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Lecture Two Chemical Equilibrium Asst .Prof.Dr. Azhar A. Ghali

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Density and specific gravity of solutions.

The numeric value for a density given as g/mL or kg/L is the same. Water with a density at

20oC of 0.998 g/mL has a density of 0.998 kg/L. The density of gases is usually given as g/L

because at normal pressures the density is about a thousand times less than that of liquids. The

density of dry air at 760 torr at 1 atm pressure is 1.185 g/L.

The specific gravity is a ratio between the mass of a given volume and the mass of an equal

volume of water at 4oC. Since the density of water at 4oC is practically 1.00 g/cc, the density

and specific gravity of aqueous solutions are almost identical. The difference needs always to

be considered for analytical work of high precision, however.

DILUTIONS

Whenever you need to go from a more concentrated solution [“stock”] to a less concentrated

one, you add solvent [usually water] to “dilute” the solution. No matter what the units of

concentration are, you can always use this one formula

C1 V1 = C2 V2

[Concentration of the stock] x [Volume of the stock] = [Concentration of the final solution] x

Volume of the final solution]

p-Functions

The p-function of a number X is written as pX and is defined as

pX = –log(X)

X= H + , Cl- , …….etc.

PH= - log [H +]


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POH = - log [OH-]

[H +] + [OH-] = 10-14= Kw

PH + POH = 14

Stoichiometric Calculations: Volumetric Analysis

In this section we look at calculations involved in titration processes as well as general

quantitative reactions. In a volumetric titration, an analyte of unknown concentration is titrated

with a standard in presence of a suitable indicator. For a reaction to be used in titration the

following characteristics should be satisfied:

1. The stoichiometry of the reaction should be exactly known. This means that we should

know the number of moles of A reacting with 1 mole of B.

2. The reaction should be rapid and reaction between A and B should occur

immediately and instantly after addition of each drop of titrant (the solution in the

burette).

3. There should be no side reactions. A reacts with B only.

4. The reaction should be quantitative. A reacts completely with B.

5. There should exist a suitable indicator which has distinct color change.

6. There should be very good agreement between the equivalence point (theoretical)

and the end point (experimental). This means that Both points should occur at the

same volume of titrant or at most a very close volume. Three reasons exist for the

disagreement between the equivalence and end points. The first is whether the

suitable indicator was selected, the second is related to concentration of reactants,

and the third is related to the value of the equilibrium constant. These factors will

be discussed in details later in the course.

Standard solutions

A standard solution is a solution of known and exactly defined concentration. Usually

standards are classified as either primary standards or secondary standards. There are not

too many secondary standards available to analysts and standardization of other substances is

necessary to prepare secondary standards. A primary standard should have the following

properties:


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1. Should have a purity of at least 99.98%

2. Stable to drying, a necessary step to expel adsorbed water molecules before

weighing

3. Should have high formula weight as the uncertainty in weight is decreased when

weight is increased

4. Should be non hygroscopic

5. Should possess the same properties as that required for a titration

NaOH and HCl are not primary standards and therefore should be standardized using a

primary or secondary standard. NaOH absorbs CO2 from air, highly hygroscopic, and usually

of low purity. HCl and other acid in solution are not standards as the percentage written on the

reagent bottle is a claimed value and should not be taken as guaranteed.

Molarity Volumetric Calculations

Volumetric calculations involving molarity are rather simple. The way this information is

presented in the text is not very helpful. Therefore, disregard and forget about all equations

and relations listed in rectangles in the text, you will not need it. What you really need is to

use the stoichiometry of the reaction to find how many mmol of A as compared to the number

of mmoles of B.

Example

A 0.4671 g sample containing NaHCO3 (FW = 84.01 mg/mmol) was dissolved and titrated

with 0.1067 M HCl requiring 40.72 mL. Find the percentage of bicarbonate in the sample.

