reaksi dalam larutan berair
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Chapter 5*Reaksi Dalam Larutan Berair
Chapter 5
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Chapter 5*
Beberapa senyawa ketika dilarutkan dalam air dapat menghantarkan arus listrik larutan elektrolit
Namun ada juga senyawa yang ketika dilarutkan dalam air tidak menghantarkan arus listrik
Senyawa di dalam AirNonelectrolyteWeak electrolyteStrong electrolyte
Chapter 5
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Chapter 5*Ionic Compounds in Water (Electrolytes)
The conductivity of the solution is due to the formation of ions when the compound dissolves in water
These ions are not the result of a chemical reaction, they are the result of a dissociation of the molecule into ions that compose the solid.Compounds in Aqueous Solution
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Chapter 5*Ionic Compounds in WaterCompounds in Aqueous Solution
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Chapter 5*Molecular Compounds in Water(Nonelectrolytes)
In this case no ions are formed, the molecules just disperse throughout the solvent.
Compounds in Aqueous Solution
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Chapter 5*Strong and Weak ElectrolytesStrong electrolytes A substance which completely ionizes in water.
For example:
Compounds in Aqueous Solution
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Chapter 5*Strong and Weak Electrolytes
Weak electrolyte: A substance which partially ionizes when dissolved in water.
For example:Compounds in Aqueous Solution
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Chapter 5*Strong and Weak ElectrolytesCompounds in Aqueous SolutionNotice that the arrow in this reaction has two heads, this indicates that two opposing reactions are occurring simultaneously.
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Chapter 5*Strong and Weak ElectrolytesCompounds in Aqueous SolutionSince both reactions occur at the same time, this is called a chemical equilibrium.
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Chapter 5*Precipitation Reaction
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Chapter 5*A reaction which forms a solid (precipitate)
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
AgCl is classified as an insoluble substance
Precipitation Reaction
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Chapter 5*Net Ionic Equation
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
AgNO3 and NaNO3 are electrolytes in solution so they actually occur as free ions.
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) AgCl(s) + Na+(aq) + NO3-(aq)
Precipitation Reaction
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Chapter 5*Net Ionic Equation
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) AgCl(s) + Na+(aq) + NO3-(aq)
Notice that NO3-(aq) and Na+(aq) occur in both the left and right side of the equation.These are called spectator ions.
Precipitation Reaction
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Chapter 5*Net Ionic Equation
Ag+(aq) + Cl-(aq) AgCl(s)
With the spectator ions removed, the resulting equation shows only the ions involved in the reaction remain.This is a net ionic equation.
Precipitation Reaction
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Chapter 5*Solubility Guidelines for Ionic CompoundsCompounds in Aqueous Solution
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Chapter 5*Solubility Guidelines for Ionic Compounds1. Predict the solubility of the following compounds:PbSO4AgCH3CO2
(NH4)3PO4
KClO4Compounds in Aqueous Solution
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Chapter 5*Solubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:PbSO4InsolubleAgCH3CO2(NH4)3PO4KClO4Compounds in Aqueous Solution
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Chapter 5*Solubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:PbSO4InsolubleAgCH3CO2Soluble
(NH4)3PO4KClO4Compounds in Aqueous Solution
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Chapter 5*Solubility Guidelines for Ionic CompoundsPredict the solubility of the following compounds:PbSO4InsolubleAgCH3CO2Soluble
(NH4)3PO4Soluble
KClO4Compounds in Aqueous Solution
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Chapter 5*Solubility Guidelines for Ionic Compounds1. Predict the solubility of the following compounds:PbSO4InsolubleAgCH3CO2Soluble
(NH4)3PO4Soluble
KClO4SolubleCompounds in Aqueous Solution2. Predict the precipitate produced by mixing an Al(NO3)3 solution with NaOH solution. Write the net ionic equation for the reaction.
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An undesireable precipitation reactionExplain the precipitation reaction mechanism and how to remove scale and explain the reaction mechanism to remove scale Chapter 5*
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Chapter 5*Acids and Bases
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Chapter 5*Acid - substance which ionizes to form hydrogen cations (H+) in solution
Examples: Hydrochloric AcidHCl Nitric AcidHNO3Acetic AcidCH3CO2HSulfuric AcidH2SO4Sulfuric acid can provide two H+s - Diprotic acid, The other acids can provide only one H+ - Monoprotic acid.Acids and Bases
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Chapter 5*Diprotic acid H2SO4 H+ + HSO4-
HSO4- H+ + SO42-Acids and BasesH2SO4 merupakan asam kuat atau elektrolit kuat (tahap ionisasi pertama berlangsung sempurna), tetapi HSO4- adalah asam lemah atau elektrolit lemah (diperlukan panah dua arah untuk menunjukkan ionisasi tak sempurna)
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Chapter 5*Triprotic acid H3PO4 H+ + H2PO4-H2PO4- H+ + HPO42-HPO42- H+ + PO43-Acids and BasesKeberadaan asam triprotik relatif sedikit, sebagai contohnya adalah asam fosfat. H3PO4 , H2PO4- , dan HPO42- merupakan asam lemah (dibutuhkan panah dua arah untuk menunjukkan tahap-tahap ionisasi)
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Chapter 5*Base - substance which ionizes to form OH- ions in solution substance which reacts with H+ ions.
