# Chapter 4 Rxns in Aqueous Solutions

## General Properties of Aqueous Solns

Solution (sol'n) = two component system.

• Solute = smaller in quantity
• Solvent = larger in quantity

Aqueous Chemical Rxns we study in this chapter

We'll examine three types of chemical reactions in aqueous solution.

1) Precipitation rxns
2) Acid-Base rxns
3) Reduction-Oxidation 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, HNO3, NaCl, NaOH, KOH, Ba(OH)2
Weak electrolyte — incomplete dissociation (usually less than 10%)
Examples include: HF, CH3CO2H
Nonelectrolyte — no dissociation occurs
Examples: H2O, glucose(C6H12O6)

## Precipitation Rxns and Solubility

When NaCl is mixed with AgNO3 in water, you might write chemical equation as,

NaCl + AgNO3 AgCl(s) + NaNO3

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 AgNO3(s) in water Ag+(aq) + NO3-(aq)

Further, if we add NaCl(s) and AgNO3(s) into water, it makes sense that Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(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) + NO3-(aq) Na+(aq) + AgCl(s) + NO3-(aq)

Since Na+ and NO3- 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,

H2SO4(aq) + MgCO3(s) H2O(l) + CO2(g) + MgSO4(aq)

Hg2(NO3)2(aq) + 2NH4Cl(aq) Hg2Cl2(s) + 2NH4NO3(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.

Solubility table of common ions
NO3-X- ClO4- CH3CO2- SO42-OH- CO32-PO43-
Gr 1 cations
NH4+
solsol solsolsolsolsolsol
Gr 2 cationssolsolsol solBaSO4 Mg(OH)2 insol insol
TMsol AgX
Hg2X2
PbX2
Ba(ClO4)2
Hg2(ClO4)2
Pb(ClO4)2
Ba(CH3CO2)2
Hg2(CH3CO2)2
Pb(CH3CO2)2
solinsol insolinsol
 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 AlCl3 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 Al2(CO3)3, since the product molecules should have no charge. It means that 2AlCl3 and 3Na2CO3 are needed. Then, we have in the solution,

2Al3+(aq) + 6Cl-(aq) + 6Na+(aq) + 3CO32-(aq) Al2(CO3)3(s) + 6Na+(aq) + 6Cl-(aq)

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,

2Al3+(aq) + 3CO32-(aq) Al2(CO3)3(s)

This is the net ionic eqn for this reaction.

b) We want to know the concentration of AlCl3 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 AlCl3. 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 Na2CO3, which has its meaning to be 0.137 molNa2CO3 = 1000 mLNa2CO3. Therefore, we set up our dimensional analysis as,

?molAlCl3 = 35.5 mLNa2CO3

$?mo{l}_{AlC{l}_{3}}=35.5m{L}_{NaC{O}_{3}}\left(\frac{0.137mo{l}_{NaC{O}_{3}}}{1000m{L}_{NaC{O}_{3}}}\right)\left(\frac{1mo{l}_{C{O}_{3}^{2-}}}{1mo{l}_{NaC{O}_{3}}}\right)\left(\frac{2mo{l}_{A{l}^{3+}}}{3mo{l}_{C{O}_{3}^{2-}}}\right)\left(\frac{1mo{l}_{AlC{l}_{3}}}{1mo{l}_{A{l}^{3+}}}\right)=3.24×{10}^{-3}mo{l}_{AlC{l}_{3}}$

Using this mole number, we can calculate the molarity of AlCl3 as,

${M}_{AlC{l}_{3}}=\frac{3.24×{10}^{-3}mo{l}_{AlC{l}_{3}}}{30.0mL\left(\frac{1L}{1000mL}\right)}=0.108{M}_{AlC{l}_{3}}$

## Acids, Bases, and Neutralization Rxns

As seen in Chapter 3,

Acid produces H+

Base produces OH-

Acid + Base → H2O — 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: HC2H3O2 ⇄ H+ + C2H3O2-
⇄ indicates that all three species are present in the solution (said to be in equilibrium)

Weak Base—Incomplete dissociation into cation and OH-

example: NH3 + H2O ⇄ NH4+ + OH-

Neutralization Rxns
Three circumstances:

1) Strong acid reacting with strong base

H+ + OH- H2O

2) Weak acid (HA) reacting with strong base

HA ⇄ H+ + A-
H+ + OH- H2O
Combining the two yields
HA + OH- A- + H2O

3) Strong acid (HA) reacting with weak base (B)

B + H2O BH+ + OH-
H+ + OH- H2O
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 H3BO3, which is a weak acid? Assume that all three H in H3BO3 are dissociable.

