General Properties of Aqueous Solns
Solution (sol'n) = two component system.
Aqueous Chemical Rxns we study in this chapter
We'll examine three types of chemical reactions in aqueous solution.
 1) Precipitation rxns
 2) AcidBase rxns
 3) ReductionOxidation rxns
Electrolytes
Electrolyte — solutions of ion Conducts electricity — hence the name
 When the compounds is dissolved in an aqueous environment,
the molecules dissociate into two ions, cation and
anion.
 Strong electrolyte — complete dissociation occurs
 Examples include: HCl, HNO_{3}, NaCl, NaOH, KOH, Ba(OH)_{2}
 Weak electrolyte — incomplete dissociation (usually less than
10%)
 Examples include: HF, CH_{3}CO_{2}H
 Nonelectrolyte — no dissociation occurs
 Examples: H_{2}O, glucose(C_{6}H_{12}O_{6})
Precipitation Rxns and Solubility
When NaCl is mixed with AgNO_{3} in water, you might write chemical equation as,
The chemical eqn above may be described as molecular equation, where all species are accounted for. However, ionic compound such as NaCl ionizes
 When NaCl(s) is placed in water →
Na^{+}(aq) + Cl^{}(aq)
 Similarly, if we dissolve AgNO_{3}(s) in water
→
Ag^{+}(aq) + NO_{3}^{}(aq)
Further, if we add NaCl(s) and AgNO_{3}(s) into water, it makes sense that → Na^{+}(aq) + Cl^{}(aq) + Ag^{+}(aq) + NO_{3}^{}(aq)
Let's say that Ag^{+} and Cl^{} ions react to form AgCl(s ), and no other ions react. Then, the reaction is written as,
Na^{+}(aq) + Cl^{}(aq) + Ag^{+}(aq) + NO_{3}^{}(aq) → Na^{+}(aq) + AgCl(s) + NO_{3}^{}(aq)
Since Na^{+} and NO_{3}^{} ions don't react, they appear on the both sides of the equation. Chemical equation is similar to mathematical equation, if the same species appears on the both sides of the equation, they cancel each other. Thus, we have,
Ag^{+}(aq) + Cl^{}(aq) → AgCl(s)
This type of equation, after cancellation, is called net ionic equation.
Example: Write net ionic equations for the following,
H_{2}SO_{4}(aq) + MgCO_{3}(s) → H_{2}O(l) + CO_{2}(g) + MgSO_{4}(aq)
Hg_{2}(NO_{3})_{2}(aq) + 2NH_{4}Cl(aq) → Hg_{2}Cl_{2}(s) + 2NH_{4}NO_{3}(aq)
Precipitate (PPT) — solid formed in solution as a result of chemical rxn
PPT rxn — rxn produces PPT
Another way to think about PPT rxn is that two soluble molecules dissolved in water in the form of cations and anions, are recombined such that the new molecule is no longer soluble in water, and becomes insoluble, thusprecipitate out.
Why? Simply put: The process is energetically favored. (Much more on this in Chapter 8!)
How can we tell what combinations of ions would form PPT in aqueous solution? The following table shows solubility of cation—anion combination.
NO_{3}^{}  X^{}  ClO_{4}^{}  CH_{3}CO_{2}^{}  SO_{4}^{2}  OH^{}  CO_{3}^{2}  PO_{4}^{3}  

Gr 1 cations NH_{4}^{+}  sol  sol  sol  sol  sol  sol  sol  sol 
Gr 2 cations  sol  sol  sol  sol  BaSO_{4}  Mg(OH)_{2}  insol  insol 
TM  sol 
AgX Hg_{2}X_{2} PbX_{2} 
Ba(ClO_{4})_{2} Hg_{2}(ClO_{4})_{2} Pb(ClO_{4})_{2} 
Ba(CH_{3}CO_{2})_{2} Hg_{2}(CH_{3}CO_{2})_{2} Pb(CH_{3}CO_{2})_{2} 
sol  insol  insol  insol 
Gr = group on P.T. X = halogens (Cl, Br, I), sol = soluble, insol = insoluble. Compounds written in red are specific compounds that are insoluble, otherwise the combination is soluble in general. 
