Introductory Chemistry Online/Acids, Bases and pH

  Chapter 8.Acids, Bases and pH

We have all heard of acids… acid indigestion, the acid content of vinegars and cheap wines, vats of sulfuric acid and arch villains. In the last chapter we learned about the molecular dipole in water and how water molecules form hydrogen bonds among themselves, and to other polar molecules. In this chapter we will see that this hydrogen bonding can be strong enough to actually break the O—H bonds in water and in other hydrogen bonded molecules, transferring the proton to water to form the hydronium ion, H₃O⁺. In its simplest sense, the formation of the hydronium ion is the “acidity” that we are all familiar with. We will see how we measure this acidity (the pH scale). We will get an introduction to the concept of equilibrium and you will finally come to appreciate the utility of the log and anti-log buttons on your scientific calculator!

  8.1 Hydrogen Bonding

In Chapter 7, we explored the unique properties of water that allow it to serve as a powerful solvent with the ability to dissolve both ionic compounds, as well as polar molecular compounds. We attributed this to the ability of water molecules to align themselves so that the polarized hydrogen-oxygen bonds could stabilize cations, anions, and virtually any compound that also contained a significantly polarized covalent bond. By this logic, it is not at all surprising that water can also react strongly with itself, and indeed water exists as a vast network of molecules aligned so that their positive and negative dipoles interact with each other. This is shown in the drawing in Figure 8.1, and the dotted lines indicate interactions between hydrogens (the positive end of the molecular dipole) and oxygens (the negative end of the dipole). A bond that is formed from a hydrogen atom, which is part of a polar covalent bond (such as the O—H bond) to another, more electronegative atom (that has at least one unshared pair of electrons in its valence shell) is called a hydrogen bond. Recall that oxygen has two unshared pairs in its valence shell, and the hydrogen-oxygen interaction in water is the classic example of a hydrogen bond. Hydrogen bonds are weak, relative to covalent bonds. The energy required to break the O—H covalent bond (the bond dissociation energy) is about 111 kcal/mole, or in more proper SI units, 464 kJ/mole. The energy required to break an O—H••••O hydrogen bond is about 5 kcal/mole (21 kJ/mole), or less than 5% of the energy of a “real” covalent bond. Even though hydrogen bonds are relatively weak, if you consider that every water molecule is participating in a least four hydrogen bonds, the total energy of hydrogen bonding interactions can rapidly become significant. Hydrogen bonding is generally used to explain the high boiling point of water (100 ˚C). For many compounds which do not possess highly polarized bonds, boiling points parallel the molar mass of the compound. Methane, CH₄, has a molar mass of 16 and a boiling point of –164 ˚C. Water, with a molar mass of 18, has a boiling point of +100 ˚C. Although these two compounds have similar molar masses, a significant amount of energy must be put into the polar molecule, water, in order to move into the gas phase, relative to the non-polar methane. The extra energy that is required is necessary to break down the hydrogen bonding network.

Hydrogen bonding is also important is DNA. According to the Watson-Crick model, the double helix of DNA is assembled and stabilized by hydrogen pairing between matching “bases”. In Figure 8.2, the hydrogen bonds between adenine and thymidine are shown as the faint dotted lines. The hydrogen bonds are formed between the oxygen atoms (red) and the adjacent N—H bonds, and between the central nitrogen (blue) and the adjacent N—H bond. It is suggested that the precise alignment of these hydrogen bonds contributes to stability of the double helix and ensures the proper alignment of the corresponding base pairs.

  8.2 Ionization of Acids in Solution

In general, acids can be thought of as molecular compounds containing at least one hydrogen which is covalently bonded to a more electronegative atom. As an example, consider the compound hydrogen fluoride (HF). As we discussed in Chapter 7, the electronegativities of hydrogen and fluorine are 2.1 and 4.0, respectively, and the hydrogen-fluoride covalent bond is very highly polarized, as shown in the electrostatic potential map shown in Figure 8.3.

