASE ISSUES
• Acid rain and coal mine run-off:– Kentucky, West Virginia, Indiana, Illinois,
Minnesota… many places.
• Industrial acid/base run-off in rivers:– Raritan River is one of many.
• Acid rain and steel manufacture Acid:– Adirondks and Great Smokys– Illinois and Indiana mills
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ACID/BASE ISSUES
• Corrosion– N and S oxides– Acidic oxides
• Particulates in the air:– Donora, Pennsylvania– Dupont “nylon” in downtown Chicago
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Lessons from History
The disadvantage of men not knowing the past is that they do not know the present.
- G. K. Chesterton
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