CHEM191 Module 1 - Chemical Reactions In Aqueous Solutions (Notes) PDF

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CHEM191 Module 1Chemical Reactions In Aqueous SolutionsLecture 1StoichiometryChemical EquationsA chemical equation tells us about the ratio in which the reactants in a chemical reaction react to give the products. It is concerned with numbers. For example:This equation tells us that 1 molecule of ni...

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CHEM191 Module 1 Chemical Reactions In Aqueous Solutions

Lecture 1 Stoichiometry Chemical Equations A chemical equation tells us about the ratio in which the reactants in a chemical reaction react to give the products. It is concerned with numbers. For example:

N2(g) + 3H2(g) → 2NH3(g) This equation tells us that 1 molecule of nitrogen will react with 3 molecules of hydrogen to give 2 molecules of ammonia. These equations must be balanced, meaning there is the same number of each type of atom on each side.

Moles & Molar Mass The mole (mol) is the SI unit of amount of substance. It is defined as the amount of substance that contains the same number of specified entities as there are atoms in exactly 12g of the carbon isotope 12C. There are 6.022 x 1023 in exactly 12g of the carbon isotope 12C, therefore 1 mole of anything contains 6.022 x 1023 entities. A mole of a bunch of different substances will contain the same number of entities, but they will all have different masses as the mass of each individual entity is different. The molar mass is what we use to describe the mass of 1 mole of a specified entity. The mass, molar mass, and number of moles are given by this equation:

n=

m M

Number of moles is measured in moles (mol), mass is measured in grams (g), and molar mass is measured in grams per mole (g mol-1). The molar mass of every atom is given on the periodic table. Mole Relationships Balance chemical equations give mole relationships between all reactants and products. Let's say we are trying to find the mass of the amount of product (B) for a particular reaction, when given the mass of the reactant (A):

A→B If we are given the grams of substance A, we use the molar mass of substance A to find the number of moles of substance A. Then using the coefficients of A and B from the balanced chemical equation, we can work out the number of moles of substance B. Then using the molar mass of substance B, we can find the grams of substance B.

Limiting Reagents Sometimes, the reactants are not present in exact stoichiometric amounts. In these cases, 1 of the reactants will be a limiting reagent. For example:

2H2(g) + O2(g) → 2H2O(l ) If 5.0g of hydrogen is reacted with 21.0g of oxygen, there is a mass of 1 of the reactants that remain after the reaction is complete, because the other reactant has been all used up. To work out what reactant has been used up, we must work out which reactant gives the smallest amount of product (from the stoichiometry of the equation).

5.0g = 2.5m ol 2.0g m ol −1 21.0g n(O2) = = 0.66m ol 32g m ol −1 n(H2) =

Because of the stoichiometry, we know 2.5mol of hydrogen will give 25mol of water, and 0.66mol of oxygen will give 1.32mol of water. Therefore, when 1.32mol of water has been produced, all of the oxygen would have been used up, so the reaction stops. This leaves an excess of hydrogen. When 1.32mol of water has been produced, 1.32mol of hydrogen would have been used up, so the remaining hydrogen is 1.18mol. The mass of hydrogen remaining is therefore:

m(H2) = m × n = 1.18m ol × 2.0g m ol −1 = 2.4g

Concentration We can define the composition of a solution in terms of its concentration, the amount of solute per volume of solvent:

c=

n v

Concentration is measured in moles per litre (mol L-1).

Lecture 2 Aqueous Solutions Water Molecules Aqueous solutions are important in biological systems. The solubility of substances in aqueous biological solutions is crucial in the transport of gases, the transport of inorganic materials, and the application and transport of drugs. To understand why some substances are soluble in water and why some are not, we need to first look at the molecule of water itself. A water molecule contains 2 O-H bonds. Because the oxygen atom is a much more electronegative element than the hydrogen atom, the pair of electrons in the bond are not equally shared. Instead, they spend more time surrounding the oxygen atom. This gives the oxygen atom a partial negative charge or dipole (δ-) and consequently the hydrogen atom gains a partial positive charge or dipole (δ+). Therefore the molecule is polar. Water molecules are attracted to each other by electrostatic interactions between these partial charges (dipole-dipole interactions). These dipole-dipole interactions are found in many molecules. However, a small number of these molecules have particularly strong interactions, which we call hydrogen bonding. The polar water molecule engages in extensive hydrogen bonding, both in solid and solution states. Other polar molecules and ionic solids tend to be soluble in water. Water Properties Due to the extensive hydrogen bonding of water, it has very high melting and boiling points compared to very similar molecules such as ammonia and methane. Water also gets denser when it melts, unlike some other molecules. This is because in liquid water, the molecules move around close to each other, continuously making and breaking hydrogen bonds with their neighbours. In solid water, the molecules are rigidly held in a structure which maximises the possible number of hydrogen bonds. This structure is more open, making solid water more denser liquid water.