Solution

We should write the equation in order to identify the stoichiometry

NaHCO3 + HCl = NaCl + H2CO3

Now it is clear that the number of mmol of bicarbonate is equal to the number of mmol HCl

mmol NaHCO3 = mmol HCl

mmol = M x VmL

mmol NaHCO3 = (0.1067 mmol/ml ) x 40.72 mL = 4.345 mmol

Now get mg bicarbonate by multiplying mmol times FW

Mg NaHCO3 = 4.345 mmol x (84.01 mg/mmol) = 365.01


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% NaHCO3 = (365.01 x 10-3 g/0.4671 g) x 100 = 78.14%

We can use dimensional analysis to calculate the mg NaHCO3 directly then get the percentage

as above.

? mg NaHCO3 = (0.1067 mmol HCl/ml) x 40.72 mL x (mmol NaHCO3/mmol HCl) x (84.01

mg NaHCO3/ mmol NaHCO3) = 365.0 mg

Example

A 0.4671 g sample containing Na2CO3 (FW = 106mg/mmol) was dissolved and titrated with

0.1067 M HCl requiring 40.72 mL. Find the percentage of carbonate in the sample.

Solution

The equation should be the first thing to formulate

Na2CO3 +2 HCl = 2NaCl + H2CO3

mmol Na2CO3 = ½ mmol HCl

Now get the number of mmol Na2CO3 = ½ x ( MHCl x VmL (HCl) )

Now get the number of mmol Na2CO3 = ½ x 0.1067 x 40.72 = 2.172 mmol

Now get mg Na2CO3 = mmol x FW = 2.172 x 106 = 230 mg

% Na2CO3 = (230 x 10-3 g/0.4671 g ) x 100 = 49.3 %

Example

How many mL of 0.25 M NaOH will react with 5.0 mL of 0.10 M H2SO4.

H2SO4 + 2 NaOH = Na2SO4 + 2 H2O

Solution

mmol NaOH = 2 mmol H2SO4

mmol NaOH = 2 {M (H2SO4) x VmL (H2SO4)}

mmol NaOH = 2 x 0.10 x 5.0 = 1.0 mmol

mmol NaOH = MNaOH x VmL (NaOH)

1.0 = 0.25 x VmL

VmL = 4.0 mL


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We can also calculate the volume in one step using dimensional analysis

? mL NaOH = (mL NaOH/0.25 mmol NaOH) x (2 mmolNaOH/mmol H2SO4) x (0.10

mmol H2SO4 / mL H2SO4) x 5.0 mL = 4.0 mL

In a previous lecture you were introduced to calculations involved in titrimetric reactions. We

agreed that you do not have to memorize relations listed in the text and the only relation you

need to remember is the mmol. A mmol is mg/FW or molarity times volume (mL). We have

also agreed that writing the equation of the reaction is essential and is the first step in solving

any problem where a reaction takes place. Let us look at more problems and develop a logic

solution using the simple mmol concept.

General Concepts of Chemical Equilibrium

analytical chemistry a branch of chemistry that deals with the identification of compounds

and mixtures (qualitative analysis) or the determination of the proportions of the

constituents (quantitative analysis): techniques commonly used are titration, precipitation,

spectroscopy, chromatography, etc.

In this chapter you will be introduced to basic equilibrium concepts and related calculations.

The type of calculations you will learn to perform in this chapter will be essential for solving

equilibrium problems in later chapters. The ease of solving equilibrium problems will depend

on the skills you will master in this chapter.

The Equilibrium Constant

For chemical reactions that do not proceed to completion, an equilibrium constant can be

written as the quotient of multiplication of molar concentrations of products divided by that of

reactants, each raised to a power equal to its number of moles.

aA + bB = cC + dD

Where, the small letters represent the number of moles of substances A, B, C, and D. The

equilibrium constant is written as:

Keq = ([C]c [D]d)/([A]a [B]b)

In a chemical reaction, as reactants start to react products start to form. Therefore, reactants

continuously decrease and products continuously increase till a point is reached where

eventually no change in concentrations can be detected. This is the point of equilibrium which

is a point where the rate of the forward reaction (product formation) equals the rate


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Of backward reaction (reactants formation). In fact equilibrium implies continuous

transformation between infinitesimally small amounts of products and reactants.

