Ch' 4: Solution Stoichiometry Types of Chemical Reactions

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Ch. 4 Solution Stoichiometry Types of Chemical
Reactions

• Water is a polar molecule the bond between the
  oxygen and the hydrogen in water is covalent but
  unlike standard covalent molecules, oxygen has a
  greater attraction for electrons, thus the oxygen
  becomes slightly negative and the hydrogen
  becomes slightly positive because of this
  unequal charge, water is a polar molecule
• Solubility because water is a polar molecule, it
  dissolves ionic substances most readily and
  covalent substances to a lesser degree because of
  their non-polar characteristics.
• Remember Like Dissolves Like.

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Strong vs Weak?

• Strong electrolytes (Acids or Bases) are those
  that dissociate 100
• Weak Electrolytes (Acids or Bases) are those that
  dissociate only partially or not at all.
• Remember Sulfuric Acid was Strong but Acetic
  Acid was weak and Sodium Hydroxide was strong
  while Ammonia was weak

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Concentration amounts of chemicals in a solution

• Molarity moles of solute per liter of solution
• M moles of solute/liter of solution
• .2M solution means that there are
• .2moles of solute/liter solution
• Calculate the molarity of a solution prepared by
  dissolving 11.5g of solid NaOH in enough water to
  make 1.5 L of solution.
• 1st convert grams to moles
• 11.5g NaOH x 1 mol NaOH 0.288 mol NaOH
• 40.00 g NaOH

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Concentration continued

• 2nd calculate molarity by dividing the moles by
  the volume of the solution.
• Molarity mol solute/L solution
• 0.288 mol NaOH/ 1.50 L solution
• 0.192 M NaOH
• Remember the volume must be in Liters

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Dilution when water is added to a solution to
produce the molarity desired

• Key Point Moles of Solute after dilution Moles
  of solute before dilution
• Example Suppose we need to prepare 500ml of 1.00
  M acetic acid from a 17.4 stock solution. What
  volume of stock solution is required?
• 1st Convert 500ml to moles
• 500ml x 1 L solution x 1.00 mol HC2H3O2 .500 mol
• 1000 ml 1 L Solution HC2H3O2

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Continued

• 2nd Since we need to know the volume that
  contains .500 moles we calculate
• V x M moles V x 17.4 mol HC2H3O2 .500 mol
• 1 L solution HC2H3O2
• Solve for V 0.0287 L or 28.7 ml solution
• Thus we can take 28.7 ml of 17.4 M HC2H3O2 and
  dilute it to a total volume of 500 ml to form a
  1M solution of acetic acid

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Types of Chemical Reactions

• Precipitation reactions what solid will be
  formed when two solutions are mixed?
• Must dissociate the ions to find what can be
  formed
• Ex Solutions of Potassium Chromate is added to
  Barium Nitrate The ions present in solution are
  K, CrO4-2, Ba2, NO3-

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Continued Net Ionic equations

• K CrO4-2 Ba2NO3- ? products
• What products can be formed? Remember that
  cations bond with anions
• So we get products of potassium nitrate and
  barium chromate
• K NO3- Ba2 CrO4-2
• But which one forms that precipitate?

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Net ionics

• Not all ionic compounds will dissolve in water
  (or other polar solvents). Review the solubility
  rules for ionic compounds dissolving in water.
• Any ionic compound whose cation is Group IA or
  ammonium is soluble, no matter what the anion is.
• All nitrates, chlorates, perchlorates, and
  acetates are soluble in water, no matter what
  their cation is.
• All chlorides, bromides and iodides are soluble
  in water except silver, lead (II) and mercury
  (I)
• All sulfates are soluble in water except silver,
  lead (II) and mercury (I), calcium, barium and
  strontium
• All phosphates, phosphites, sulfites, chromates,
  dichromates, fluorides, permanganates, oxalates,
  oxides, nitrides, sulfides, cyanides, carbonates
  and hydroxides are INSOLUBLE in water unless
  their cation is Group IA Group IIA are only
  slightly soluble hydroxides

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What makes an insoluble precipitate?

