Chapter 4 Chemical Rxns in Aqueous Solution

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

cation

anion

Strong electrolyte

₃

₂

Weak electrolyte

₃

₂

Nonelectrolyte

₂

₆

₁₂

₆

In the figure above, the solution in the beaker is incorporated as a part of electrical circuit. The solutions in the beaker are: (a) nonelectrolyte, (b) weak electrolyte, and (c) strong electrolyte

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

→

⁺

₃

→

⁺

₃

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.

molecular eqn	NaCl + AgNO₃ → AgCl(s) + NaNO₃
ionic eqn	Na⁺(aq) + Cl^-(aq) + Ag⁺(aq) + NO₃^-(aq) → Na⁺(aq) + AgCl(s) + NO₃^-(aq)
net ionic eqn	Ag⁺(aq) + Cl^-(aq) → AgCl(s)

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, thus precipitate 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^-	SO₄^2-	OH^-	CO₃^2-	PO₄^3-
Gr 1 cations NH₄⁺	sol	sol	sol	sol	sol	sol	sol	sol
Gr 2 cations	sol	sol	sol	sol	BaSO₄	Mg(OH)₂	insol	insol
TM	sol	AgX Hg₂X₂ PbX₂	Ba(ClO₄)₂ Hg₂(ClO₄)₂ Pb(ClO₄)₂	Ba(CH₃CO₂)₂ Hg₂(CH₃CO₂)₂ Pb(CH₃CO₂)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: Write molecular, ionic and net ionic equations for adding the following compounds in water: a) HCl and AgNO₃ b) K₃PO₃ and CaNO₃ c) AgI and NH₄ClO₄

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)

Acids, Bases, and Neutralization Rxns

As seen in Chapter 3,

Acid produces H⁺

Base produces OH^-

Acid + Base → H₂O — Neutralization rxn

If you write down molecular equation for acid and base reaction, you would get:

acid + base → water + salt Salt is a product of conjugate base and conjugate acid of the starting acid and base.

Strong Acid—Complete dissociation into H⁺ and anion

⁺

Strong Base—Complete dissociation into cation and OH^-

⁺

Weak Acid—Incomplete dissociation into H⁺ and anion

₂

₃

₂

⁺

₂

₃

₂

said to be in equilibrium

Weak Base—Incomplete dissociation into cation and OH^-

₃

₂

₄

⁺

Neutralization Rxns
Three circumstances:

1) Strong acid reacting with strong base

⁺

→

₂

2) Weak acid (HA) reacting with strong base

⁺

→

₂

Combining the two yields

→

₂

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

₂

→

⁺

→

₂

Combining the two, yields

⁺

→

⁺

The following lists the common acids and bases that you encounter in this course.

Common Acids and Bases with their strength

Example: Write net ionic equation for adding Sr(OH)₂ and HF in aqueous environment.

Since Sr(OH)₂ is a strong base, and HF is a week acid, we have:

HF ⇄ H⁺ + F^-

H⁺ + OH^- → H₂O

HF + OH^- → H₂O + F^-

Example: Write net ionic equation for adding Ba(OH)₂(aq) and CH₃CO₂H(aq).

Acetic acid CH₃CO₂H is a weak acid, thus, CH₃CO₂H ⇄ H⁺ + CH₃CO₂^-

H⁺ + OH^- → H₂O

CH₃CO₂H + OH^- → H₂O + CH₃CO₂^-

Example: Write net ionic equation for adding NH₃(aq) and HCl(aq).

Ammonia NH₃ is a weak base, thus, NH₃ + H₂O ⇄ NH₄⁺ + OH^-

H⁺ + OH^- → H₂O

NH₃ + H⁺ → NH₄⁺ + OH^-

Example: 4.33(b) Balance the following equation and write the ionic and net ionic equation. Ba(OH)₂(aq) + H₃PO₄(aq) →

Ba(OH)₂ is strong base, H₃PO₄ is weak acid. Thus, H₃PO₄ ⇄ H⁺ + H₂PO₄^- ⇄ 2H⁺ + HPO₄^2- ⇄ 3H⁺ + PO₄^3-

H⁺ + OH^- → H₂O

For net ionic equation we add these two equations, nad we get

H₃PO₄ + 3OH^- → PO₄^3- + 3H₂O In order to obtian ionic equation you need to put Ba²⁺ ions in. So, you can do: H₃PO₄ + 1.5Ba²⁺ + 3OH^- → 1.5Ba²⁺ + PO₄^3- + 3H₂O Or you can make 1.5 into whole number 2H₃PO₄ + 3Ba²⁺ + 6OH^- → 3Ba²⁺ + 2PO₄^3- + 6H₂O

Reduction—Oxidation Rxns (Redox rxns)

Electron transfer between two different species

Oxidation

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.

oxidation number

2) Separate the equation into reduction and oxidation half rxns

a) Balance all atoms other than H and O

₂

⁺

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.