Calculation of Equilibrium Constants

We know from thermodynamics that a reaction occurs spontaneously if it has a negative G

(Gibbs free energy) where:

G = H - TS

Where, H is the change in enthalpy of the reaction and S is the change in entropy. At

standard conditions of temperature and pressure (standard state) we have the standard free

energy Go where:

Go = Ho - TSo

The standard free energy is related to equilibrium constant by the relation:

Go = - RT ln K

K = e-Go/RT

R is the gas constant (8.314 deg K-1mol-1)

It should be clear that Go gives us good information about the spontaneity of the reaction but

it offers no clue on the rate at which the reaction may occur.

Le Chatelier's Principle

The equilibrium concentrations of reactants and products can be changed by applying an

external stress to the system, e.g. increasing or decreasing the concentration of a reactant or

product, changing temperature or pressure. The change occurs in a direction which tends to

counterpart the applied stress. For example:

a. Increasing the temperature of an exothermic reaction will shift the reaction to

left (more reactants and less products). The opposite is observed if the reaction is

exothermic or if heat is removed from an exothermic reaction.

b. In a reaction where the number of gaseous molecules produced is more than

the number of reacting gaseous molecules, increasing the pressure of the system

will shift the reaction toward the reactants in an attempt to decrease the number

of moles and vice versa. Reactions in solutions are usually insensitive to changes

in pressure.

c. Increasing the concentration of a reactant or removing a product will result in

a shift of reaction towards more products and vice versa.

d. Some reactions can be facilitated by addition of a catalyst (a substance that is

not part of reactants or products but its presence makes the reaction faster). The


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catalyst does not change the position of equilibrium but makes the time required to

reach this equilibrium point shorter.

Le Chatelier's principle can be advantageously used to force reactions that are close to

completion to proceed to completion. This is usually done by addition of more reagent in a

gravimetric procedure, allowing a gas to escape if one of the products is a gas, etc..

Calculations Using Equilibrium Constants

Here, we are faced with three situations which should be considered separately:

1. When the equilibrium constant is very small. Most reactants do not undergo

a reaction and very little products are produced. Usually the concentration of

products can be neglected as compared to reactants concentrations.

2. When the equilibrium constant is very high. In this case, most reactants

disappear and the reaction container contains mainly the products. In calculations,

one can neglect the concentration of remaining reactants as compared to

concentrations of products.

3. A situation where the equilibrium constant is moderate. Appreciable

amounts of reactants are left and appreciable amounts of products are also formed

in the reaction mixture. One can not neglect the reactants or the products

concentration. Following are examples on each case.

Example

Calculate the equilibrium concentration of A and B in a 0.10 M solution of weak electrolyte

AB if the equilibrium constant of the reaction is 3.0x10-6.

Solution

AB A + B

First we should look at the value of the equilibrium constant to have an appreciation of

what is going on. It is clear that we have a small equilibrium constant which suggests

that very little products may have been formed and thus we build our solution on the

assumption that AB will mainly remain unreacted except for a very little concentration

x. Therefore, AB concentration will decrease by x and A and B will be formed in a

concentration equals x for each. The following table represents what is happening.


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Before Equilibrium 0.10 0 0

Equation

AB

A

B

At Equilibrium 0.10 -x x x

K = [A][B]/[AB]

Substitution of equilibrium concentration gives

K = (x)(x)/(0.1-x) = 3.0 x 10-6

In order to solve this equation, we should make a justified assumption which we have

discussed above; that is x is very small as compared to 0.10 M and later we will check whether

our assumption is valid or not. Now assume 0.10 >>x. We have

3.0 x 10-6 = x2/0.10, thus we have

x = 5.5 x 10-4 M

Now we should check the validity of our assumption that 0.10 >>x by calculating the

relative error

Relative error = (5.5x10-4/0.10) x 100 = 0.55%

In this course, we will consider our assumption valid if the relative error is below 5%. Therefore, our assumption is valid and we have