Because of this, when hydrogen fluoride is dissolved in water, water molecules orient themselves around HF so that the water dipoles interact with, and stabilize, the highly polarized H—F bond. Important to this stabilization is the hydrogen bond that is formed between the hydrogen of HF and the oxygen of an adjacent water. This hydrogen bond not only stabilizes the HF molecular dipole, but it also weakens the H—F covalent bond. As a result of this weakening, the H—F bond stretches (the bond length increases) and then fully breaks. The hydrogen that was hydrogen-bonded to the water molecule now becomes fully bonded to the oxygen, forming the species H₃O⁺ (the hydronium ion) and the fluorine now exists as a fluoride anion. This is known as the process of acid dissociation and the steps in this process are shown using “ball and stick” molecular models in Figure 8.4.

The chemical equation describing the acid dissociation reaction of HF is given in Equation 8.1a. The products of the reaction, fluoride anion and the hydronium ion, are oppositely charged ions, and it is reasonable to assume that they will be attracted to each other. If they do come in contact, it is also reasonable to suggest that the process of hydrogen transfer that we described above can also happen in reverse. That is, H₃O⁺ can hydrogen bond to the fluoride ion and the hydrogen can be transferred back, to form HF and water. The chemical equation describing this process is shown in Equation 8.1b. In fact, these two reaction do occur simultaneously (and very rapidly) in solution. When we speak of a set of forward- and back-reactions that occur together on a very fast time-scale, we describe the set of reactions as an equilibrium and we use a special double arrow in the chemical reaction to show this (Equation 8.1c). Equation 8.1c can be said to represent the equilibrium dissociation of HF in water.

HF (aq) + H₂O → H₃O⁺ (aq) + F^– (aq)Eq. 8.1a

HF (aq) + H₂O←H₃O⁺ (aq) + F^– (aq)Eq. 8.1b

HF (aq) + H₂O ⇄ H₃O⁺ (aq) + F^– (aq) Eq. 8.1c

For any equilibrium, an equilibrium constant can be written that describes whether the products or the reactants will be the predominant species in solution. We will address this fully in Chapter 10, but according to the Law of Mass Action, the equilibrium constant, K for this reaction, is simply given by the ratio of the concentrations of the products and the reactants (concentrations from each side are multiplied together if there is more than one). Note that any solid or liquid reactants or products (such as water) are not included in the expression. Thus for the ionization of HF;

$K_{a}={\frac {\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{F}}^{-}\right]}{\left[{\text{HF}}\right]}}$

.

When you are dealing with acids, the equilibrium constant is generally called an acid dissociation constant, and is written as K_a. The larger the value of K_a, the greater the extent of ionization and the higher the resulting concentration of the hydronium ion. Because the concentration of the hydronium ion is directly correlated with acidity, acids with a large value of K_a are termed strong acids. We will introduce “weak acids” in Chapter 10, but for now the important thing to remember is that strong acids are virtually 100% ionized in solution. That doesn’t mean that the back-reaction does not occur, is simply means that much more favorable and that 99.9999999999% of the acid is present in its ionized form. Because this exceeds the number of significant figures that we typically work with, strong acids are generally described as 100% ionized in solution. Table 8.1 lists the common strong acids that we will study in this text.

Table 8.1. Common Strong Acids

HCl	hydrochloric acid
HNO3	nitric acid
H2SO4	sulfuric acid
HBr	hydrobromic acid
HI	hydroiodic acid
HClO4	perchloric acid

  8.3 Conjugate Acid-Base Pairs

Acid dissociation reactions are often described in terms of the concepts of conjugate acids and their corresponding conjugate bases. The description of “acids and bases” that we will deal with in this text will be limited to simple dissociation reactions, like those shown above, where a hydronium ion is produced. This description is referred to as the Brønsted-Lowery Acid-Base Theory, and in the Brønsted theory, the conjugate acid is defined as the species that donates a hydrogen in the forward reaction, and the conjugate base is the species that accepts a hydrogen the reverse reaction. Thus for the ionization of HCl, HCl is the conjugate acid and Cl^– is the conjugate base.

HCl (aq) + H₂O ⇄ H₃O⁺ (aq) + Cl^– (aq)

In the discussion of Brønsted acid-base behavior, the hydrogen atom that is transferred is generally referred to as a proton, because it is transferred as a hydrogen atom without its electron. Thus for the ionization of HCl, HCl (the conjugate acid) is a proton donor and Cl^– (the conjugate base) is a proton acceptor. In General Chemistry you will learn that acid-base behavior can also be described in terms of electron donors and electron acceptors (the Lewis Acid-Base Theory in which an acid is an electron acceptor and a base is an electron donor), but here we will limit our discussion to simple, strong, Brønsted acids and bases.

