Density and Specific Gravity of Solutions

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

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 +]

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

POH = - log [OH-]

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

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

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 11

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

-3 % NaHCO3 = (365.01 x 10 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

-3 % Na2CO3 = (230 x 10 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 12

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

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:

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

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

O f 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 - TS 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 - TSo 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

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

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.

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

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 + − acceptor allows the formation of hydronium (H3O ) and hydroxide ions (OH ). In pure water, the concentration of hydronium ions equals that of hydroxide ions. At 25 oC, the concentrations of both hydronium and hydroxide ions equal 1.0×10−7. The ion product of water, Kw, is the equilibrium condition for the self-ionization of water and is express as follows:

+ − −14 Kw=[H3O ][OH ]=1.0×10 .

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

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

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:

+ − −14 Kw=[H3O ][OH ]=1.0×10

pKw=pH+pOH=14 .

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

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.

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

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

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

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

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

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

+ -7 However, [H ]from water = 10 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

Lecture Two Chemical Equilibrium Asst .Prof.Dr. Azhar A. Ghali 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 - -13 + Therefore, the[OH ] = 10 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:

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

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.

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

Acid Dissociation Constants, Ka, for Common Weak Acids

Weak Acid Equation Ka + - -5 acetic acid HC2H3O2 H + C2H3O2 1.8 × 10 -5 C6H5CO2H 6.4 × 10 benzoic acid + - H + C6H5CO2 + - -2 chlorous acid HClO2 H + ClO2 1.2 × 10 + - -4 formic acid HCHO2 H + CHO2 1.8 × 10 + - -10 hydrocyanic HCN H + CN 6.2 × 10 acid + - -4 Hydrofluoric HF H + F 7.2 × 10

+ - -9 hypobromous HOBr H + OBr 2 × 10 acid + - -8 hypochlorous HOCl H + OCl 3.5 × 10 acid + - -11 hypoiodous HOI H + OI 2 × 10 acid -4 CH3CH(OH)CO2H 1.38 × 10 lactic acid + - H + CH3CH(OH)CO2 + - -4 nitrous acid HNO2 H + NO2 4.0 × 10 + - -10 phenol HOC6H5 H + OC6H5 1.6 × 10 -5 CH3CH2CO2H 1.3 × 10 propionic + - acid H + CH3CH2CO2

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.

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

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.

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

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