• Based on these rules the Potassium nitrate is
  soluble, why?
• Rule 1 and rule 2
• The Barium Chromate is insoluble, why?
• Rule 5
• So we write the net ionic equation as
• K CrO4-2 Ba2NO3- ? K NO3- BaCrO4 (s)
• Crossing out the like ions we get
• CrO4-2 Ba2 ? BaCrO4 (s)

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Molecular Equations Complete Ionic Equations
Net Ionic Equations

• Aqueous potassium chloride is added to aqueous
  silver nitrate to form a silver chloride
  precipitate and aqueous potassium nitrate
• Molecular Equation
• KCl (aq) AgNO3 (aq) ? AgCl (s) KNO3(aq)
• Complete Ionic Equation
• K Cl- Ag NO3 ? AgCl (s) K NO3
• Net Ionic Equation
• Cl- Ag ? AgCl (s)

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Calculations with precipitation reactions

• When aqueous solutions of sodium sulfate and lead
  II Nitrate are mixed, a precipitate forms,
  determine the precipitate. Write the reaction and
  predict the products.
• Na2SO4 Pb(NO3)2 ? NaNO3 PbSO4
• SO42- Pb2? PbSO4
• Is it balanced?

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Calculate the mass of lead sulfate formed if 1.25
L of 0.05M lead nitrate and 2.00 L of .025M
sodium sulfate are mixed

• Calculate the moles of reactants present
• 1.25 L Pb2 x .05 mols Pb2 0.0625mol
  Pb2 1 L Pb2
• 2.00 L SO42- x .025 mols SO42- 0.05 mol SO42-
• 1 L SO42-
• Which is the limiting reactant?
• 11 ratio making the SO42- limiting reactant
• Calculate the mols of product.
• Ratio 11 so mols of product produced is 0.05 mol
• 0.05 mol PbSO4 x 303.3 g PbSO4 15.2 g PbSO4
• 1 mol PbSO4

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Acid Base Reactions an acid is a proton donor a
base is a proton acceptor

• Often called a neutralization reaction
• Most strong acid/ strong base reactions have this
  net ionic equation
• H (aq) OH- (aq) ? H2O
• You must be careful when working with weak
  acid/base reactions

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Weak Acid/ Strong Base Reactions

• Recall acetic acid is a weak acid, when mixed
  with potassium hydroxide, a strong base, the
  hydroxide ions have enough strength to rip the H
  right off of the non-dissociated molecules, and
  it can be assumed this will occur completely. The
  net ionic equation is
• OH- (aq) HC2H3O2 (aq) ? H2O (l) C2H3O2- (aq)

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Neutralization Calculations when just enough
base has been added to an acid to react exactly
and produce a neutral solution

• 28.0 ml of .250 M HNO3 53.0 ml of .320M KOH are
  mixed. Calculate the amount of water formed in
  the resulting reaction.
• HNO3 KOH ? KNO3 H2O
• H OH- ? H2O (l)
• Calculate mols next
• 28.0 ml HNO3 x 1 L x .25 mol H 7.00 x 10-3 mol
  H
• 1000 ml 1 L
• 53.0 ml KOH x 1L x .32 mol OH- 1.70 x 10-2
  mol OH-
• 1000 ml 1L
• Ratio 11 so HNO3 is limiting so there will 7.00
  x 10 -3 mol of water formed

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What is the concentration of ions in excess after
the reaction has finished?

• Since KOH is excess
• Original amount- amount used amount in
  excess
• 1.70 x 10-2 7.00 x 10-3 1.00 x 10-2 mols
  water
• The volume is needed add the two solutions 28.0
  ml 53.0 ml 81.0 ml .081 L
• Molarity of excess OH- ions 1.00x10-2 mol
• .081 L
• 0.123 M OH- ions

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Titration Calculations

• You titrate a 25.00 ml amount of your NaOH with
  0.1067 M HCl. The sample turns clear after the
  addition of 42.95 ml of the HCl. What is the
  molarity of a NaOH Solution?
• H OH- ? H2O
• 0.167 H x .04295 L H 4.583 x 10 -3mol H
• 1L
• From the equation mols of OH- H for
  neutralization
• M NaOH mol/L 4.583 x 10 -3mol OH- .1833 m
  NaOH
• .025 L