Title: Ch' 4: Solution Stoichiometry Types of Chemical Reactions
1Ch. 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.
2Strong 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
3Concentration 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
4Concentration 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
5Dilution 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
6Continued
- 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
7Types 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-
8Continued 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?
9Net 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
10What 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)
11Molecular 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)
12Calculations 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?
13Calculate 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
14Acid 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
15Weak 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)
16Neutralization 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
17What 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
18Titration 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