Example: Aluminum chloride reacts with sodium carbonate
to form insoluble compound, aluminum carbonate. a) What is the net ionic eqn?
b) What is the molarity of a solution of AlCl_{3} if 30.0 mL is
required to react with 35.5 mL of 0.137 M sodium carbonate?
a) One way to think about this problem is to think about the products first. One of the products is Al_{2}(CO_{3})_{3}, since the product molecules should have no charge. It means that 2AlCl_{3} and 3Na_{2}CO_{3} are needed. Then, we have in the solution,
From the Solubility Table above, we know that Na^{+} and Cl^{} ions stay dissolved in water. Thus, they become spectator ions, and they are cancelled out from the equation. Thusly,
This is the net ionic eqn for this reaction.
b) We want to know the concentration of AlCl_{3} for which we know the
volume (30 mL). From the definition of molarity, mol/vol, you can see that
we do not know the number of moles of AlCl_{3}. So, that is what
we are after. The given here should be 35.5mL simply because this is
the only number that are not conversion factor. 0.137 M is molarity of
Na_{2}CO_{3}, which has its meaning to be 0.137
mol_{Na2CO3} = 1000
mL_{Na2CO3}. Therefore, we set up our
dimensional analysis as,
$$?mo{l}_{AlC{l}_{3}}=35.5m{L}_{NaC{O}_{3}}\left(\frac{0.137mo{l}_{NaC{O}_{3}}}{1000m{L}_{NaC{O}_{3}}}\right)\left(\frac{1mo{l}_{C{O}_{3}^{2}}}{1mo{l}_{NaC{O}_{3}}}\right)\left(\frac{2mo{l}_{A{l}^{3+}}}{3mo{l}_{C{O}_{3}^{2}}}\right)\left(\frac{1mo{l}_{AlC{l}_{3}}}{1mo{l}_{A{l}^{3+}}}\right)=3.24\times {10}^{3}mo{l}_{AlC{l}_{3}}$$
Using this mole number, we can calculate the molarity of AlCl_{3} as,
$${M}_{AlC{l}_{3}}=\frac{3.24\times {10}^{3}mo{l}_{AlC{l}_{3}}}{30.0mL\left(\frac{1L}{1000mL}\right)}=0.108{M}_{AlC{l}_{3}}$$
Acids, Bases, and Neutralization Rxns
As seen in Chapter 3,
Acid produces H^{+}
Base produces OH^{}
Acid + Base → H_{2}O — Neutralization rxn
Strong Acid—Complete dissociation into H^{+} and anion
 example: HCl → H^{+} + Cl^{}
Strong Base—Complete dissociation into cation and OH^{}
 example: NaOH → Na^{+} + OH^{}
Weak Acid—Incomplete dissociation into H^{+} and anion
 example: HC_{2}H_{3}O_{2} ⇄ H^{+}
+ C_{2}H_{3}O_{2}^{}
⇄ indicates that all three species are present in the solution (said to be in equilibrium)
Weak Base—Incomplete dissociation into cation and OH^{}
 example: NH_{3} + H_{2}O ⇄ NH_{4}^{+}
+ OH^{}
Neutralization Rxns
Three circumstances:
1) Strong acid reacting with strong base
 H^{+} + OH^{} → H_{2}O
2) Weak acid (HA) reacting with strong base
 HA ⇄ H^{+} + A^{}
 H^{+} + OH^{} → H_{2}O
 Combining the two yields
 HA + OH^{} → A^{} +
H_{2}O
3) Strong acid (HA) reacting with weak base (B)
 B + H_{2}O → BH^{+} +
OH^{}
 H^{+} + OH^{} → H_{2}O
 Combining the two, yields
 B + H^{+} → BH^{+}
The following lists the common acids and bases that you encounter in this course.
