Chemistry: Acids and Bases

Question 1

Litmus Paper

Answer 1

Turns red in acidic solution and blue in basic solution.

Question 2

Acids Properties

Answer 2

• Have a sour taste. - Aqueous solutions can conduct electricity. - React with bases to form water and a “salt.” - Nonoxidizing acids react with metals to produce hydrogen gas. - Cause color changes in plant dyes, turn litmus paper red.

Question 3

Bases Properties

Answer 3

• Have a bitter taste. - Aqueous solutions can conduct electricity. - React with acids to form water and a “salt.” - Feel slippery to the touch. - Cause color changes in plant dyes, turn litmus paper blue.

Question 4

Arrhenius Definition

Answer 4

Swante Arrhenius formed this toward end of 19th century. He defined an acid as a species that produces H+ (a proton) in an aqueous solution and a base as a species that produces OH- (a hydroxide ion) in an aqueous solution. These definitions, though useful, fail to describe acidic and base behavior in nonaqueous media. An example of an Arrhenius acid, base, and acid-base reaction, respectively, are: HCl(aq) –> H+(aq) + Cl-(aq) NaOH(aq) –> Na+(aq) + OH-(aq) HCl(aq) + NaOH(aq) –> NaCl(aq) + H2O(l)

Question 5

Bronsted-Lowry Definition

Answer 5

More general definition proposed by Johannes Bronsted and Thomas Lowry in 1923. An acid is a species that donates protons, while a base is a species that accepts protons. Advantage to this definition is it’s not limited to aqueous solutions. Acids and bases always occur in pairs, called conjugate acid-base pairs. The two members of a conjugate pair are related by the transfer of a proton. For example, H3O+ is the conjugate acid of the base H2O and NO2- is the conjugate base of HNO2: H3O+(aq) H2O(aq) + H+(aq) HNO2(aq) NO2-(aq) + H+(aq)

Question 6

Lewis Definition

Answer 6

At same time as Bronsted and Lowry, Gilbert Lewis also proposed a definition. He defined an acid as an electron-pair acceptor and a base as an electron-pair donor. Lewis’s are the most inclusive definitions. Just as every Arrhenius acid is a Bronted-Lowry acid, every Bronsted-Lowry acid is also a Lewis acid. However, the Lewis definition encompasses some species not included within the Bronsted-Lowry definition. For example, BCl3 and AlCl3 can each accept an electron pair and are therefore Lewis acids, despite their inability to donate protons.

Question 7

Nomenclature of Arrhenius Acids

Answer 7

The name of an acid is related to the name of the parent anion (the anion that combines with H+ to form the acid). Acids formed from anions whose names end in -ide have the predfix hydro- and the ending -ic. F Flouride = HF Hydrofluoric acid Br Bromide = HB Hydrobromic acid Acids formed from oxyanions are called oxyacids. If the anion ends in -ite (less oxygen), then the acid will end with -ous acid. If the anion ends in -ate (more oxygen), then the acid will end with -ic acid. Prefixes in the names of anions are retained. ClO- Hypochlorite –> HClO Hypochlorous Acid ClO2- Chlorite –> HClO2 Chlorous Acid CLO3- Chlorate –> HClO3 Chloric Acid ClO4- Perchlorate –> HClO4 Perchloric Acid NO2- Nitrite –> HNO2 Nitrous Acid NO3- Nitrate –> HNO3 Nitric Acid

Question 8

Hydrogen Ion Equilibria (pH and pOH)

Answer 8

Hydrogen ion concentration, [H+], is generally measured as pH, where: pH = -log[H+] = log(1/[H+]) Likewise, hydroxide ion concentration, [OH-], is measured as pOH where: pOH = -log[OH-] = log(1/[OH-]) In any aqueous solution, the H2O solvent dissociates slightly: H2O(l) H+(aq) + OH-(aq) This dissociation is an equilibrium reaction and is therefore described by a contsant, Kw, the water dissociation constant. Kw = [H+][OH-] = 10^-14 Rewriting this equation is log form gives: pH + pOH = 14 In pure H2O, [H+] is equal to [OH-] because for every mole of H2O that dissociates, one mole of H+ and one mole of OH- are formed. A solution with equal concentration of H+ and OH- is neutral and has a pH of 7 (-log 10^-7 = 7). A pH below 7 indicates a relative excess of H+ ions and, therefore, an acidic solution; a pH above 7 indicates a relative excess of OH- ions and, therefore, a basic solution.