Examples: ammoniaNH3sodium hydroxideNaOHAcids and Bases
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Chapter 5*Acid-Base Reaction
H+ + OH- H2O
It is clear that the metal hydroxides (NaOH for example) provide OH- by disassociation.Bases like ammonia make OH- by reacting with water (ionization)NH3 + H2O NH4+ + OH-
Acids and Bases
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Chapter 5*Strong and Weak Acids and Bases
Strong acids and bases are strong electrolytes.
Weak acids and bases are weak electrolytes.Acids and Bases
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Chapter 5*Strong Acids
The strength of acids and bases are concerned with the ionization (or dissociation) of the substance, not its chemical reactivity.
Example:Hydrofluoric acid (HF) is a weak acid, but it is very chemically reactive.this substance cant be stored in glass bottles because it reacts with glass (silicon dioxide).Acids and Bases
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Chapter 5*Common Strong Acids and Bases
Common Strong AcidsHydrochloric AcidHClHydrobromic AcidHBrHydroiodic acidHINitric AcidHNO3Perchloric AcidHClO4Sulfuric AcidH2SO4Acids and Bases
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Chapter 5*Common Strong Acids and Bases
Common Strong BasesLithium HydroxideLiOHSodium HydroxideNaOHPotassium HydroxideKOH
Acids and Bases
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Chapter 5*Neutralization Reaction Reaction between an acid and a base.
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
The neutralization between acid and metal hydroxide produces water and a salt
Salt an ionic compound whose cation comes from a base and anion from an acid.
Acids, Bases, and Salts
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Chapter 5*Neutralization Reaction
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
despite the appearance of the equation, the reaction actually takes place between the ions.
Total Ionic EquationH+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) H2O(l) + Na+(aq) + Cl-(aq)
Acids, Bases, and Salts
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Chapter 5*Neutralization Reaction
Total Ionic EquationH+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) H2O(l) + Na+(aq) + Cl-(aq)
Net Ionic EquationH+(aq) + OH-(aq) H2O(l)Acids, Bases, and Salts
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Chapter 5*Metal Carbonates and AcidGas-Forming Reactions2 HCl(aq) + Na2CO3(aq) 2 NaCl(aq) + H2O(l) + CO2(g)
Metal carbonates (or bicarbonates) always form a salt, water and carbon dioxide
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Chapter 5*Metal Sulfide and AcidGas-Forming Reactions2 HCl(aq) + Na2S(s) H2S(g) + 2 NaCl(aq)
Metal sulfides form a salt and hydrogen sulfide.
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Chapter 5*Metal Sulfite and AcidGas-Forming Reactions2 HCl(aq) + Na2SO3(s) SO2(g) + 2 NaCl(aq) + H2O(l)
Metal sulfites form a salt, sulfur dioxide and water.
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Chapter 5*Ammonium Salt and Strong BaseGas-Forming ReactionsNH4Cl(s) + NaOH(aq) NH3(g) + NaCl(aq) + H2O(l)
This reaction forms ammonia, salt and water
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Chapter 5*Oxidation-Reduction ReactionsReaction where electrons are exchanged.2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)
Na(s) Na+(aq) + 1 e-oxidation loss of electrons
2 H+(g) + 2 e- H2(g)reduction gain of electrons
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Chapter 5*Oxidation-Reduction ReactionsReaction where electrons are exchanged.2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)
An alternate approach is to describe how one reagent effects another.Reducing Agent, a substance that causes another substance to be reduced. Na(s) Na+(aq) + 1 e-Oxidizing Agent, a substance that causes another substnace to be oxidized. 2 H+(g) + 2 e- H2(g)
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Chapter 5*Oxidation-Reduction ReactionsOxidation NumbersEach atom of a pure element has an oxidation number of zero(0).For monatomic ions, the oxidation number equals the charge on the ion.Fluorine always has an oxidation state of -1 in compounds.Cl, Br, and I always have oxidation numbers of -1, except when combined with oxygen or fluorine.The oxidation number of H is +1 (exception in binary compounds of metal-hidrogen, LiH, NaH, CaH2 -1) and O is -2 in most compounds (exception in H2O2 and O22- -1).The sum of the oxidation numbers must equal the charge on the molecule or ion.