Common Acids and Bases with their strength
Example: What volume of 2.16 M Ba(OH)_{2} can be neutralized by 10.00 g of H_{3}BO_{3}, which is a weak acid? Assume that all three H in H_{3}BO_{3} are dissociable.
Sicne H_{3}BO_{3} is a weak acid, we can use the net ionic equation for neutralization rxn between weak acid and strong base, which is
In our example, the reaction is
H_{3}BO_{3} + 3OH^{} → 3H_{2}O + BO_{3}^{3}
There are 3 H of H_{3}BO_{3} reacting, therefore we need 3 mol of OH^{} to neutralize all H. Then we ask, now becoming to be routine, dimensional analysis question,
$$?m{L}_{Ba{(OH)}_{2}}=10.00{g}_{{H}_{3}B{O}_{3}}\left(\frac{1mo{l}_{{H}_{3}B{O}_{3}}}{61.834{g}_{{H}_{3}B{O}_{3}}}\right)\left(\frac{3mo{l}_{O{H}^{}}}{1mo{l}_{{H}_{3}B{O}_{3}}}\right)\left(\frac{1mo{l}_{Ba{(OH)}_{2}}}{2mo{l}_{O{H}^{}}}\right)\left(\frac{1000m{L}_{Ba{(OH)}_{2}}}{2.16mo{l}_{Ba{(OH)}_{2}}}\right)=112m{L}_{Ba{(OH)}_{2}}$$
Example: What volume of 0.285 M Sr(OH)_{2} is required to neutralize 25.00 mL of 0.275 M HF?
Sr(OH)_{2} is a strong base, while HF is a weak acid. So the net ionic equation is,
Then, the question becomes,
Reduction—Oxidation Rxns (Redox rxns)
Electron transfer between two different species
 One loses an electron — Oxidation
 One gains an electron — Reduction
So,
Zn^{2+} 2e^{} → Zn reduction
This eqn can be interpretted to have neutral Cu giving 2 electrons to Zn^{2+}, therefore, Cu becomes Cu^{2+} and Zn^{2+} becomes Zn. Another words, oxidation and reduction ocurr simultaneously.
Many biological catalysts are Redox in nature:
 Fe in hemoglobin
 Co in vitamin B12
 Cu in cytochrome
 Mo in nitrogenase
The following shows how to balance redox equation. Yes, it is tedius!
Steps involved in balancing the redox eqn. 

If the solution is basic, add enough OH^{} on the both sides of the equation so that all H^{+} becomes H_{2}O 
Oxidation number, ox#
Ox# is a fictitious charge of atoms in molecule. We can easily assign the ox# for each atom in a molecule by using the following priority table. Priority becomes lower as you go down the list.
Priority  Atom types  Assigned Ox#  Examples 

1  Atoms in their element state  0  Br_{2} Br is 0 
2  Monatomic ion  charge  Cl^{}, Ox# = 1 
3  Total of ox# in a molecule in an ion  sum = 0 sum = charge  
4  In their compounds, Gr IA Gr IIA  +1  KCl,
ox#_{K}=+1 MgCl_{2}, ox#_{Mg}=+2 
5  F in compounds  1  CH_{2}F_{2}, ox#_{F}=1 
6  H in compounds  +1  H_{2}O each H, ox#=+1 
7  O in compounds  2  H_{2}O each O, ox#=2 
8  In their compounds, Gr 17A Gr 16A Gr 15A  1 2 3  HCl, ox#_{Cl}
= 1 H_{2}Se, ox#_{Se} = 2 PH_{3}, ox#_{P} = 3 
Example: What are the oxidation numbers for N and O in NO_{3}^{}?
N is a Group 15A element. So, oxygen has higher priority than N. It means that we assign O to have ox#=2. Then, let's call ox# for N as x. Using the this fact as well as the item #3 where the total of the ox#'s in an ion is ionic charge, we can set up the following,
Sovling for x, giving x = +5. Therefore, ox#_{N} = +5.
Calculate the ox# for the following.