Example 8.1 Identifying Conjugate Acids and Conjugate Bases

For each of the reactions given below, identify the conjugate acid and the conjugate base. For example (d), also identify the conjugate acid and the conjugate base in the reverse reaction.

a.HClO4 (aq) + H2O  ⇄  H3O+ (aq) + ClO4– (aq) 

b.H2SO4 (aq) + H2O  ⇄  H3O+ (aq) + HSO4– (aq) 

c.HSO4– (aq) + H2O  ⇄  H3O+ (aq) + SO22- (aq) 

d.HNO3 (aq) + NH3  ⇄  NH4+ (aq) + NO3– (aq) 

Exercise 8.1 Identifying Conjugate Acids and Conjugate Bases

For each of the reactions given below, identify the conjugate acid and the conjugate base.

a) H2PO4- + H3O+    ⇄    H2O + H3PO4

b) NH3(g) + H2O   ⇄   NH4+(aq) + OH-(aq)

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

  8.4 Acids-Bases Reactions: Neutralization

In Chapter 5, we examined a special case of a double replacement reaction in which an acid reacted with a base to give water and a pair of ions in solution. In the context of the Brønsted Theory, a base can be thought of as an ionic compound that produces the hydroxide anion in solution. Thus, sodium hydroxide, NaOH, ionizes to form the sodium cation and the hydroxide anion (HO^–).

NaOH (aq) → Na⁺ (aq) + HO^– (aq)

We have written this equation using a single arrow because sodium hydroxide is a strong base and is essentially 100% ionized in solution. The hydroxide anion (HO^–) reacts with the hydronium ion (H₃O⁺) to form two moles of water, as shown in the equation given below.

H₃O⁺ + HO^– → 2 H₂O

Thus, if you have aqueous solutions of HCl and NaOH, the following process occur:

• HCl ionizes to form the hydronium ion:

HCl (aq) + H₂O ⇄ H₃O⁺ (aq) + Cl^– (aq)

• NaOH ionizes to form the hydroxide anion:

NaOH (aq) → Na⁺ (aq) + HO^– (aq)

• HO^– reacts with H₃O⁺ to form two moles of water:

H₃O⁺ + HO^– → 2 H₂O

If we add up the set of three equations, we see that hydronium and hydroxide ions appear on both sides of the arrow and cancel, leaving:

HCl (aq) + NaOH (aq) → H₂O + NaCl (aq)

In this equation, we have not shown the additional water from the hydronium ion and we have grouped the sodium and chloride ions as NaCl (aq), with the understanding that it will be fully ionized in aqueous solution. This is an example of a neutralization reaction; an acid and a base have reacted to form water. When we write neutralization equations we generally do not show hydronium or hydroxide ions and we generally show ionic species as distinct compounds. Neutralization equations therefore look very much like the other double replacements that we studied in Chapter 5.

Example 8.2 Writing and Balancing Neutralization Equations

For each of the following, write a balanced neutralization equation:

a.The reaction of calcium hydroxide with hydrochloric acid. 

b.The reaction of sodium hydroxide with sulfuric acid (both ionizations). 

c.The reaction of barium hydroxide with nitric acid. 

Exercise 8.2 Writing and Balancing Neutralization Equations

Write a balanced neutralization equation for the reaction of calcium hydroxide
 with sulfuric acid.

  8.5 The Meaning of Neutrality: The Autoprotolysis of Water

In the previous section, we described the reaction of an acid and a base to form water. When all of the acid and base have been consumed, we are left with water and an aqueous solution containing an ionic compound. Another way to think of this would be to say that, the acid we began with had a high concentration of hydronium cations (H₃O⁺), the base had a high concentration of hydroxide anions (HO^–) and the neutral solution contains only water. This, however, is a little too simplistic. Just like water can promote the ionization of acids, water can also promote the ionization of itself. Picture two water molecules sharing a hydrogen bond. Just like we saw in Figure 8.4 for HF , the partially bonded hydrogen can transfer along the hydrogen bond to form a hydronium cation and a hydroxide anion (Figure 8.5). This process occurs very rapidly in pure water, thus, any sample of pure water will always contain a small concentration of hydronium and hydroxide ions. How small is “small”? Very small! In pure water at 25 ^oC, the concentration of hydronium ions ([H₃O⁺]) and hydroxide ions ([HO^–]) will both be equal to exactly 1 × 10^-7 M. Based on this, we can expand upon our definitions of acidic, basic and neutral solutions:

• A solution is acidic if [H₃O⁺] > 1 × 10^-7 M.
• A solution is basic if [H₃O⁺] < 1 × 10^-7 M.
• A solution is neutral if [H₃O⁺] = 1 × 10^-7 M.