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Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers
Examples
PCl5P 1( ) =Cl 5( ) = ________ 0
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Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers
Example
PCl5P 1( ? ) = ?Cl 5(-1) = __-5____ 0
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Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers
Example
PCl5P 1(+5) = +5Cl 5(-1) = __-5____ 0
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Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers
Example
CO32-C 1( ) = O 3( ) = _____ -2
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Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers
Example
CO32-C 1(?) = ?O 3(-2) = __-6__ -2
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Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers
Example
CO32-C 1(+4) = +4O 3(-2) = __-6__ -2
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Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers
Example
K2CrO4K 2( ) =O 4( ) =Cr 1( ) = ________ 0
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Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers
Example
K2CrO4K 2(+1) = +2O 4(-2) = -8Cr 1( ? ) = ___?____ 0
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Chapter 5*Oxidation-Reduction ReactionsOxidation Numbers
Example
K2CrO4K 2(+1) = +2O 4(-2) = -8Cr 1(+6) = ___+6__ 0
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Chapter 5*Oxidation-Reduction Reaction Types
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Chapter 5*Oxidation-Reduction Reactions Types
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Chapter 5*Oxidation-Reduction Reactions TypesHydrogen displacementMetal displacementHalogen displacement
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Chapter 5*Oxidation-Reduction Reactions Types
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Chapter 5*Oxidation-Reduction Reactions Types
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Chapter 5*Oxidation-Reduction Reactions Types
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Chapter 5*
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Reaksi dalam Larutan BerairAgNO3NaCl
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Reaksi dalam larutanAgNO3(aq) + NaCl (aq) AgCl (aq) + NaNO3 (s) (mengendap)
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Solvent/zat pelarut: komponen yang keadaan fisiknya tidak berubah selama larutan dibentuk Solute/zat terlarut: komponen lain yang dilarutkan dalam solventKonsentrasi molar: banyaknya mol tiap satuan volume dari larutanTerminologi yg digunakan dlm larutan
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Contoh:Ingin dibuat larutan NaCl dengan konsentrasi 1M dalam 1L larutan. Berapa gram NaCl (murni) yang diperlukan untuk membuat larutan ini?Konsentrasi larutan (Molaritas)
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Chapter 5*Membuat Larutan
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Chapter 5*Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
Solutions
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Chapter 5*Example: What is the molarity(M) of a solution which contains 17.51 g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
Solutions
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Chapter 5*Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
Solution volume
Solutions
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Chapter 5*Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
Solution volume
Solutions
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Chapter 5*Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?Solutions
Chapter 5
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Chapter 5*Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
Solutions
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Chapter 5*Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
Solutions
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SolutionsChapter 5*
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Molaritas dari densitas dan persen beratJika suatu senyawa A yang terlarut membentuk suatu larutan yang memiliki kerapatan a g/l dengan persen berat larutan w%. Maka konsentrasi dapat dinyatakan sebagai:
CA = A %w/ MR
Di mana:CA: konsentrasi A dalam suatu larutan tiap satuan volume (misal mol/liter) atau MA: kerapatan atau massa jenis zat A (misal g/l)MR: berat molekul zat A%w: persen berat larutan
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Chapter 5*Dilution
Solutions
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Chapter 5*DilutionExample: What volume of 6.00M NaOH is required to make 500mL of 0.100M NaOH?
Mconcentrated = 6.00MMdilute = 0.100MVconcentrated = ?Vdilute = 500mL
0.100M(500mL) = 6.00M(Vconcentrated)
Vconcentrated = 8.33mLSolutions
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Chapter 5*
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Chapter 5*pH ScaleConcentration scale for acids and bases.The square brackets around the H+ indicate that the concentration of H+ is in molarity.So, a change of 1 pH unit indicates a 10X change in H+ concentration.
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Chapter 5*Solution StoichiometryWe can now use molarity to determine stoichiometric quantities.
ExampleHow many grams of hydrogen gas are produced when 20.0 mL of 1.75M HCl is allowed to react with 15.0g of sodium metal?
2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
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Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
Convert quantities to moles
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Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
Convert quantities to moles
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Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
Convert quantities to moles
Determine limiting reagent
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Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
Convert quantities to moles
Determine limiting reagent
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Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
Calculate moles of H2
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Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
Calculate moles of H2
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Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
Calculate moles of H2
Calculate grams of H2
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Chapter 5*Solution Stoichiometry2 HCl(aq) + 2 Na(s) H2(g) + 2 NaCl(aq)
Calculate moles of H2
Calculate grams of H2
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Chapter 5*Solution StoichiometryGravimetric Analysis
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Chapter 5*Solution StoichiometryGravimetric Analysis
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Chapter 5*Solution StoichiometryAcid Base Titrations
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Chapter 5*Solution StoichiometryAcid Base Titrations
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Chapter 5*Solution StoichiometryAcid Base Titrations
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Chapter 5*4,10, 16, 20, 24, 30, 36, 38, 44, 60, 62, 68Practice Problems
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