H_{2}S, CH_{4}, SF_{4}, CO_{2}, FeO_{4}^{2}
When ox# in the rxn becomes smaller, it is reduction.
When ox# in the rxn becomes larger, it is oxidation.
Balancing HalfRxns
The 4 steps to balance halfrxn:
 a) Balance all atoms other than H and O
 b) Balance O by adding H_{2}O
 c) Balance H by adding H^{+}
 d) Balance charge by adding H^{+}
Combining the two HalfRxns
To combine the balanced halfrxns, we need to cancel out all the free electrons, denoted as e^{}.
Example: Balance the following rxn:
First, we need to assign ox# so that we know which compound is oxidized or reduced. Using the priority table, we assign ox# as,
As you can see, the ox# on Mn in the reactant is +7, and that became +2 in the product side (ox# is reduced in magnitude, hence 'reductions'), therefore the process is reduction. N was the unknown ox# for NO_{2}^{} and NO_{3}^{}, and the magnitude of ox# had increased, which is oxidation process.
Now we can divide the above equation into two halfrxns as,
ox NO_{2}^{} → NO_{3}^{}
Let's balance each of these now. First, we need to balance atoms other than H and O. In the reduction case, it is Mn, and in the oxidation case, it is N. In the reduction, there is only one Mn on the both sides of the equation, therefore, they are balanced. In the oxidation the same is true; N are balanced.
The next step is to balance O by adding H_{2}O. Then,
ox NO_{2}^{} + H_{2}O → NO_{3}^{}
In the reduction, there are 4 O's of the lefthand side of the equation, therefore 4H_{2}O were added. Similarly for oxidation one H_{2}O on the lefthand side was needed.
We need to balance H by adding H^{+}. In the reduction side, we added 4 H_{2}O, therefore, we have 8 H altogether. So, we add 8H^{+} on the lefthand side. In the oxidation, we need to add 2H^{+} on the righthand side. Then, we have
ox NO_{2}^{} + H_{2}O → NO_{3}^{} + 2H^{+}
The last step is to balance the charge by adding e^{}. In the reduction, the total of the charge is +7 on the lefthand side (1 from MnO_{4}^{} and +8 from H^{+}), and +2 on the righthand side. So, if 5e^{} is added to the lefthand side, the total charge becomes +2, which is the same as on the righthand side. Similarly, in the oxidation, if 2e^{} is added to the righthand side, both sides become 1 charge.
ox NO_{2}^{} + H_{2}O → NO_{3}^{} + 2H^{+} + 2e^{}
We now need to combine the two halfrxns to make into an overall equation. You notice in the two halfrxns that electrons appear on the opposite sides of the equations. When combined, and if the same number of electrons appear they cancel over and across the arrow of the chemical equation. Currently, we have 5e^{} on the left and 2e^{} on the right. So, if we multiply the entire reduction equation by 2 and multiply the entire oxidation equation by 5, we find 10 electrons in each of the equations. That is,
ox 5( NO_{2}^{} + H_{2}O → NO_{3}^{} + 2H^{+} + 2e^{})
By distributing the factors by combining the two halfrxns, we get
Finally, we arrive at,
2MnO_{4}^{} + 6H^{+} + 5NO_{2}^{} → 2Mn^{2+} + 3H_{2}O + 5NO_{3}^{} 
This is the balanced redox equation.
As seen above, there are 6H^{+} in this equation. This implies that the reaction is in acidic condition. If in case, you desire to carry out the reaction in basic condition, you must add 6OH^{} to the both sides of the equation. Therefore,
Then, some of the water cancel out, and leads to the balanced redox equation in basic condition is,
2MnO_{4}^{} + 3H_{2}O + 5NO_{2}^{} → 2Mn^{2+} + 5NO_{3}^{} + 6OH^{} 
Example: Blood alcohol level can be determined by a titration with potassium dichromate according to the following (not balanced):
What is the blood alcohol level in mass percent if 8.76 mL of 0.04988 M K_{2}Cr_{2}O_{7} is required for titration of a 10.002 g sample of blood?
