Chapter 4 Aqueous RXNS Solution Chemistry Part II
4.7 Stoichiometry of PPT Rxns
In Chapter 4,The principles of stoichiometry apply to reactions in solution, not in solid (s) form like we saw in Chapter 3.
Instead of dividing grams by molar mass to get moles, we calculate moles of a compound in solution by manipulating the molarity equation to solve for moles, thus you multiply volume by molarity. The problem solving strategy for solution stoichiometry problems is as follows: 1. Identify the species present in the solution and which ones react.(solubility Table, the ones that form the solid ppt.) 2. Write the balanced net ionic equation for the reaction. 3. Calculate the moles of the reactants (V x M). 4. Determine, if necessary, which is the limiting reactant. V x M = moles 5. Calculate the moles of product(s) formed based on the limiting reactant. 6. Convert moles to grams or other units, as required. Calculate the mass of solid NaCl that must be added to 1.50L of a 0.100 M AgNO3 solution to precipitate all the Ag+ in the form of AgCl. 1. Write the rxn: NaCl(s) + AgNO3(aq)
NaNO3(aq) + AgCl(s)
Check to se if it is balanced, if not do so. This one is.
2. Solve for moles of given sol'n using the molarity equation
Moles of AgNO3 = 1.50L
x 0.100 mol AgNO3
= 0.150 sol'n
L
mol AgNO3
3. Use Mole Ratio from balanced equation to get moles of what you are trying to find
0.150 mol AgNO3 x 1 mol NaCl
= 0.150 mol NaCl
1 mol AgNO3
4. Convert moles to grams
0.150 mol NaCl x 58.45 g NaCl
= 8.78 g NaCl
1 mol NaCl
BOOK SOLUTION:Calculate the mass of solid NaCl that must be added to 1.50L of a 0.100 M AgNO3 solution to precipitate all the Ag+ in the form of AgCl.
Because Ag+ and Cl­ react in a 1:1 ratio,0.150 mol of Cl­ ions completely react with all of the Ag+ in sol'n.
NaCl
0.150
Na
+
0.150
+ Cl­
0.150 1. Identify species in sol'n mixed solution contains: Ag
+
NO3
Na
­
+
Cl
­
NaNO3 and AgCl would form, solubility table tells us that NaNO3 is soluble and AgCl is insoluble. Therefore solid AgCl forms according to the following net ionic equation.
Thus. 0.150 mol NaCl are required.
0.150 mol NaCl x 58.45 g NaCl = 8.77 g NaCl
1 mol NaCl
2. Write the balanced net ionic equation for the reaction that happens
+
­
Ag (aq) + Cl (aq)
AgCl(s)
3. Calculate the number of moles of reactants We must add enough Cl­ ions to react with all the Ag+ present. Remember that a 0.100M AgNO3 contains 0.100 M Ag+ ions and 0.100 M NO3­ ions. (AgNO3
Ag+ + NO3­)
0.100 M
0.100M
0.100 M
Moles of Ag+ = (1.50 L)(0.100 moles/L) = 0.150 mol Ag+
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Chapter 4 Aqueous RXNS Solution Chemistry Part II
When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates out. Calculate the mass of PbSO4 formed 2. Solve for moles of each solution using the molarity equation
moles of Pb(NO3)2 = 1.25L
x 0.0500 mol Pb(NO3)2
when 1.25L of 0.0500 M Pb(NO3)2 and 2.00L of 0.0250 M Na2SO4
1L
= 0.0625 mol P(NO3)2
are mixed.
When two solutions are mixed that have their volumes and molarites reported, you basically have a limiting reactant problem!!!
moles of Na2SO4 = 2.00L x 0.0250 mol Na2SO4
1L
limiting reactant
1. Write the reaction: Na2SO4(aq) + Pb(NO3)2(aq) PbSO4(s) + 2NaNO3
3. Use Mole Ratio from balanced equation
0.0500 mol Na2SO4 x 1 mol Pb(NO3)2
= 0.0500 mol Pb(NO3)2
= 0.0500 mol Na2SO4 BOOK SOLTUTION: When aqueous solutions of Na2SO4 and Pb
(NO3)2 are mixed, PbSO4 precipitates out. Calculate the mass of PbSO4 formed when 1.25L of 0.0500 M Pb(NO3)2 and 2.00L of 0.0250 M Na2SO4 are mixed. 1 mol Na2SO4 4. Convert moles to grams
Na2SO4 Na+ +
Pb(NO3)2
Pb2+ +
1 mol Pb(NO3)2
NO3­
So, Na
SO4 Pb
NO3 ions are in soluiton. NaNO3 and PbSO4 will bond, sodium nitrate is soluble, lead sulfate is not. The resulting net ionic equation is:
+
0.0500 mol Pb(NO3)2 x 303.30 g Pb(NO3)2 = 15.2 g Pb(NO3)2
SO42­ Pb2+(aq)
2­ +
2+
­
SO42­(aq)
PbSO4(s)
Next, calculate number of moles of the reactants.