Working with these definitions, if you have a solution with [H₃O⁺] = 4.5 × 10^-4 M, it will be acidic (4.5 × 10^-4 > 1 × 10^-7). If you have a solution with [H₃O⁺] = 1 × 10^-4 M, it will be basic (1 × 10^-4 < 1 × 10^-7). Finally, a neutral solution is one in which [H₃O⁺] and [HO^–] are both 1 × 10^-7 M.

Recalling our discussion of acid dissociation constants from Section 8.2, we can write the ionization equilibrium for water and the expression for the dissociation constant, K_a, as shown below:

2 H₂O ⇄ H₃O⁺ + HO^– and [

$K_{a}={\frac {\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{HO}}^{-}\right]}{\left[{\text{H}}_{\text{2}}{\text{O}}\right]^{2}}}$

Because the molar concentration of water (in pure water) is 55.5 M, the term for [H₂O] is essentially a constant and can be neglected (recall the law of mass action). When we do this, we call the dissociation constant K_W, where K_W is defined as $K_{W}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{HO}}^{-}\right]$. At neutrality and 25 ^oC, [H₃O⁺] and [HO^–] are both 1 × 10^-7 M, therefore:

$K_{W}=\left[1\times 10^{-7}\right]\left[1\times 10^{-7}\right]=1\times 10^{-14}$

.

This simple relationship is actually quite powerful. Because K_W is a constant, if we know either a hydronium ion or a hydroxide ion concentration, we can directly calculate the concentration of the other species. For example, if you are given that [H₃O⁺] is 1 × 10^-4 M, [HO^–] can be calculated as:

$K_{W}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{HO}}^{-}\right]$

$K_{W}=1\times 10^{-14}=\left[1\times 10^{-4}\right]\left[{\text{HO}}^{-}\right]$

$\left[{\text{HO}}^{-}\right]={\frac {{\text{1}}\times {\text{10}}^{-{\text{14}}}}{{\text{1}}\times {\text{10}}^{-{\text{4}}}}}=1\times 10^{-10}{\text{M}}$

Example 8.3 Calculating [H₃O⁺] and [HO^-] using K_W

a. A solution at 25 ^oC, is known to have a hydronium ion concentration of 4.5 × 10^-5 M;
what is the concentration of hydroxide ion in this solution?

b. A solution at 25 ^oC, is known to have a hydroxide ion concentration of 7.5 ×
10^-2 M; what is the concentration of hydronium ion in this solution?

Exercise 8.3 Calculating [H₃O⁺] and [HO^-] using K_W

A solution is known to have a hydronium ion concentration of 9.5 × 10^-8 M;
what is the concentration of hydroxide ion in this solution?

  8.6 pH Calculations

One thing that you should notice about the numbers in the previous examples is that they are very small. In general, chemists find that working with large negative exponents like these (very small numbers) is cumbersome. To simplify the process, calculations involving hydronium ion concentrations are generally done using logarithms. Recall that a logarithm is simply the exponent that some base number needs to be raised to in order to generate a given number. In these calculations, we will use a base of 10. A number such as 10,000 can be written as 10⁴, so by the definition, the logarithm of 10⁴ is simply 4. For a small number such as 10^-7, the logarithm is again simply the exponent, or -7. Before calculators became readily available, taking the logarithm of a number that was not an integral power of 10 meant a trip to “log tables” (or even worse, using a slide rule). Now, pushing the LOG button on an a scientific calculator makes the process trivial. For example, the logarithm of 14,283 (with the push of a button) is 4.15482. If you are paying attention, you should have noticed that the logarithm contains six digits, while the original number (14,283) only contains five significant figures. This is because a logarithm consists of two sets of numbers; the digits to the left of the decimal point (called the characteristic) simply reflect the integral power of 10, and are not included when you count significant figures. The numbers after the decimal (the mantissa) should have the same significance as your experimental number, thus a logarithm of 4.15482 actually represents five significant figures.