We first need to balance the equation above. We need to determine what's been reduced or oxidized. Let's assign ox#.
It means,
ox C_{2}H_{5}OH → CO_{2}
Balance atoms other than O and H.
ox C_{2}H_{5}OH → 2CO_{2}
Balance O by adding H_{2}O.
ox C_{2}H_{5}OH + 3H_{2}O → 2CO_{2}
Balance H by adding H^{+}.
ox C_{2}H_{5}OH + 3H_{2}O → 2CO_{2} + 12H^{+}
Balance the charge.
ox C_{2}H_{5}OH + 3H_{2}O → 2CO_{2} + 12H^{+} + 12e^{}
Combining by multiply the reduction equation by 2, and add together,
Now we can do the dimensional analysis. We need to know the mass of C_{2}H_{5}OH in the sample. We start our calculation by using the given as 8.76 mL_{K2Cr2O7}.
$$?{g}_{{C}_{2}{H}_{5}OH}=8.76m{L}_{{K}_{2}C{r}_{2}{O}_{7}}\left(\frac{0.04988mo{l}_{{K}_{2}C{r}_{2}{O}_{7}}}{1000m{L}_{{K}_{2}C{r}_{2}{O}_{7}}}\right)\left(\frac{1mo{l}_{C{r}_{2}{O}_{7}^{2}}}{1mo{l}_{{K}_{2}C{r}_{2}{O}_{7}}}\right)\left(\frac{1mo{l}_{{C}_{2}{H}_{5}OH}}{2mo{l}_{C{r}_{2}{O}_{7}^{2}}}\right)\left(\frac{46.07{g}_{{C}_{2}{H}_{5}OH}}{1mo{l}_{{C}_{2}{H}_{5}OH}}\right)=0.010065{g}_{{C}_{2}{H}_{5}OH}$$
Therefore, we have $${\%}_{alcohol}=\frac{0.010065{g}_{{C}_{2}{H}_{5}OH}}{10.002{g}_{blood}}100\%=0.100631\%=0.101\%$$
Molarity, unit of concentration
Solution
A solution is a two component system where the smaller component is called solute and the larger component is called solvent.
Concentration is a measure of amount of solute in the solution. The most common concentration unit is Molarity, M.
$$M\equiv \frac{mo{l}_{solute}}{{L}_{solution}}=\frac{mo{l}_{solute}}{1000m{L}_{solution}}$$
By knowing the concentration of the solution, you can predict the theoretical yield of products, and others. Molarity contains two units, mol_{solute } and L_{solution}. So, molarity can be used as a conversion factor in the dimensional analysis!
Example: Molarity
You have 50.00 mL solutioin of 3.50x10^{2}M NaOH. How many mol of OH^{} ion are there?
$$?mo{l}_{O{H}^{}}=50.00m{L}_{NaOH}\left(\frac{3.50\times {10}^{2}mo{l}_{NaOH}}{1000m{L}_{NaOH}}\right)\left(\frac{1mo{l}_{O{H}^{}}}{1mo{l}_{NaOH}}\right)=1.75\times {10}^{3}mo{l}_{O{H}^{}}$$
Dilution
Often you encounter a situation that you need to make a solution from the concentrated solution. What do you do?
Dilution can be easily dealt with using dimensional analysis. Two example should illustrate the importance of using dimensional analysis.
Example: Dilution #1
How many mL of 0.500 M HCl do you need to make a 200.0 mL of 3.00 x 10^{2} M HCl?
$$?m{L}_{HC{l}_{conc}}=200.0m{L}_{HCl}\left(\frac{3.00\times {10}^{2}mo{l}_{HCl}}{1000m{L}_{HCl}}\right)\left(\frac{1000m{l}_{HC{l}_{conc}}}{0.500mo{l}_{HC{l}_{conc}}}\right)=12.0m{L}_{HC{l}_{conc}}$$
Example: Dilution #2
How many mL of 0.500 M Ba(OH)_{2} do you need to make a 200.0 mL of 3.00 x 10^{2} M OH^{} solution?