Aqueous Solubility Before getting into detail about solubility in aqueous solutions, let’s define some terms: - Solution - formed when 1 or more chemical species dissolves in a liquid.

-

Solute - a substance that dissolves in a liquid (can be solid, liquid, or gas). Solvent - a liquid in which the solute dissolves. Dilute solution - a solution which much more moles of solvent than there is solute. Electrolyte - a substance that dissolves to give ions in solution (such as sodium chloride) Non-electrolyte - a substance that dissolves without the formation of ions in solution (such as glucose).

Water is an excellent solvent for ionic solids, for example, the dissolution of sodium chloride: + + Cl − NaCl(s) → Na(aq) (aq)

This dissolution occurs spontaneously, despite the fact that the electrostatic forces holding sodium chloride together are very strong. Ions by themselves are a very high energy species, but solvation helps to stabilise the ions. Solvation involves interactions of the charged ions with the dipoles of the water molecules. All ions in water are surrounded by a number of water molecules. When water is the solvent, this process is called hydration. Water is also a good solvent for polar molecules (ones that also have polar bonds due to large differences in electronegativity between atoms in a bond). This includes solids and liquids such as:

Ionic solids, acids, and bases, are all electrolytes, as they produce ions when dissolved. Polar molecules which do not produce ions when dissolved, instead have their intermolecular hydrogen bonds broken, without any change to the bonding within the molecule itself. Non-polar molecules such as oxygen and nitrogen molecules are generally not very soluble in water. However, they do have some solubility. This is due to the fact that the permanent dipole of water can induce a dipole in the nonpolar molecule. The magnitude of the induced dipole depends on the number of electrons and the shape of the non-polar molecule. Strong & Weak Electrolytes Strong electrolytes undergo complete dissociation in water. When the compound dissolves, all the dissolved species are ions, so in other words, the reaction goes to completion. Even ionic solids which are not very water-soluble (like silver chloride), are classified as strong electrolytes, as all the solid that dissolves is completely dissociated into ions. Weak electrolytes undergo incomplete dissociation in water. When the compound dissolves, most of the dissolved species are molecules, some of which then dissociate to give ions. Here, the dissociation reaction does not go to completion. Weak acids are classic examples of weak electrolytes.

Lecture 3 Equilibrium Reaction Quotient Weak electrolytes undergo incomplete dissociation in water. When the compound dissolves, most of the dissolved species are molecules, some of which then dissociate to give ions. Weak acids, such as acetic acid, are good examples:

CH3COOH(l) → CH3COOH(aq) + H2O(l) ⇌ CH3COO(−aq) + H3O(+aq)

This is the case with many chemical reactions. We need a way to quantify the amounts of the reactants and products present in the reaction mixture over time. This can be done with the reaction quotient. For the reaction:

a A + bB → cC + d D [C ]c[D]d Q= [A ]a[B ]b The square brackets represent concentrations. The concentrations are then raised to the power of the stoichiometric coefficients. At some point during these reactions, the value of Q will stop changing. This does not mean that the reaction has gone to completion, as there are still reactant molecules present. What it tells us is the reaction is going both ways at the same time. We say that this is at equilibrium (denoted by the equilibrium arrow sign).

Equilibrium Constant We can quantify a system at equilibrium by using the equilibrium constant (K). At equilibrium, Q is equal to K. Using the concentrations when the system is at equilibrium, we can find the equilibrium constant:

N2O2(g) ⇌ 2NO2( g) [NO2]2 K= [N2O4] = 2.7 When Q of a reaction mixture is less than the value of K, reactants are converted into products until equilibrium is reached. The opposite is also true, so if Q of a reaction mixture is more than the value of K, products are converted into reactants until equilibrium is reached. Therefore, no matter what concentrations we start with, the equilibrium mixture will always be the same, unless there is a temperature change. The value of K (usually given at 25°C) gives information about the extent of the reaction at equilibrium. If: - The value of K is very high, then the reaction essentially goes to completion, and the position of equilibrium lies far to the right. - The value of K is very low, then the reaction barely begins, and the position of equilibrium lies far to the left. It is important to note that pure solids and pure liquids don’t appear in an equilibrium constant expression. For example:

2Na HCO3(s) ⇌ Na2CO3(s) + H2Og + CO2(g) K = [H2O][CO2] Both sodium bicarbonate and sodium carbonate are solids so they are not included in the expression. Water is a gas here, so it is included, but often it is in liquid form, so would not be included. Concentration Tables It is very useful to be able to construct a concentration table in order to solve problems. For example:

H2(g) + I2(g) ⇌ 2HI(g) 0.100mol of H2(g) and 0.100mol of I2(g) were placed in a 1.00L flask. When equilibrium was established, [I2] = 0.020mol L-1 so what is K for this reaction? Concentration

H2(g)

I2(g)

HI(g)

Initial

0.100

0.100

0.00

Change

-0.080 (0.100 - 0.010)

-0.080 (0.100 - 0.010)

0.160 (2 x 0.080)

Equilibrium

0.020

0.020

0.160

From the stoichiometry, we are able to work out the equilibrium concentrations of each of the reactants and products. From there, we can substitute these concentrations into our equilibrium constant expression:

(0.160)2 (0.020)(0.020) = 64

K=

Lecture 4 Equilibria & Solubility Response To Change We need to know how a system at equilibrium responds to change in amounts of products or reactants, and changes in pressure. We can do this by comparing values of Q and K.

A⇌B [B] Q= [A] If more A is added, then Q will decrease below K. This means that more B will be produced and more A will be used up to restore the system back to equilibrium. If more B, is added, then Q will increase above K. This means that more A will be produced and more B will be used up to restore the system back to equilibrium. Note that K does not change, only Q changes. We can change the pressure by changing the volume, or adding an inert gas. Take for example:

CaCO3(s) ⇌ CaO(s) + CO2(g) Remembering the equation for concentration, we can write a reaction quotient expression:

Q = [CO2] n(CO2) = V = n(CO2) ×

1 V

This shows us that Q is proportional to 1 over the volume. So if volume increases, Q will decrease below K. This causes equilibrium to move to the right, and so more carbon dioxide is produced. Changing the pressure by changing the volume will have no effect on an equilibrium system with equal numbers of moles of gas on either side, as the volume terms cancel out in the expression for Q. Changing the pressure by adding an inert gas (like helium) will have no effect on the position of equilibrium, as there is no term for the inert gas in the expression for Q (it is not involved in the reaction).

Solubility Solubility (s) is a measure of how much solute will dissolve in a given volume of solvent. A saturated solution is formed when the maximum amount of solute is dissolved in a particular volume of solvent at a temperature T. The following equilibrium exists in a saturated solution:

sol ute(s) ⇌ sol ute(aq) The solubility of a solute is the amount of solute that will dissolve in 1L of water. The solubility depends on:

- The chemical nature of solute and solvent. - Temperature. - Pressure (for gaseous solutes). Solubility Product We can find an equilibrium constant for the dissolution of a slightly soluble salt in water, called the solubility product. For example: + + Cl − A gCl(s) ⇌ A g(aq) (aq)

Ksp = [A g +][Cl −] Remember that pure solids do not appear in equilibrium constant expressions, therefore solubility products are always just the products of the ions, raised to the power of their stoichiometric coefficients. We can also derive a relationship between the solubility product and the value of solubility itself. The solubility is equal to the amount of solid that can dissolve in 1L of water. So if “s” mole of silver chloride dissolves, there will be “s” moles each of silver ions and chloride ions, so:

[A g +] = [Cl −] = s So for this electrolyte:

Ksp = s × s = s 2 A comparison of Q and K can tell us whether or not a precipitate will form on mixing aqueous solutions of ions. If: - Q < K, then no precipitate will form. - Q > K, then a precipitate will occur. Common Ion Effect The presence of a common ion will always decrease the solubility of an ionic solid. Take for example: 2+ + SO 2− BaSO4(s) ⇌ Ba(aq) 4(aq)

The presence of either barium or sulphate ions shifts equilibrium to the left, meaning that the solubility of the ionic solid is reduced. To work out problems to do with the common ion effect, it is helpful to use concentration tables. What is the solubility of BaSO4 in 0.30mol L-1 aqueous Na2SO4 solution? Ksp (BaSO4) = 1.1 x 10-10. Concentration

Ba2+

BaSO4

SO42-

Initial

0.00

0.30

Change

s

s

Equilibrium

s

0.30 + s

Ksp = [Ba2+][SO42− ] = (s)(0.30 + s) = 1.1 × 10−10 We substitute the equilibrium concentrations in the solubility product expression. Then we can assume that:

0.30 + s ≈ 0.30 This is because s is very small. We can then go on to solve for s.