Sicne H3BO3 is a weak acid, we can use the net ionic equation for neutralization rxn between weak acid and strong base, which is

HA + OH- A- + H2O

In our example, the reaction is

H3BO3 + 3OH- 3H2O + BO33-

There are 3 H of H3BO3 reacting, therefore we need 3 mol of OH- to neutralize all H. Then we ask, now becoming to be routine, dimensional analysis question,

?mLBa(OH)2 = 10.00gH3BO3

$?m{L}_{Ba{\left(OH\right)}_{2}}=10.00{g}_{{H}_{3}B{O}_{3}}\left(\frac{1mo{l}_{{H}_{3}B{O}_{3}}}{61.834{g}_{{H}_{3}B{O}_{3}}}\right)\left(\frac{3mo{l}_{O{H}^{-}}}{1mo{l}_{{H}_{3}B{O}_{3}}}\right)\left(\frac{1mo{l}_{Ba{\left(OH\right)}_{2}}}{2mo{l}_{O{H}^{-}}}\right)\left(\frac{1000m{L}_{Ba{\left(OH\right)}_{2}}}{2.16mo{l}_{Ba{\left(OH\right)}_{2}}}\right)=112m{L}_{Ba{\left(OH\right)}_{2}}$

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,

HF + OH- F- + H2O

Then, the question becomes,

? mLSr(OH)2 = 25.00 mLHF
$?m{L}_{Sr{\left(OH\right)}_{2}}=25.00m{L}_{HF}\left(\frac{0.275mo{l}_{HF}}{1000m{L}_{HF}}\right)\left(\frac{1mo{l}_{O{H}^{-}}}{1mo{l}_{HF}}\right)\left(\frac{1mo{l}_{Sr{\left(OH\right)}_{2}}}{2mo{l}_{O{H}^{-}}}\right)\left(\frac{1000m{L}_{Sr{\left(OH\right)}_{2}}}{0.285mo{l}_{Sr{\left(OH\right)}_{2}}}\right)=12.1m{L}_{Sr{\left(OH\right)}_{2}}$

## Reduction—Oxidation Rxns (Redox rxns)

Electron transfer between two different species

One loses an electron — Oxidation
One gains an electron — Reduction

So,

Cu Cu2+ + 2e- oxidation

Zn2+ 2e- Zn reduction

By combining the oxidation eqn and reduction eqn, we get redox eqn.

Cu + Zn2+ Cu2+ + Zn

This eqn can be interpretted to have neutral Cu giving 2 electrons to Zn2+, therefore, Cu becomes Cu2+ and Zn2+ 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 red-ox equation. Yes, it is tedius!

 Steps involved in balancing the redox eqn. 1) Assign the so-called oxidation number, ox# 2) Separate the equation into reduction and oxidation half rxns 3) Balance each half rxn as follows: a) Balance all atoms other than H and O b) Balance O by adding H2O c) Balance H by adding H+ d) Balance the charge by adding e- 4) Combine the two half-rxns such that there is no free electron. If the solution is basic, add enough OH- on the both sides of the equation so that all H+ becomes H2O

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.

PriorityAtom typesAssigned Ox#Examples
1Atoms in their element state 0Br2 Br is 0
2Monatomic ionchargeCl-, Ox# = -1
3Total of ox# in a molecule
in an ion
sum = 0
sum = charge
4In their compounds, Gr IA
Gr IIA
+1KCl, ox#K=+1
MgCl2, ox#Mg=+2
5F in compounds-1CH2F2, ox#F=-1
6H in compounds+1H2O each H, ox#=+1
7O in compounds-2H2O each O, ox#=-2
8In their compounds, Gr 17A
Gr 16A
Gr 15A
-1
-2
-3
HCl, ox#Cl = -1
H2Se, ox#Se = -2
PH3, ox#P = -3

Example: What are the oxidation numbers for N and O in NO3-?

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,

x + 3(-2) = -1

Sovling for x, giving x = +5. Therefore, ox#N = +5.

Calculate the ox# for the following.

H2S, CH4, SF4, CO2, FeO42-

When ox# in the rxn becomes smaller, it is reduction.

When ox# in the rxn becomes larger, it is oxidation.