$$?m{L}_{Ba{(OH)}_{2}}=200.0m{L}_{O{H}^{}}\left(\frac{3.00\times {10}^{2}mo{l}_{O{H}^{}}}{1000m{L}_{O{H}^{}}}\right)\left(\frac{1mo{l}_{Ba{(OH)}_{2}}}{2mo{l}_{O{H}^{}}}\right)\left(\frac{1000m{l}_{Ba{(OH)}_{2}}}{0.500mo{l}_{Ba{(OH)}_{2}}}\right)=6.00m{L}_{Ba{(OH)}_{2}}$$
Do you see the difference between the two examples above? So, how do you actually make the solution? Here is how.
Gravimetric Analysis
Gravimetric Analysis — Quantitative analysis based on measurement mass of compounds.In the ppt rxn section, we discussed adding AgNO_{3} solution to NaCl solution.
We came up with its net ionic equation to be,
Let's use the example above to solve a problem.
Example: How many grams of Cl is in the sample if 25.00 mL of 0.1975 M AgNO_{3} is required to convert all Cl to AgCl?
$$?{g}_{C{l}^{}}=25.00m{L}_{AgN{O}_{3}}\left(\frac{0.1975mo{l}_{AgN{O}_{3}}}{1000m{L}_{AgN{O}_{3}}}\right)\left(\frac{1mo{l}_{A{g}^{+}}}{1mo{l}_{AgN{O}_{3}}}\right)\left(\frac{1mo{l}_{C{l}^{}}}{1mo{l}_{A{g}^{+}}}\right)\left(\frac{35.45{g}_{C{l}^{}}}{1mo{l}_{C{l}^{}}}\right)=0.175034{g}_{C{l}^{}}=0.175{g}_{C{l}^{}}$$
Example: A 40.00 mL of 0.0348 M Pb(NO_{3})_{2} is mixed with 0.00940 M aluminum iodide. a) Write net ionic equation. b) What is the minimum volume of AlI_{3} needed to complete the reaction? c) How many g of precipitate (product) form?
a)
b) $$?m{L}_{Al{I}_{3}}=40.00m{L}_{Pb{(N{O}_{3})}_{2}}\left(\frac{0.0348mo{l}_{Pb{(N{O}_{3})}_{2}}}{1000m{L}_{Pb{(N{O}_{3})}_{2}}}\right)\left(\frac{1mo{l}_{P{b}^{2+}}}{1mo{l}_{Pb{(N{O}_{3})}_{2}}}\right)\left(\frac{2mo{l}_{{I}^{}}}{1mo{l}_{P{b}^{2+}}}\right)\left(\frac{1mo{l}_{Al{I}_{3}}}{3mo{l}_{{I}^{}}}\right)\left(\frac{1000m{L}_{Al{I}_{3}}}{0.0094mo{l}_{Al{I}_{3}}}\right)=98.723mL=98.7m{L}_{Al{I}_{3}}$$
c)
$$?{g}_{Pb{I}_{2}}=40.00m{L}_{Pb{(N{O}_{3})}_{2}}\left(\frac{0.0348mo{l}_{Pb{(N{O}_{3})}_{2}}}{1000m{L}_{Pb{(N{O}_{3})}_{2}}}\right)\left(\frac{1mo{l}_{P{b}^{2+}}}{1mo{l}_{Pb{(N{O}_{3})}_{2}}}\right)\left(\frac{1mo{l}_{Pb{I}_{2}}}{1mo{l}_{P{b}^{2+}}}\right)\left(\frac{461.01{g}_{Pb{I}_{2}}}{1mo{l}_{Pb{I}_{2}}}\right)=0.641726g=0.642{g}_{Pb{I}_{2}}$$
AcidBase Titration
Titration is a procedure to analyze the concentration of unknown quantity of some substance using known concentration of the other substance. These two different substances are related by chemical reaction. It means that you must know the balanced chemical equation to find out unknown.