Since 0.0500 M Pb(NO3)2 contains 0.0500 M Pb2+ ions, we can use the Molarity equations to solve for moles.
1.25 L x 0.0500 mol Pb(NO3)2 x 303.30 g Pb(NO3)2 = 15.2 g Pb(NO3)2
1 mol Pb(NO3)2
0.0500 mol Pb2+ = 0.0625 mol Pb2+
L
2.00 L x 0.0250 mol SO42­
= 0.0500 mol SO42­
L
The limiter, # of moles of pdt (Pb(NO3)2) produced!
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Chapter 4 Aqueous RXNS Solution Chemistry Part II
4.8 Acid­Base Rxns
According the Arrhenius concept of acids and bases, Traditionally, acids produce H+ ions in solution: • HCl(g) dissolves in water HCl(aq) H+(aq)+ Cl­(aq) An Acid base rxn is often called a neutralization reaction. Its when just enough base is added to react exactly with the acid in solution, we say the acid has been neutralized.
Neutralization rxn: Acid + Base a salt + water (**All ionic compounds are considered salts.) • H2SO4 (aq)
H+(aq)+ HSO4 (aq) and
• HCl(aq) + NaOH(aq) • HSO4 (aq)
H+(aq) + SO4 (aq) A more modern and useful definition of acids and bases is the Brønsted­
Lowry definition. 2­
NaCl(aq) + H2O(l)
Bases traditionally produce OH­(hydroxide) ions in solution: Acids are proton (H+ ion) donors. • NaOH(s) in water Bases are proton acceptors
Na+(aq) + OH­(aq)
Ammonia is also a base, even though it has no OH­ in its chemical formula: • NH3 (g) in water
NH3(aq) + H2O(l)
NH4+ (aq) + OH(aq)
Performing Calculations for Acid Base Rxns
1. Identify the species present in the solution and which ones react.(solubility Table, the ones that form the solid ppt.) 2. Write the balanced net ionic equation for the reaction. 3. Calculate the moles of the reactants Use the volumes of the original soln's and their molarities 4. Determine, if necessary, which is the limiting reactant where appropriate. 5. Calculate the moles of required reactant or pdt.
What volume of 0.100M HCl solution is needed to neutralize 25.0 ml of 0.350M NaOH.
What's in solution? H+ Cl­ Na+ OH­
What rxn will occur? 2 possibilities
NaCl is soluble, so Na+ + Cl­(aq) NaCl (s)
H+(aq) + OH­(aq)
Na+ and Cl­ are H2O(l)
spectator ions
So,
H+(aq) + OH­(aq)
equation
H2O(l) is the balanced net ionic 6. Convert moles to grams or volume (of solution), as required. Calculate moles of reactants
25.0 ml NaOH x 1L
= 0.025 L
Now, lets find out how much H+ is required.
1000ml
0.025 L
x 0.350 mol OH­
Since H+ and OH­ ions react in 1:1 ratio, 8.75 x 10­3 mol of H+ are needed to neutralize the OH­ ions present.
= 8.75 x 10­3 mol OH­
moles of H+ ions = V x 0.100 mol H+ = 8.75 x 10­3 mol of H+
L
L This problem requires the addition of just enough H+ to react exactly with the OH­ ions present. Thus we don't need to worry about a limiting reactant.
solving for V =
8.75 x 10­3 mol of H+
0.100 mol H+
= 8.75 x 10­2 L
L
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Chapter 4 Aqueous RXNS Solution Chemistry Part II
In a certain experiment, 28.0 mL of 0.250 M HNO3 and 53.0 mL of 0.320 M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H+ or OH­ ions in excess after the reaction goes to completion?