Balancing Half-Rxns

The 4 steps to balance half-rxn:

a) Balance all atoms other than H and O
b) Balance O by adding H2O
c) Balance H by adding H+
d) Balance charge by adding H+

Combining the two Half-Rxns

To combine the balanced half-rxns, we need to cancel out all the free electrons, denoted as e-.

Example: Balance the following rxn:

MnO4- + NO2- Mn2+ + NO3-

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 NO2- and NO3-, and the magnitude of ox# had increased, which is oxidation process.

Now we can divide the above equation into two half-rxns as,

red       MnO4- Mn2+

ox       NO2- NO3-

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 H2O. Then,

red       MnO4- Mn2+ + 4H2O

ox       NO2- + H2O NO3-

In the reduction, there are 4 O's of the left-hand side of the equation, therefore 4H2O were added. Similarly for oxidation one H2O on the left-hand side was needed.

We need to balance H by adding H+. In the reduction side, we added 4 H2O, therefore, we have 8 H altogether. So, we add 8H+ on the left-hand side. In the oxidation, we need to add 2H+ on the right-hand side. Then, we have

red       MnO4- + 8H+ Mn2+ + 4H2O

ox       NO2- + H2O NO3- + 2H+

The last step is to balance the charge by adding e-. In the reduction, the total of the charge is +7 on the left-hand side (-1 from MnO4- and +8 from H+), and +2 on the right-hand side. So, if 5e- is added to the left-hand side, the total charge becomes +2, which is the same as on the right-hand side. Similarly, in the oxidation, if 2e- is added to the right-hand side, both sides become -1 charge.

red       MnO4- + 8H+ 5e- Mn2+ + 4H2O

ox       NO2- + H2O NO3- + 2H+ + 2e-

We now need to combine the two half-rxns to make into an overall equation. You notice in the two half-rxns 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,

red       2( MnO4- + 8H+ + 5e- Mn2+ + 4H2O)

ox       5( NO2- + H2O NO3- + 2H+ + 2e-)

By distributing the factors by combining the two half-rxns, we get

2MnO4- + 16H+ + 10e- + 5NO2- + 5H2O 2Mn2+ + 8H2O + 5NO3- + 10H+ + 10e-

Finally, we arrive at,

 2MnO4- + 6H+ + 5NO2- → 2Mn2+ + 3H2O + 5NO3-

This is the balanced red-ox 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,

2MnO4- + 6H2O + 5NO2- 2Mn2+ + 3H2O + 5NO3- + 6OH-

Then, some of the water cancel out, and leads to the balanced red-ox equation in basic condition is,

 2MnO4- + 3H2O + 5NO2- → 2Mn2+ + 5NO3- + 6OH-

Can you do the following problem?

Example: Blood alcohol level can be determined by a titration with potassium dichromate according to the following (not balanced):

C2H5OH(aq) + Cr2O72-(aq) CO2(g) + Cr3+(aq)

What is the blood alcohol level in mass percent if 8.76 mL of 0.04988 M K2Cr2O7 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,

red       Cr2O72- Cr3+

ox       C2H5OH CO2

Balance atoms other than O and H.

red       Cr2O72- 2Cr3+

ox       C2H5OH 2CO2

red       Cr2O72- 2Cr3+ + 7H2O

ox       C2H5OH + 3H2O 2CO2

red       Cr2O72- + 14H+ 2Cr3+ + 7H2O

ox       C2H5OH + 3H2O 2CO2 + 12H+

Balance the charge.

red       Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O

ox       C2H5OH + 3H2O 2CO2 + 12H+ + 12e-

Combining by multiply the reduction equation by 2, and add together,

2Cr2O72- + 16H+ + C2H5OH 4Cr3+ + 11H2O + 2CO2

Now we can do the dimensional analysis. We need to know the mass of C2H5OH in the sample. We start our calculation by using the given as 8.76 mLK2Cr2O7.

?gC2H5OH = 8.76 mLK2Cr2O7

$?{g}_{{C}_{2}{H}_{5}OH}=8.76m{L}_{{K}_{2}C{r}_{2}{O}_{7}}\left(\frac{0.04988mo{l}_{{K}_{2}C{r}_{2}{O}_{7}}}{1000m{L}_{{K}_{2}C{r}_{2}{O}_{7}}}\right)\left(\frac{1mo{l}_{C{r}_{2}{O}_{7}^{2-}}}{1mo{l}_{{K}_{2}C{r}_{2}{O}_{7}}}\right)\left(\frac{1mo{l}_{{C}_{2}{H}_{5}OH}}{2mo{l}_{C{r}_{2}{O}_{7}^{2-}}}\right)\left(\frac{46.07{g}_{{C}_{2}{H}_{5}OH}}{1mo{l}_{{C}_{2}{H}_{5}OH}}\right)=0.010065{g}_{{C}_{2}{H}_{5}OH}$

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, molsolute and Lsolution. So, molarity can be used as a conversion factor in the dimensional analysis!