Priority	Atom types	Assigned Ox#	Examples
1	Atoms in their element state	0	Br₂ 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 MgCl₂, ox#_Mg=+2
5	F in compounds	-1	CH₂F₂, ox#_F=-1
6	H in compounds	+1	H₂O each H, ox#=+1
7	O in compounds	-2	H₂O each O, ox#=-2
8	In 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, FeO₄^2-

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

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

The following section, to balance the red-ox equation, has been moved to CHM4. Thus, you're no longer responsible for balancing the red-ox equation!

I'll give you the balanced red-ox equation for this section. However, you are responsible for which species is reduced or oxidized according to the oxidation number (ox#) and are responsible to deduce the unbalanced half-reactions.

Balancing Half-Rxns

The 4 steps to balance half-rxn:

a) Balance all atoms other than H and O

₂

⁺

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) + Cr₂O₇^2-(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 Cr₂O₇^2- → Cr³⁺

ox C₂H₅OH → CO₂

Balance atoms other than O and H.

red Cr₂O₇^2- → 2Cr³⁺

ox C₂H₅OH → 2CO₂

Balance O by adding H₂O.

red Cr₂O₇^2- → 2Cr³⁺ + 7H₂O

ox C₂H₅OH + 3H₂O → 2CO₂

Balance H by adding H⁺.

red Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7H₂O

ox C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺

Balance the charge.

red Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

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

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

2Cr₂O₇^2- + 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 mL_{K₂Cr₂O₇}.

?g_C₂H₅OH = 8.76 mL_{K₂Cr₂O₇}

$? g_{C_{2} H_{5} O H} = 8.76 m L_{K_{2} C r_{2} O_{7}} (\frac{0.04988 m o l_{K_{2} C r_{2} O_{7}}}{1000 m L_{K_{2} C r_{2} O_{7}}}) (\frac{1 m o l_{C r_{2} O_{7}^{2 -}}}{1 m o l_{K_{2} C r_{2} O_{7}}}) (\frac{1 m o l_{C_{2} H_{5} O H}}{2 m o l_{C r_{2} O_{7}^{2 -}}}) (\frac{46.07 g_{C_{2} H_{5} O H}}{1 m o l_{C_{2} H_{5} O H}}) = 0.010065 g_{C_{2} H_{5} O H}$

Therefore, we have $%_{a l c o h o l} = \frac{0.010065 g_{C_{2} H_{5} O H}}{10.002 g_{b l o o d}} 100 % = 0.100631 % = 0.101 %$

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.

$M \equiv \frac{m o l_{s o l u t e}}{L_{s o l u t i o n}} = \frac{m o l_{s o l u t e}}{1000 m L_{s o l u t i o n}}$

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 #1

You have a 30 mL solution of 1.0 M NaCl. How many moles of NaCl are in the solution?

$? m o l_{N a C l} = 30 m L_{N a C l} (\frac{1.0 m o l_{N a C l}}{1000 m L_{N a C l}}) = 0.03 m o l_{N a C l}$

Example: Molarity #2

You have a 1.0 M solution of NaCl. How many mL are there if the solution contains 0.02 mol of NaCl? $? m L_{N a C l} = 0.02 m o l_{N a C l} (\frac{1000 m L_{N a C l}}{1.0 m o l_{N a C l}}) = 20 m L_{N a C l}$

Example: Molarity #3

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

$? m o l_{O H^{-}} = 50.00 m L_{N a O H} (\frac{3.50 \times 10^{- 2} m o l_{N a O H}}{1000 m L_{N a O H}}) (\frac{1 m o l_{O H^{-}}}{1 m o l_{N a O H}}) = 1.75 \times 10^{- 3} m o l_{O H^{-}}$

Example: Molarity #4

You have 50.00 mL solutioin of 2.0x10^-2M NaOH. How many g of OH^- ion are there in the solution?

$? g_{O H^{-}} = 50.0 m L_{N a O H} (\frac{2.0 \times 10^{- 2} m o l_{N a O H}}{1000 m L_{N a O H}}) (\frac{1 m o l_{O H^{-}}}{1 m o l_{N a O H}}) (\frac{17.008 g_{O H^{-}}}{1 m o l_{O H^{-}}}) = 0.17008 g_{O H^{-}} = 0.17 g_{O H^{-}}$

Molarity can be used in conjunction with the chemical equation to deduce, for example, how much product produced, how much reactant is needed, etc. Let us look at few.

Example: Molarity #5

Given the following reaction,

AgNO₃(aq) + NaBr(s) → AgBr(s) + NaNO₃(aq) if a 20.0 mL of 5.0 X 10^-2M AgNO₃ is reacted with NaBr, how many g of AgBr is produced?

Example: Molarity #6

Given the following reaction,

AgNO₃(aq) + NaBr(s) → AgBr(s) + NaNO₃(aq) if a 0.20 g of AgBr is produced, what is the volume in mL of AgNO₃ required, if the AgNO₃ solution has a concentration of 5.0 X 10^-2M?

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?

$? m L_{H C l_{c o n c}} = 200.0 m L_{H C l} (\frac{3.00 \times 10^{- 2} m o l_{H C l}}{1000 m L_{H C l}}) (\frac{1000 m l_{H C l_{c o n c}}}{0.500 m o l_{H C l_{c o n c}}}) = 12.0 m L_{H C l_{c o n c}}$

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?