Question 9

P-Scale Values

Answer 9

A useful skill for various problems involving acids and bases, as well as their corresponding buffer solutions, is the ability to quickly convert pH, pOH, pKa, and pKb into nonlogarithmic form and vice versa. When the original value is a power of 10, the operation is relatively simple; changing the sign on the exponent gives the corresponding p-scale value directly. Example: If [H+] = 0.001, or 10^-3, then pH = 3. If Kb = 1,0 x 10^-7, then pKb = 7. More difficulty arises when the original value is not an exact power of 10; exact calculation would be excessively onerous, but a simple method or approximation exists. If the nonlogarithmic value is written in proper scientific notation, it will look like n x 10^-m, where n is a number between 1 and 10. The log of this product can be written as log(n x 10^-m) = -m + log n, and the negative log is thus m - log n. Now, since n is a number between 1 and 10, its logarithm will be a fraction between 0 and 1, thus, m - log n will be between m - 1 and m. Further, the larger n is, the larger the fraction log n will be, and therefore the closer to m -1 our answer will be.

Question 10

Relative Strengths of Acids and Bases

Answer 10

The strength of an acid or base will depend largely upon its ability to ionize. The strength of an acid, for example, can be measured by the fraction of the molecules of that acid undergoing ionization. Subsequently, the acid strength can be expressed by the following equation. % Ionization = (ionized acid concentration at equilibrium/initial concentration of acid) x 100% When an acid or base is strong, its conjugate base and acid will be weak. The stronger the acid, the weaker the conjugate base. Furthermore, within a series of weak acids, the stronger the acid, the weaker its conjugate base for all acids and bases included.

Question 11

Strong Acids and Bases

Answer 11

Strong acids and bases are those that completely dissociate into their component ions in aqueous solution. For example, when NaOH is added to water, it dissociates completely. Hence, in a 1-M solution of NaOH, complete dissociation gives 1 mole of OH- ions per liter of solution. pH = 14 - (-log[OH-]) = 14 + log[1] = 14 Virtually, no undissociated NaOH remains.

Note that the [OH-] contributed by the dissociation of H₂O is considered to be negligible in this case.

Question 12

Strong Acids & Bases: H2O Contribution

Answer 12

The contribution of OH- and H+ ions from the dissociation of H2O can be neglected only if the concentration of the acid or base is greater than 10^-7 M.

For example, the pH of a 1 x 10^-8 M HCl solution might appear to be 8 because -log(1x10^-8) = 8. However, a pH of 8 is in the basic pH range, and an HCl solution is not basic. This discrepancy arises from the fact that at low HCl concentrations, H+ from the dissociation of water does contribute significantly to the total [H+]. The [H+] from the dissociation of water is less than 1x10^-7 M due to the common ion effect.

The total concentration of H+ can be calculated from Kw = (x+1x10^-8)(x) = 1.0x10^-14, where x = [H+] = [OH-] (both from the dissociation of water molecules). Solving for x gives x = 9.5x10^-8 M, so [H+]total = (9.5x10^-8 + 1x10^-8) = 1.05x10^-7 M, and pH -log(1.05x10^-7) = 6.98, slightly less than 7, as should be expected for a very dilute, yet acidic solutions.

Question 13

Strong Acids & Bases: Common Ones

Answer 13

Strong acids commonly encountered in the lab include HClO₄ (perchloric acid), HNO₃ (nitric acid), H₂SO₄ (sulfuric acid), and HCl (hydrochloric acid). Commonly encountered strong bases include NaOH (sodium hydroxide), KOH (potassium hydroxide), and other soluble hydroxides of Group IA and IIA metals.

Question 14

Strong Acids & Bases: Calculation of pH and pOH

Answer 14

Calculation of the pH and pOH of strong acids and bases assumes complete dissociation of the acid or base in solution: [H+] = normality of strong acid and [OH-] = normality of strong base.

Question 15

Weak Acids and Bases

Answer 15

Those that only partially dissociate in aq solutions. A weak monoprotic acid (HA) in aqueous solution will achieve the following equilibrium after dissociation (H3O+ is equivalent to H+ in aqueous solution): HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Question 16

Acid Dissociation Constant (Ka)

Answer 16

Measure of the degree to which an acid dissociates.

HA(aq) + H₂O(l) <–> H₃O⁺(aq) + A^-(aq)

K_a = [H₃O⁺][A^-] / [HA]

The weaker the acid, the smaller the K_a. Note that K_a does not contain an expression for the pure liquid, water.

Question 17

Base Dissociation Constant (K_b)

Answer 17

Measure of the degree to which a base dissociates. The weaker the base, the smaller its K_b.

For a monovalent base, K_b is defined as follows: K_b = [B⁺][OH^-] / [BOH]

Question 18

Conjugate Acid

Answer 18

The acid formed when a base gains a proton.

Question 19

Conjugate Base

Answer 19

Base formed when an acid loses a proton.