[A] =5.5x10-4 M, [B] = 5.5x10-4 M, and [AB] = (0.10 – 5.5x10-4) ~ 0.10 M

Self-Ionization of Water In the self-ionization of water, the amphiprotic ability of water to act as a proton donor and

). In pure water, −OH) and hydroxide ions (+O3Hacceptor allows the formation of hydronium (

C, the o the concentration of hydronium ions equals that of hydroxide ions. At 25

. The ion product of 7−10×1.0 concentrations of both hydronium and hydroxide ions equal

ter and is express as ionization of wa-, is the equilibrium condition for the selfwK water,

follows:

Kw=[H3O+][OH−]=1.0×10−14 .

pH The term pH refers to the "potential of hydrogen ion." It was proposed by Danish biochemist

Soren Sorensen in 1909 so that there could be a more convenient way to describe hydronium

and hydroxide ion concentrations in aqueous solutions since both concentrations tend to be


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extremely small. Sorensen defined pH as the negative of the \logarithm of the concentration of

hydrogen ions. In terms of hydronium ion concentration, the equation to determine the pH of

an aqueous solution is:

pH=−log[H3O+]

pOH The pOH of an aqueous solution, which is related to the pH, can be determined by the

following equation:

pOH =−log[OH−] .

This equation uses the hydroxide concentration of an aqueous solution instead of the

hydronium concentration.

Relating pH and pOH Another equation can be used that relates the concentrations of hydronium and hydroxide

concentrations. This equation is derived from the equilibrium condition for the self-ionization

of water, Kw. It brings the three equations for pH, pOH, and Kw together to show that they are

all related to each other and either one can be found if the other two are known. The following

equation is expressed by taking the negative \logarithm of the Kw expression for the self-

ionization of water:

Kw=[H3O+][OH−]=1.0×10−14

pKw=pH+pOH=14 .


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Acid-Base Equilibria

We study equilibrium problems as related to acids and bases as well as their salts. The main

point is to know how to calculate the hydrogen ion concentration, [H+]. However, let us start

with definitions of acids and bases and then look at their equilibria.

Acid-Base Theories

Four main attempts to define acids and bases are common in the literature of chemistry.

Development of these attempts or theories usually followed a desire to explain the behavior of

substances and account for their properties as related to having acidic or basic characteristics.

Theories of acidity or basicity can be outlined below from oldest to most recent:

1. Arrhenius theory: This theory is limited to water as a solvent where an acid is defined as

a substance which ionizes in water and donates a proton. A base is a substance that

ionizes in water to give hydroxide ions. The hydrogen ion reacts with water to give a

hydronium ion while the base reacts with water to yield a hydroxide ion.

HA + H2O H3O+ + A-

B + H2O BH+ + OH-

2. Franklin Theory: This theory introduced the solvent concept where an acid was defined

as a substance that reacts with the solvent to produce the cation of the solvent . The base

is a substance that reacts with the solvent to yield the anion of the solvent.

HA + EtOH EtOH2+ + A-

B + EtOH BH+ + EtO-

3. Bronsted-Lowry Theory: Some solvents like hexane or benzene are non ionizable and the

Franklin theory can not be used to explain acidic or basic properties of substances. In

Bronsted-Lowry theory, an acid is defined as a substance that can donate a proton while a

base is a substance that can accept a proton. Also, an acid is composed of two components;

a proton and a conjugate base. For example

HOAc H+ + OAc-

Acetic acid is an acid which donates a proton and its proton is associated with a base that can

accept the proton; this base is the acetate.