Moles of H+ = 0.0280 L x 0.250 mol H+ ions
Moles of OH­ = 0.0530 L x 0.320 moles of OH­ ions
1. Identify species species available for rxn: H+ , NO3­, K+, OH­
The solubility chart tells us that KNO3 is soluble so K+ and NO3­ are spectator ions.
2. Net ionic equation = H+(aq) + OH­(aq)
H2O(l)
3. Calculate moles of reactant ions. Use volumes and molarities of the solutions being mixed.
= 0.00700 mol H+ ions
L
= 0.0170 mol OH­ ions
L
4. Determine the limiter ­ H+ is the limiter (0.00700 mol or 7.00 x 10­3 mol H+)
H+(aq) + OH­(aq)
H2O(l)
(0.00700)
(0.00700)
(0.00700)
Amount of water formed = 0.00700 mol Acid Base Titrations
OH­ is in excess,
0.0170 ­ 0.00700 = 0.0101 or 1.01 x 10­2 mol of OH­ left over (starting)
(amount used up)
Concentration of left over OH­
Volume of combined sol'n = 53 + 28 = 81 ml or 0.081 L
M = 1.01 x 10­2 mol of OH­
=0.123 M OH­
0.081 L
Titrations: A special and common laboratory type of acid­
base reaction is the titration. Its purpose is to determine the unknown concentration of an acid or base. A solution of precisely known concentration (the titrant) is added to the solution of unknown concentration (the analyte). Titrant is added until the equivalence point(or the stoichiometric point) is reached, as indicated by an indicator color change, or a rapid shift in pH.
Part II
The three requirement of a successful titration
1. the exact reaction must be known.
2. the equivalence point must be marked accurately.
3. the volumes must be known and accurately measured.
4
Chapter 4 Aqueous RXNS Solution Chemistry Part II
Sample Exercise 4.14 p. 153
A student wants to figure out the concentration of a NaOH solution (analyte). He weighs out 1.3009g of KHC8H4O4(KPH). It has a molar mass of 204.22g/mol. He dissolves the KPH in distilled water (titrant) and adds an indicator (Phenolphthalein). He titrates to the endpoint (Color Change). The difference between the final and initial buret readings indicates that 41.20 ml of NaOH was used to react (neutralize) with 1.3009 g of KPH. Calculate the concentration of the NaOH solution.
Solution: We are looking for the concentration of the NaOH we used to titrate the KHP. So, the primary equation is M=moles/L.
In a neutralization rxn the number of moles to neutralize an acid or base are often the same unless you have more than one acidic hydrogen (H+).
KHP ­ has one, so amount (# of moles) of base to neutralize it will be the same as number of moles of acid in solution.
1.3009g of KHC8H4O4
x mol KHC8H4O4 (ACID)
204.22 g KHC8H4O4
= 6.3701 x 10­3 mol KHC8H4O4
(ACID)
H+ = OH­ so, [OH­] also = 6.3701 x 10­3 mol
M of NaOH = 6.3701 x 10­3 mol of NaOH
= 0.1546 M
0.04120 L
We will use KHP again for this example. It has a molar mass of 204.22g/mol and one acidic hydrogen. In the titration, a student titrates an unknown amount of KHP with 20.46 ml of a 0.1000 M NaOH sol'n. What mass of KHP was neutralized by the NaOH sol'n?
RXN: NaOH(aq) + KHP(aq)
NaKP(aq) + H2O(l)
moles of NaOH = 0.02046 L x 0.1000 moles NaOH x 1 mol KHP
1L
1 mol NaOH
= 0.002046 mol KHP
4.9 Oxidation ­ Reduction RXNs (Redox)
0.002046 mol KHP
x
204.22g KHP
1 mol KHP
=0.4178g KHP
Reactions in which one or more electrons are transferred from one specie to another are called oxidation/reduction reactions, or redox for short. Example: 0 (+1­1) (+2­1) 0 Fe(s) + 2HCl(g) →FeCl2(s) + H2(g) e­ transferred from Fe to H
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Chapter 4 Aqueous RXNS Solution Chemistry Part II
Half Reactions 0 Fe →Fe2+ + 2e­ Oxidation 0 Assigning Oxidation Numbers
Any element or ion, whether alone or part of a compound has an oxidation state, and can be assigned an oxidation number. Sometimes these numbers represent real ionic charges, but frequently they do not. 2H + 2e →H2 Reduction +
­
**LEO goes GER Losing electrons is oxidation, and gaining electrons is reduction.
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