Question 20

Equilibrium Constant

Answer 20

Kw = Ka x Kb = 1 x 10^-14

Question 21

Calculate concentration of H+ in a 2.0 M aqueous solution of acetic acid, CH₃COOH (K_a = 1.8x10^-5).

Answer 21

First write the equlibrium reaction.

CH₃COOH(aq) <–> H⁺(aq) + CH₃COOO^-(aq)

Next, write the expression for the acid dissociation constant.

Ka = [H+][CH₃OOOO^-] / [CH₃OOOH] = 1.8x10^-5

Because acetic acid is a weak acid, the concentration of CH₃COOH at equilibrium is equal to its initial concentration, 2.0 M, less the amount dissociated, x. Likewise, [H⁺] = [CH₃COO^-] = x, since each molecule of CH₃COOH dissociates into one H⁺ ion and one CH₃COO^- ion. Thus, the equation can be rewritten as follows:

Ka = [x][x] / [2.0 - x] = 1.8x10^-5

We can approximate that 2.0 - x = 2.0 since acetic acid is a weak acid, and only slightly dissociates in water. This simplifies the calculation of x:

Ka = [x][x] / [2.0] = 1.8x10^-5

x = 6.0x10^-3 M

The fact that [x] is so much less than the initial concentration of acetic acid (2.0 M) validates the approximation; otherwise, it would have been necessary to solve for x using the quadratic formula. (A rule of thumb is that the approximation is valid as long as x is less than 5% of the initial concentration.)

Question 22

Neutralization Reaction

Answer 22

Acids and bases may react with each other, forming a salt and (often, but not always) water, in this reaction. For example:

HA + BOH –> BA + H₂O

The salt may precipitate out or remain ionized in solution, depending on its solubility and the amount produced. Neutralization reactions generally go to completion.

Question 23

Hydrolysis

Answer 23

The salt ions react with water to give back the acid or base. Reverse of neutralization.

Question 24

Products Between Strong and Weak Acids and Bases

Answer 24

The products of a reaction between equal concentrations of a strong acid and base are a salt and water. The acid and base neutralize each other, so the resulting solution is neutral (pH = 7), and the ions formed in the reaction don’t react with water.

HCl + NaOH –> NaCl +H₂O

The product of a reaction between a strong acid and a weak base is also a salt, but usually no water is formed since weak bases are usually not hydroxides. However in some cases, the cation of salt will react with the water solvent, reforming the weak base. This reaction constitute hydrolysis.

HCl (aq) + NH₃ (aq) <–> NH₄⁺ (aq) + Cl^- (aq)
NH₄⁺ (aq) + H₂O (aq) <–> NH₃ (aq) + H₃O⁺ (aq)

When a weak acid reacts with a strong base, the solution is basic due to the hydrolysis of the salt to reform the acid with the concurrent formation of hydroxide ion from the hydrolyzed water molecules.

HClO + NaOH –> NaClO + H₂O

The pH of a solution containing a weak acid and a weak base depends on the relative strengths of the reactants.

HClO + NH₃ <–> NH₄ClO

Question 25

Polyvalence and Normality

Answer 25

The relative acidity or basicity of an aqueous solution is determined by the relative concentrations of acid and base equivalents. An acid equivalent is equal to one mole of H⁺ (or H₃O⁺) ions; a base equivalent is equal to one mole of OH^- ions.

Some acids and bases are polyvalent, that is, each mole of acid or base liberates more than one acid or base equivalent.

For example, the diprotic acid H₂SO₄ undergoes the following dissociation in water:

H₂SO₄(aq) –> H⁺(aq) + HSO₄^-(aq)
HSO₄^-(aq) –> H⁺(aq) + SO₄^-2(aq)

One mole of H₂SO₄ can thus produce two acid equivalents (two moles of H⁺). The acidity or basicity of a solution depends upon the concentration of acidic or basic equivalents that can be liberated. The quantity of acidic or basic capacity is directly indicated by the solution’s normality. Because each mole of H₃PO₄ can liberate 3 moles (equivalents) of H⁺, a 2 M H₃PO₄ solution would be 6 N (6 normal).

Question 26

Equivalent Weight

Answer 26

Another useful measurement is equivalent weight. For example, the gram molecular weight of H₂SO₄ is 98 g/mol. Because each mole liberates two acid equivalents, the gram equivalent weight of H₂SO₄ would be 98/2 = 49 g; that is, the dissociation of 49 g of H₂SO₄ would release one acid equivalent.