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4. Lewis Theory: Lewis introduced the electronic theory for acids and bases where in Lewis

theory an acid is defined as a substance that accepts electrons while a base is a substance

that donates electrons. Therefore, ammonia is a base because it donates electrons as in the

reaction

H+ + :NH3 H:NH3+

AlCl3 is an acid because it accepts electrons from a base such as :OR2

AlCl3 + : OR2 = Cl3Al:OR2

Acid-Base Equilibria in Water

Fortunately, we will only deal with aqueous solutions which means that water will always be

our solvent. Water itself undergoes self ionization as follows

2 H2O H3O+ + OH-

K = [H3O+][OH-]/[H2O]2

However, only a very small amount of water does ionize and the overall water concentration

will be constant. Therefore, one can write

Kw = [H3O+][OH-]

Kw is called autoprotolysis constant of water or ion product of water, we will also refer to

[H3O+] as simply [H+] although this is not strictly correct due to the very reactive nature of H+

Kw = [H+][OH-] = 10-14 at 25 oC.

The pH Scale

In most cases, the hydrogen ion concentration is very small which makes it difficult to

practically express a meaningful concept for such a small value. Currently, the pH scale is

used to better have an appreciation of the value of the hydrogen ion concentration where:

pH = - log [H+]

We also know that kw = [H+][OH-] = 10-14 or

pH + pOH = 14

Therefore, calculation of either pH or pOH can be used to find the other.


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We are faced with different types of solutions that we should know how to calculate the pH or

pOH for. These include calculation of pH for

1. Strong acids and strong bases

2. Weak acids (monoprotic) and weak bases (monobasic)

3. Salts of weak acids and salts of weak bases

4. Mixtures of weak acids and their salts (buffer solutions)

5. Polyprotic acids and their salts and polybasic bases and their salts

We shall also look at pH calculations for mixtures of acids and bases as well as pH

calculations for very dilute solutions of the abovementioned systems.

pH calculations

1. Strong Acids and Strong Bases

Strong acids and strong bases are those substances which are completely dissociated in water

and dissociation is represented by one arrow pointing to right. Examples of strong acids

include HCl, HNO3, HClO4, and H2SO4 (only first proton). Examples of strong bases include

NaOH, KOH, Ca(OH)2, as well as other metal hydroxides.

Example

Find the pH of a 0.1 M HCl solution.

Solution

HCl is a strong acid that completely dissociates in water, therefore we have

HCl H+ + Cl-

H2O H+ + OH-

[H+]Solution = [H+]from HCl +[H+]from water

However, [H+]from water = 10-7 in absence of a common ion, therefore it will be much less in

presence of HCl and can thus be neglected as compared to 0.1 ( 0.1>>[H+]from water)

[H+]solution = [H+]HCl = 0.1

pH = -log 0.1 = 1


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We can always look at the equilibria present in water to solve such questions, we have only

water that we can write an equilibrium constant for and we can write:

Before Equilibrium

0.1

0

Equation H2O H+ OH-

At Equilibrium 0.1 + x x

Kw = (0.1 + x)(x)

However, x is very small as compared to 0.1 (0.1>>x)

10-14 = 0.1 x

x = 10-13 M

Therefore, the[OH-] = 10-13 M = [H+]from water

Relative error = (10-13/0.1) x 100 = 10-10%

[H+] = 0.1 + x = 0.1 + 10-13 ~ 0.1

pH = 1

Weak Acid Equilibrium

When an uncharged weak acid is added to water, a homogeneous equilibrium

forms in which aqueous acid molecules, HA(aq), react with liquid water to form

aqueous hydronium ions and aqueous anions, A-(aq). The latter are produced when

the acid molecules lose H+ ions to water.

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

In writing an equilibrium constant expression for this homogeneous equilibrium,

we leave out the concentration of the liquid water. The equilibrium constant for this

expression is called the acid dissociation constant, Ka.

= acid dissociation constant

When the equilibrium in question occurs in solution, the chemical formulas

enclosed in brackets in the equilibrium constant expression represent the molarities

of the substances (moles of solute per liter of solution).

Remember that H+ can be used to represent H3O+, thus simplifying our depiction of

the reaction between a weak acid and water and its acid dissociation constant

expression:


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HA(aq) H+(aq) + A-(aq)

= acid dissociation constant

For example, acetic acid is a weak acid, because when it is added to water, it reacts

with the water in a reversible fashion to form hydronium and acetate ions.