Example: Molarity

You have 50.00 mL solutioin of 3.50x10-2M NaOH. How many mol of OH- ion are there?

$?mo{l}_{O{H}^{-}}=50.00m{L}_{NaOH}\left(\frac{3.50×{10}^{-2}mo{l}_{NaOH}}{1000m{L}_{NaOH}}\right)\left(\frac{1mo{l}_{O{H}^{-}}}{1mo{l}_{NaOH}}\right)=1.75×{10}^{-3}mo{l}_{O{H}^{-}}$

### 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?

$?m{L}_{HC{l}_{conc}}=200.0m{L}_{HCl}\left(\frac{3.00×{10}^{-2}mo{l}_{HCl}}{1000m{L}_{HCl}}\right)\left(\frac{1000m{l}_{HC{l}_{conc}}}{0.500mo{l}_{HC{l}_{conc}}}\right)=12.0m{L}_{HC{l}_{conc}}$

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?

$?m{L}_{Ba{\left(OH\right)}_{2}}=200.0m{L}_{O{H}^{-}}\left(\frac{3.00×{10}^{-2}mo{l}_{O{H}^{-}}}{1000m{L}_{O{H}^{-}}}\right)\left(\frac{1mo{l}_{Ba{\left(OH\right)}_{2}}}{2mo{l}_{O{H}^{-}}}\right)\left(\frac{1000m{l}_{Ba{\left(OH\right)}_{2}}}{0.500mo{l}_{Ba{\left(OH\right)}_{2}}}\right)=6.00m{L}_{Ba{\left(OH\right)}_{2}}$

Do you see the difference between the two examples above? So, how do you actually make the solution? Here is how.

In the above example (#2), place 6.00 mL of 0.500 M Ba(OH)2 solution into a 200 mL volumetric flask, then add water to the calibration mark (in our example 200 mL--not 100 mL as in the picture).

More difficult question is:

Example: Dilution #3

What volume of 0.416 M Mg(NO3)2 should be added to 255 mL of 0.102 M KNO3 to produce a solution with a concentration of 0.278 M NO3- ions? Assume volumes are additive.

The situation is described as shown in the following diagram.

The first thing you must notice is that in Beaker 2 the concentration of NO3- = 0.832M, as a consequence of the chemical formula. Since we don't know the final volume of the solution in Beaker 3, we have to think a little.

In order to obtain the concentration specified in Beaker 3, we add contributions from Beakers 1 and 2, of course, which means:

 ${n}_{Bea\mathrm{ker}1}+{n}_{Bea\mathrm{ker}2}={n}_{Bea\mathrm{ker}3}$ 1
and we know the definition of molarity, $M= n V$ and it also means that
 $n=MV$ 2
Let's substitute Equation 2 into Equation 1:
 $M 1 V 1 + M 2 V 2 = M 3 V 3$ 3
for which we know M1, V1, M2, and M3. We don't know V2 and V3. The subscripts indicate which beaker. But, we also know the volume is additive, i.e.
 $V 1 + V 2 = V 3$ 4
for which we only know V1. But, hey!, we have two equations with 2 unknowns! It means that we can substitute Equation 4 into Equation 3, $M 1 V 1 + M 2 V 2 = M 3 ( V 1 + V 2 )$ to arrive at the expression for V2, $V 2 = M 3 V 1 − M 1 V 1 ( M 2 − M 3 )$

Derivation of the equation above is shown below, if you care to know.

+ Derivation of above
$M 1 V 1 + M 2 V 2 = M 3 ( V 1 + V 2 )$ Distribute on the right-hand side. $M 1 V 1 + M 2 V 2 = M 3 V 1 + M 3 V 2$ Collect terms containing V2 to the left-hand side. $M 2 V 2 − M 3 V 2 = M 3 V 1 − M 1 V 1$ Factor out V2. $( M 2 − M 3 ) V 2 = M 3 V 1 − M 1 V 1$ Divide both sides with the terms in the parenthesis $V 2 = M 3 V 1 − M 1 V 1 ( M 2 − M 3 )$

Substitute the numbers in, keeping in mind that the concentration, [NO3-]=0.832M. $V 2 = ( 0.278M⋅0.225L−0.102M⋅0.225L ) 0.832M−0.278M =0.07148L=71.5mL$

## Gravimetric Analysis

Gravimetric Analysis — Quantitative analysis based on measurement mass of compounds.