Question 27

Amphoteric Species

Answer 27

An amphoteric, or amphiprotic, species is one that can act either as an acid or base, depending on its chemical environment. In the Bronsted-Lowry sense, an amphoteric species can either gain or lose a proton. Water is the most common example. When water reacts with a base, it behaves as an acid:

H₂O + B^- <–> HB + OH^-

When water reacts with an acid, it behaves as a base:

HA + H₂O <–> H₃O⁺ + A^-

The partially dissociated conjugate base of a polyprotic acid is usually amphoteric (e.g., HSO₄^- can either gain an H⁺ to form H₂SO₄ or lose an H⁺ to form SO₄^-2). The hydroxides of certain metals (e.g., Al, Zn, Pb, and Cr) are also amphoteric. Furthermore, species that can act as either oxidizing or reducing agents are considered to be amphoteric as well, since by accepting or donating electron pairs they act as Lewis acids or bases, respectively.

Question 28

Titration and Buffers

Answer 28

Titration is procedure used to determine the molarity of an acid or base. This is accomplished by reacting a known volume of a solution of unknown concentration with a known volume of a solution of known concentration. When the number of acid equivalents equal the number of base equivalents added, or vice versa, the equivalence point is reached.

It’s important to emphasize that, while a strong acid/strong base titration will have an equivalence point at pH 7, the equivalence point need not always occur at pH 7. Also, when titrating polyprotic acids or bases, there are several equivalence points, as each different acidic or basic species is titrated separately.

The equivalence point in a titration is estimated in two common ways: either by using a graphical method, plotting the pH of the solution as a function of added titrant by using a pH meter, or by watching for a color change of an added indicator.

Question 29

Indicator

Answer 29

Weak organic acids or bases that have different colors in their undissociated and dissociated states. Used in low concentrations and therefore don’t significantly alter the equivalence point.

The point at which the indicator changes color is not the equivalence point but is called the end point. If the titration is performed well, the volume difference (and therefore the error) between the end point and the equivalence point is usually small and may be corrected for or ignored.

Question 30

Titration: Strong Acid and Strong Base

Answer 30

Consider the titration of 10mL or a 0.1 N solution of HCl with a 0.1 N solution of NaOH. Plotting the pH of the reaction solution vs. the quantity of naOH added gives a curve.

Because HCl is a strong acid and NaOH is a strong base, the equivalence point of the titration will be at pH 7, and the solution will be neutral. Note that the endpoint is close to, but not exactly equal to pH 7; selection of a better indicator, say one that changes color at pH 8, would have given a better approximation.

In the early part of the curve (when little base has been added), the acidic species predominates, and so the addition of small amounts of base will not appreciably change either the [OH^-] or the pH. Similarly, in the last part of the titration curve (when an excess of base has been added), the addition of small amount of base will not change the [OH^-] significantly, and the pH remains relatively constant. The addition of base most alters the concentrations of H⁺ and OH^- near the equivalence point, and thus the pH changes most drastically in that region.

Question 31

Titration: Weak Acid and Strong Base

Answer 31

The initial pH of the weak acid is greater than it is in the Strong Acid and Strong Base titration. The pH changes most significantly early on in the titration, and the equivalence point is in the basic range.

Question 32

Buffer Solution

Answer 32

Consists of a mix of a weak acid and its salt (which consists of its conjugate base and a cation) or a mixture of a weak base and its salt (which consists of its conjugate acid and an anion).

Two examples of buffers are a solution of acetic acid (CH₃COOH) and its salt, sodium acetate (CH₃COONa), and a solution of ammonia (NH₃), and its salt, ammonium chloride (NH₃Cl).

Buffer solutions have the useful property of resisting changes in pH when small amounts of acid or base are added.

Consider a buffer solution of aceitc acid and sodium acetate.

CH₃COOH <–> H⁺ + CH₃COO^-
CH₃COONa <–> Na⁺ + CH₃COO^-

When a small amount of NaOH is added to the buffer, the OH^- ions from the NaOH react with the H⁺ ions present in the solution; subsequently, more acetic acid dissociates (equilibrium shifts to the right), restoring the [H+]. Thus, an increase in [OH^-] doesn’t appreciably change pH. Likewise, when a small amount of HCl is added to the buffer, H⁺ ions from the HCl react with the acetate ions to form acetic acid. Thus, [H⁺] is kept relatively constant, and the pH of the solution is relatively unchanged.

Question 33

Henderson-Hasselbalch Equation

Answer 33

Used to estimate the pH of a solution in the buffer region where the concentrations of the species and its conjugate are present in approximately equal concentrations. For a weak acid buffer solution:

pH = pKa + log[(conjugate acid) / (weak base)]

Note that when [conjugate base] = [weak acid] (in a titration, halfway to the equivalence pont), the pH = pKa, because the log 1 = 0.

Likewise, for a weak base buffer solution:

pOH = pKb + log([conjugate acid] / [weak base])

pOH = pKb when [conjugate acid] = [weak base].