$? m L_{B a {(O H)}_{2}} = 200.0 m L_{O H^{-}} (\frac{3.00 \times 10^{- 2} m o l_{O H^{-}}}{1000 m L_{O H^{-}}}) (\frac{1 m o l_{B a {(O H)}_{2}}}{2 m o l_{O H^{-}}}) (\frac{1000 m l_{B a {(O H)}_{2}}}{0.500 m o l_{B a {(O H)}_{2}}}) = 6.00 m L_{B a {(O H)}_{2}}$

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).

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^{-}} = 25.00 m L_{A g N O_{3}} (\frac{0.1975 m o l_{A g N O_{3}}}{1000 m L_{A g N O_{3}}}) (\frac{1 m o l_{A g^{+}}}{1 m o l_{A g N O_{3}}}) (\frac{1 m o l_{C l^{-}}}{1 m o l_{A g^{+}}}) (\frac{35.45 g_{C l^{-}}}{1 m o 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?

Pb²⁺(aq) + 2I^-(aq) → PbI₂(s)

b) $? m L_{A l I_{3}} = 40.00 m L_{P b {(N O_{3})}_{2}} (\frac{0.0348 m o l_{P b {(N O_{3})}_{2}}}{1000 m L_{P b {(N O_{3})}_{2}}}) (\frac{1 m o l_{P b^{2 +}}}{1 m o l_{P b {(N O_{3})}_{2}}}) (\frac{2 m o l_{I^{-}}}{1 m o l_{P b^{2 +}}}) (\frac{1 m o l_{A l I_{3}}}{3 m o l_{I^{-}}}) (\frac{1000 m L_{A l I_{3}}}{0.0094 m o l_{A l I_{3}}}) = 98.723 m L = 98.7 m L_{A l I_{3}}$

$? g_{P b I_{2}} = 40.00 m L_{P b {(N O_{3})}_{2}} (\frac{0.0348 m o l_{P b {(N O_{3})}_{2}}}{1000 m L_{P b {(N O_{3})}_{2}}}) (\frac{1 m o l_{P b^{2 +}}}{1 m o l_{P b {(N O_{3})}_{2}}}) (\frac{1 m o l_{P b I_{2}}}{1 m o l_{P b^{2 +}}}) (\frac{461.01 g_{P b I_{2}}}{1 m o l_{P b I_{2}}}) = 0.641726 g = 0.642 g_{P b I_{2}}$

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

$? m L_{N a O H} = 50.0 m L_{H C l} (\frac{0.50 m o l_{H C l}}{1000 m L_{H C l}}) (\frac{1 m o l_{H^{+}}}{1 m o l_{H C l}}) (\frac{1 m o l_{O H^{-}}}{1 m o l_{H^{+}}}) (\frac{1 m o l_{N a O H}}{1 m o l_{O H^{-}}}) (\frac{1000 m L_{N a O H}}{0.70 m o l_{N a O H}}) = 35.714 m L = 36 m L_{N a O H}$

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

$? m L_{H F} = 50.0 m L_{B a {(O H)}_{2}} (\frac{0.0224 m o l_{B a {(O H)}_{2}}}{1000 m L_{B a {(O H)}_{2}}}) (\frac{2 m o l_{O H^{-}}}{1 m o l_{B a {(O H)}_{2}}}) (\frac{1 m o l_{H F}}{1 m o l_{O H^{-}}}) (\frac{1000 m L_{H F}}{1.72 \times 10^{- 2} m o l_{H F}}) = 130.23 m L = 130 m L$

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^- + S^2- → S + Cl^- Ag⁺ + Cl^- → AgCl

But the first one is red-ox 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!

red: ClO^- + 2H⁺ + 2e^- → Cl^- + H₂

ox: S^2- → S + 2e^-

overall: ClO^- + S^2- + 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^- + S^2- + H₂O → Cl^- + S + 2OH^-

From 4.95 g of AgCl, we can calculate the amount of NaClO since we have the balanced chemical equations. $? g_{N a C l O} = 4.95 g_{A g C l} (\frac{1 m o l_{A g C l}}{143.32 g_{A g C l}}) (\frac{1 m o l_{C l^{-}}}{1 m o l_{A g C l}}) (\frac{1 m o l_{N a C l O}}{1 m o l_{C l^{-}}}) (\frac{77.44 g_{N a C l O}}{1 m o l_{N a C l O}}) = 2.67463 g_{N a C l O}$ but, first convert 50 mL to g $? g_{b l e a c h} = 50 m L (\frac{1.02 g}{1 m L}) = 51 g$ Then, $\frac{2.67463 g_{N a C l O}}{51 g_{b l e a c h}} \times 100 % = 5.2444 % = 5.24 %$

Chapter 4 Rxns in Aqueous Solutions

Aqueous solution chemistry

General Properties of Aqueous Solns

Electrolytes

Precipitation Rxns and Solubility

Acids, Bases, and Neutralization Rxns

Reduction—Oxidation Rxns (Redox rxns)

Molarity, unit of concentration

Dilution

Gravimetric Analysis

Acid-Base Titration

Red-Ox Titration