There is one other convention that chemists apply when they are dealing with logarithms of hydronium ion concentrations, that is, the logarithm is multiplied by (-1) to change its sign. Why would we do this? In most aqueous solutions, [H₃O⁺] will vary between 10^-1 and 10^-13 M, giving logarithms of –1 to –13. To make these numbers easier to work with, we take the negative of the logarithm (-log[H₃O⁺]) and call it a pH value. The use of the lower-case “p” reminds us that we have taken the negative of the logarithm, and the upper-case “H” tells us that we are referring to the hydronium ion concentration. Converting a hydronium ion concentration to a pH value is simple. Suppose you have a solution where [H₃O⁺] = 3.46 × 10^-4 M and you want to know the corresponding pH value. You would enter 3.46 × 10^-4 into your calculator and press the LOG button. The display should read “-3.460923901”. First, we multiply this by (-1) and get 3.460923901. Next, we examine the number of significant figures. Our experimental number, 3.46 × 10^-4 has three significant figures, so our mantissa must have three digits. We round our answer and express our result as, pH = 3.461.

The reverse process is equally simple. If you are given a pH value of 7.04 and are asked to calculate a hydronium ion concentration, you would first multiply the pH value by (-1) to give –7.04. Enter this in your calculator and then press the key (or key combination) to calculate “10^x”; your display should read “9.120108 × 10^–8”. There are only two digits in our original mantissa (7.04) so we must round this to two significant figures, or [H₃O⁺] = 9.1 × 10^-8.

Example 8.4 Calculating [H₃O⁺] and pH Values

a. A solution is known to have a hydronium ion concentration of 4.5 ×
 10-5 M; what is the pH this solution?

b. A solution is known to have a pH of 9.553; what is the concentration of
 hydronium ion in this solution?

Exercise 8.4 Calculating [H₃O⁺] and pH Values

a. A solution is known to have a hydronium ion concentration of 9.5 ×
 10-8 M; what is the pH this solution?

b. A solution is known to have a pH of 4.57; what is the hydronium ion
 concentration of this solution?

There is another useful calculation that we can do by combining what we know about pH and expression $K_{W}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{HO}}^{-}\right]$. We know that K_W = 10^-14 and we know that (-log [H₃O⁺]) is pH. If we define (-log [HO^–]) as pOH, we can take our expression for K_W and take the (-log) of both sides (remember, algebraically you can perform the same operation on both sides of an equation) we get:

$K_{W}=10^{-14}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{HO}}^{-}\right]$

$-\log \left(10^{-14}\right)=\left(-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\right)+\left(-\log \left[{\text{HO}}^{-}\right]\right)$

$14=p{\text{H}}+p{\text{OH}}$

Which tells us that the values of pH and pOH must always add up to give 14! Thus, if the pH is 3.5, the pOH must be 14 – 3.5 = 11.5. This relationship is quite useful as it allows you to quickly convert between pH and pOH, and therefore between [H₃O⁺] and [HO^–].

We can now re-address neutrality in terms of the pH scale:

• A solution is acidic if pH < 7.
• A solution is basic if pH > 7.
• A solution is neutral if pH = 7.

The simplest way to determine the pH of a solution is to use an electronic pH meter (Figure 8.6). A pH meter is actually a sensitive millivolt meter that measures the potential across a thin, sensitive glass electrode that is immersed in the solution. The voltage that develops is a direct function of the pH of the solution and the circuitry is calibrated so that the voltage is directly converted into the equivalent of a pH value. You will most likely use a simple pH meter in the laboratory. The thing to remember is that the sensing electrode has a very thin, fragile, glass membrane and is somewhat expensive to replace. Be careful!

A simple way to estimate the pH of a solution is by using an indicator. A pH indicator is a compound that undergoes a change in color at a certain pH value. For example, phenolphthalein (Figure 8.7) is a commonly used indicator that is colorless at pH values below 9, but is pink at pH 10 and above (at very high pH it becomes colorless again). In the laboratory, a small amount of phenolphthalein is added to a solution at low pH and then a base is slowly added to achieve neutrality. When the phenolphthalein changes from colorless to pink, you know that enough base has been added to neutralize all of the acid that is present. In reality, the transition occurs at pH 9.2, not pH 7, so the resulting solution is actually slightly alkaline, but the additional hydroxide ion concentration at pH 9 (10^-5 M) is generally insignificant relative to the concentrations of the solutions being tested.