Ksp = (s)(0.30) = 1.1 × 10−10 s = 3.67 × 10−10 The solubility of barium sulphate in pure water is 1.0 x 10-5 mol L-1 which shows us that the solubility is much lower when extra sulphate ions are present.

Lecture 5 Acids & Bases Lewis Acids & Bases One of the definitions for acids and bases come from G.N Lewis. He says: - A Lewis acid is an electron-pair acceptor. - A Lewis base is an electron-pair donor. The reaction of boron trifluoride and ammonia is a Lewis acid-base reaction.

Brønsted-Lowry Acids & Bases Proposed by Johannes Brønsted and Martin Lowry in 1923: - An acid is a proton (H+) donor. - A base is a proton (H+) acceptor. This definition is mostly used to describe acid-base behaviour in aqueous solutions. The simplest example of all is water reacting with itself:

2H2O ⇌ H3O + + OH − One water molecules (acting as an acid) donates a proton to the other water molecule (acting as a base) to give a hydronium ion and a hydroxide ion. Because water acts as an acid and a base, it is described as amphiprotic.

Conjugate Acids & Bases Note that both the forward and reverse reactions in the reaction above involve the transfer of a proton and are thus both acid-base reactions. Acid-base reactions will always involve two sets of species which differ only by a proton. These species are called conjugate acid-base pair. Other examples of conjugate acid-base pairs include: - HCl / Cl- H2SO4 / HSO4- NH4+ / NH3

Making Acids & Bases Methane (CH4) contains four protons yet it is not acidic. Hydrochloric acid (HCl) however is an extremely strong acid. So what gives a molecule the ability to act as an acid? It must have a proton attached to an electronegative element via a polar bond. The strong acids generally have a proton attached to a halogen in group 17 or oxygen.

Because the base needs to make a new bond to the proton when it accepts it, it needs to have at least one lone pair of electrons. Therefore, bases usually contain group 15, 16 or 17 elements. The strongest bases are usually formed from deprotonation of molecules containing H-X bonds, where X is a group 15 or 16 element. For example: - OH- which is formed by deprotonation of water. - NH2- which is formed by deprotonation of ammonia.

Auto Ionisation Of Water The reaction of water to form hydronium and hydroxide ions is an equilibrium process and has an associated equilibrium constant (Kw), called the auto-ionisation constant of water:

K w = [H3O +][OH −] = 1 × 10−14 In neutral water, the concentration of hydronium and hydroxide ions are equal, therefore:

K w = [H3O +][H3O +] = 1 × 10−14 [H3O +] = 1.0 × 107 = [OH −] Note that we have written hydronium ions rather than protons. This is because protons have no finite lifetime in water. It is highly polarising so it will immediately bind to a water molecule to form hydronium.

The pH Scale We use the pH scale for measuring the concentration of hydronium ions in aqueous solution:

pH = − l og [H3O +] As the concentration of hydronium ions in neutral water is 1.0 x 10-7, the pH of neutral water must be:

pH = − l og (1.0 × 10−7) =7 To convert back into concentration, use the formula:

[H3O +] = 10−pH Similar to pH, we can also define pOH as:

pOH = − l og [OH −] Neutral water will also have a pOH of 7. As well as this, pKw can be defined as:

pK w = − l ogK w = 14 Since we know that:

[H3O +][OH −] = K w This tells us that always:

pH + p OH = 14

Acid Strengths Some acids donate their protons essentially completely to water. We say they undergo complete dissociation (strong electrolytes):

HCl + H2O → H3O + + Cl The end solution only consists of hydronium and chloride ions. Acids that donate their protons on reactions which go to completion are called strong acids. Examples of this include hydrochloric acid, sulphuric acid, nitric acid, hydrobromic acid, hydroiodic acid, and perchloric acid. We can calculate the concentration of hydronium ions in strong acid solutions from the stoichiometry of the dissociation reaction. For example: In the stomach, HCl has a concentration of 0.16mol L-1. Calculate the concentration of hydronium ions and the pH of the HCl.

[H3O +] = [HCl ] = 0.16 ph = − l og [H3O +] = − l og (0.16) = 0.80

Base Strengths Some bases react with water to give essentially complete formation of the hydroxide ion:

NH2− + H2O → NH3 + OH − The resultant solution only consists of ammonia and hydroxide ions. Alternatively, an ionic c...