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)

or HC2H3O2(aq) H+(aq) + C2H3O2-

(aq)

= 1.8 × 10-5

EXAMPLE 1 - Writing an Acid Dissociation Constant: Write the equation for the

reaction between the weak acid nitrous acid and water, and write the expression for

its acid dissociation constant.

Solution:

HNO2(aq) + H2O(l) H3O+(aq) + NO2-(aq)

or HNO2(aq) H+(aq) + NO2-(aq)

The table below lists acid dissociation constants for some common weak acids.

These Ka values can be used to describe the relative strength of the acids. A

stronger acid will generate more hydronium ions in solution. A larger Ka indicates

a greater ratio of ions (including hydronium ions) to uncharged acid. Therefore, a

larger Ka indicates a stronger acid. For example, the larger Ka for chlorous acid

(1.2 × 10-2) compared to acetic acid (1.8 × 10-5) tells us that chlorous acid is

stronger than acetic acid.


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Acid Dissociation Constants, Ka, for Common Weak Acids

Weak Acid Equation Ka

acetic acid HC2H3O2 H+ + C2H3O2- 1.8 × 10-5

benzoic acid C6H5CO2H

H+ + C6H5CO2-

6.4 × 10-5

chlorous acid HClO2 H+ + ClO2- 1.2 × 10-2

formic acid HCHO2 H+ + CHO2- 1.8 × 10-4

hydrocyanic

acid

HCN H+ + CN- 6.2 × 10-10

Hydrofluoric HF H+ + F- 7.2 × 10-4

hypobromous

acid

HOBr H+ + OBr- 2 × 10-9

hypochlorous

acid

HOCl H+ + OCl- 3.5 × 10-8

hypoiodous

acid

HOI H+ + OI- 2 × 10-11

lactic acid CH3CH(OH)CO2H

H+ + CH3CH(OH)CO2-

1.38 × 10-4

nitrous acid HNO2 H+ + NO2- 4.0 × 10-4

phenol HOC6H5 H+ + OC6H5- 1.6 × 10-10

propionic

acid

CH3CH2CO2H

H+ + CH3CH2CO2-

1.3 × 10-5

The following study sheet describes one procedure for calculating the pH of

solutions of weak acids. If you take other chemistry courses, you will find that

there are variations on this procedure for some weak acid solutions.


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Calculating pH for Weak Acid Solutions

You are given the concentration of a weak acid solution and asked to calculate its

pH.

General Steps -

STEP 1 Write the equation for the ionization of the weak acid in water.

HA(aq) H+(aq) + A-(aq)

STEP 2 Write the Ka expression for the weak acid.

STEP 3 Describe each equilibrium concentration in terms of x.

x = [H+]equilibrium = [A-]equilibrium

[HA]equilibrium = [HA]initial - x

STEP 4 Assume that the initial concentration of weak acid is approximately equal

to the equilibrium concentration. (Weak acids are rarely ionized to a large degree.

We can most often assume that the initial concentration added, [HA]initial is much

larger than x. Thus, the equilibrium concentration is approximately equal to the

concentration added. You may learn how to deal with weak acid solutions for

which this approximation is not appropriate in other chemistry courses.)

[HA]equilibrium = [HA]initial

STEP 5 Plug the concentrations described in terms of x into the Ka expression, and

solve for x.


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EXAMPLE 2 - pH Calculations for Weak Acid Solutions: Vinegar is a dilute

water solution of acetic acid with small amounts of other components. Calculate

the pH of bottled vinegar that is 0.667 M HC2H3O2, assuming that none of the other

components affect the acidity of the solution.

HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

We get the value for the acid dissociation constant for this reaction from the table

above.

x2 = 1.2 × 10-5 x = 3.5 × 10-3

[H+] = 3.5 × 10-3 M H+ pH = -log(3.5 × 10-3) = 2.46

Density and specific gravity of solutionsqu.edu.iq/ph/wp-content/uploads/2015/03/L-2-.pdf ·...

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