The solution of known concentration is called standard solution, is placed in a buret. Slowly the standard solution is added to the solution containing unknown amount in a flask. When the stoichiometric amount of the standard solution is added, called the equivalence point, the reaction is complete and the titration is finished. In many reactions, the equivalence point is hard to detect, therefore one may add an indicator to the unknown to change color of the solution when some condition is met. Design of indicator that matches the equivalence point becomes, then, important.
Example: AcidBase Titration #1
If a 50.0 mL of 0.50 M HCl is titrated with 0.20 M NaOH, how many mL of NaOH solution is needed to reach equivalence point?
The net ionic equation for this acidbase reaction is:
$$?m{L}_{NaOH}=50.0m{L}_{HCl}\left(\frac{0.50mo{l}_{HCl}}{1000m{L}_{HCl}}\right)\left(\frac{1mo{l}_{{H}^{+}}}{1mo{l}_{HCl}}\right)\left(\frac{1mo{l}_{O{H}^{}}}{1mo{l}_{{H}^{+}}}\right)\left(\frac{1mo{l}_{NaOH}}{1mo{l}_{O{H}^{}}}\right)\left(\frac{1000m{L}_{NaOH}}{0.70mo{l}_{NaOH}}\right)=35.714mL=36m{L}_{NaOH}$$
Example: AcidBase Titration #2
If a 50.0 mL of 0.0224 M Ba(OH)_{2} is titrated with 1.72 x 10^{2} M HF, how many mL of HF is required to reach equivalence point? Write a net ionic equation, first.
HF is a weak acid and Ba(OH)_{2} is a strong base. Therefore, the net ionic equation is:
$$?m{L}_{HF}=50.0m{L}_{Ba{(OH)}_{2}}\left(\frac{0.0224mo{l}_{Ba{(OH)}_{2}}}{1000m{L}_{Ba{(OH)}_{2}}}\right)\left(\frac{2mo{l}_{O{H}^{}}}{1mo{l}_{Ba{(OH)}_{2}}}\right)\left(\frac{1mo{l}_{HF}}{1mo{l}_{O{H}^{}}}\right)\left(\frac{1000m{L}_{HF}}{1.72\times {10}^{2}mo{l}_{HF}}\right)=130.23mL=130mL$$
RedOx Titration
Example: RedOx Titration #1
Bleach is a solution of sodium hypochlorite (NaClO). To determine the hypochlorite ion (ClO^{}) content, sulfide ion is added in basic solution and producing S(s) and aqueous chloride ion. Subsequently, the chloride ion is precipitated with AgNO_{3} to produce AgCl(s). When 50.0 mL of bleach (d = 1.02 g/mL) is treated as described above, 4.95 g of AgCl is obtained. What is the mass percent of NaClO in the bleach. First write two net ionic equations.
But the first one is redox equation because S^{2} → S is an oxidation process since the former has ox# = 2 and the latter ox# = 0. Then, you need to do what you need to do!
but the question stated that it is in basic solution, therefore add 2OH^{} to get rid of H^{+}. Then,
From 4.95 g of AgCl, we can calculate the amount of NaClO since we have the balanced chemical equations. $$?{g}_{NaClO}=4.95{g}_{AgCl}\left(\frac{1mo{l}_{AgCl}}{143.32{g}_{AgCl}}\right)\left(\frac{1mo{l}_{C{l}^{}}}{1mo{l}_{AgCl}}\right)\left(\frac{1mo{l}_{NaClO}}{1mo{l}_{C{l}^{}}}\right)\left(\frac{77.44{g}_{NaClO}}{1mo{l}_{NaClO}}\right)=2.67463{g}_{NaClO}$$ but, first convert 50 mL to g $$?{g}_{bleach}=50mL\left(\frac{1.02g}{1mL}\right)=51g$$ Then, $$\frac{2.67463{g}_{NaClO}}{51{g}_{bleach}}\times 100\%=5.2444\%=5.24\%$$