In the ppt rxn section, we discussed adding AgNO3 solution to NaCl solution. We came up with its net ionic equation to be,

Ag+(aq) + Cl-(aq) AgCl(s)

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 AgNO3 is required to convert all Cl to AgCl?

$? g C l − =25.00m L AgN O 3 ( 0.1975mo l AgN O 3 1000m L AgN O 3 )( 1mo l A g + 1mo l AgN O 3 )( 1mo l C l − 1mo l A g + )( 35.45 g C l − 1mo l C l − )=0.175034 g C l − =0.175 g C l −$

Example: A 40.00 mL of 0.0348 M Pb(NO3)2 is mixed with 0.00940 M aluminum iodide. a) Write net ionic equation. b) What is the minimum volume of AlI3 needed to complete the reaction? c) How many g of precipitate (product) form?

a)

Pb2+(aq) + 2I-(aq) PbI2(s)

b) $?m L Al I 3 =40.00m L Pb (N O 3 ) 2 ( 0.0348mo l Pb (N O 3 ) 2 1000m L Pb (N O 3 ) 2 )( 1mo l P b 2+ 1mo l Pb (N O 3 ) 2 )( 2mo l I − 1mo l P b 2+ )( 1mo l Al I 3 3mo l I − )( 1000m L Al I 3 0.0094mo l Al I 3 )=98.723mL=98.7m L Al I 3$

c)

$? g Pb I 2 =40.00m L Pb (N O 3 ) 2 ( 0.0348mo l Pb (N O 3 ) 2 1000m L Pb (N O 3 ) 2 )( 1mo l P b 2+ 1mo l Pb (N O 3 ) 2 )( 1mo l Pb I 2 1mo l P b 2+ )( 461.01 g Pb I 2 1mo l Pb I 2 )=0.641726g=0.642 g Pb I 2$

## Acid-Base 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: Acid-Base 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 acid-base reaction is:

H+ + OH- → H2O

$?m{L}_{NaOH}=50.0m{L}_{HCl}\left(\frac{0.50mo{l}_{HCl}}{1000m{L}_{HCl}}\right)\left(\frac{1mo{l}_{{H}^{+}}}{1mo{l}_{HCl}}\right)\left(\frac{1mo{l}_{O{H}^{-}}}{1mo{l}_{{H}^{+}}}\right)\left(\frac{1mo{l}_{NaOH}}{1mo{l}_{O{H}^{-}}}\right)\left(\frac{1000m{L}_{NaOH}}{0.70mo{l}_{NaOH}}\right)=35.714mL=36m{L}_{NaOH}$

Example: Acid-Base 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:

HF + OH F- + H2O

$?m L HF =50.0m L Ba (OH) 2 ( 0.0224mo l Ba (OH) 2 1000m L Ba (OH) 2 )( 2mo l O H − 1mo l Ba (OH) 2 )( 1mo l HF 1mo l O H − )( 1000m L HF 1.72× 10 −2 mo l HF )=130.23mL=130mL$

## Red-Ox Titration

Example: Red-Ox 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 AgNO3 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.

ClO- + S2- S + Cl-
Ag+ + Cl- AgCl

But the first one is red-ox equation because S2- → 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!

red: ClO- + 2H+ + 2e- Cl- + H2

ox: S2- S + 2e-

overall: ClO- + S2- + 2H+ Cl- + S + H2O

but the question stated that it is in basic solution, therefore add 2OH- to get rid of H+. Then,

overall: ClO- + S2- + H2O Cl- + S + 2OH-

From 4.95 g of AgCl, we can calculate the amount of NaClO since we have the balanced chemical equations. $? g NaClO =4.95 g AgCl ( 1mo l AgCl 143.32 g AgCl )( 1mo l C l − 1mo l AgCl )( 1mo l NaClO 1mo l C l − )( 77.44 g NaClO 1mo l NaClO )=2.67463 g NaClO$ but, first convert 50 mL to g $? g bleach =50mL( 1.02g 1mL )=51g$ Then, $2.67463 g NaClO 51 g bleach ×100%=5.2444%=5.24%$