A convenient way to estimate the pH of a solution is to use pH paper. This is simply a strip of paper that has a mixture of indicators embedded in it. The indicators are chosen so that the paper takes on a slightly different color over a range of pH values. The simplest pH paper is litmus paper that changes from pink to blue as a solution goes from acid to base. Other pH papers are more exotic. Figure 8.8 shows a common commercial pH paper that changes colors gradually over a pH range of 1 to 12. In the laboratory, you will use both indicators, like phenolphthalein, and pH papers in neutralization experiments called titrations, as described in the following section.

  8.7 Titrations: Neutralization and Stoichiometry

One of the standard laboratory exercises in General Chemistry is an acid-base titration. In order to perform an acid-base titration, you must have a solution of acid or base with a known concentration. You then slowly add a known volume of this solution, using a volumetric burette, to an acid or base solution with an unknown concentration until neutrality has been achieved. At that point, you know the volume and concentration of the reactant you have added, which means that you can calculate the number of moles that you added. Based on the stoichiometry of your neutralization reaction, you then know how many moles of acid or base were in the unknown sample. How do you know when you have reached neutrality? Generally an indicator or a pH meter is used (as described in Section 8.5). For example, if we had a solution of NaOH that was exactly 0.100 M and we had a beaker containing an unknown concentration of HCl. To perform the titration we would add a few drops of a stock phenolphthalein solution to our HCl, and then slowly add a measured amount of the NaOH solution until all of the acid had been consumed and the indicator changed from colorless to pink (Figure 8.9).

HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O

Working with the above example, if the volume of base that we added (as measured on the buret) was 12.6 mL, we could calculate the number of moles present in the unknown acid solution. This is equal to the known concentration of our NaOH (0.100 M) multiplied by the volume required for neutrality (0.0126 L), or:

$\left(0.100{\text{ }}{}^{\text{moles}}\!\!\diagup \!\!{}_{\text{L}}\;\right)\left(0.0126{\text{ L}}\right)={\text{1}}{\text{.26}}\times {\text{10}}^{\text{-3}}{\text{ moles}}$

If we had used exactly 100.0 mL of our unknown acid in our titration, the concentration of our acid would be:

${}^{\left({\text{1}}{\text{.26}}\times {\text{10}}^{\text{-3}}{\text{ moles}}\right)}\!\!\diagup \!\!{}_{\left(0.1000{\text{ L}}\right)}\;={\text{1}}{\text{.26}}\times {\text{10}}^{\text{-2}}{\text{ M}}$

.

Example 8.5 Acid-Base Titrations

a. You are given a solution containing an unknown concentration of HCl. You
 carefully measure 50.0 mL of this solution into a flask and then add a few drops of
 phenolphthalein solution. You prepare a buret containing 0.055 M NaOH and
 note that the initial level of the solution in the buret is 12.6 mL. You
 slowly add the NaOH solution to the acid until the color change just
 occurs (as evidence of the color change becomes visible, you carefully
 stir the solution after each drop has been added). When the acid
 solution turns (and remains) pink, you note that the volume in the buret is now
 28.9 mL. What is the concentration of the unknown acid solution?

Exercise 8.5 Acid-Base Titrations

If 25.00 mL of HCl solution with a concentration of 0.1234 M is neutralized by
 23.45 mL of NaOH, what is the concentration of the base?

  Study Points

• A bond that is formed from a hydrogen atom, which is part of a polar covalent bond (such as the O—H bond) to another, more electronegative atom (that has at least one unshared pair of electrons in its valence shell) is called a hydrogen bond. Hydrogen bonds are weak, partially covalent bonds. The bond dissociation energy of the O—H covalent bond 464 kJ/mole; the bond dissociation energy an O—H••••O hydrogen bond is about 21 kJ/mole.
• Even though hydrogen bonds are relatively weak, the vast network of hydrogen bonds in water makes the energy significant, and hydrogen bonding is generally used to explain the high boiling point of water (100 ˚C), relative to molecules of similar mass that cannot hydrogen bond. The extra energy represents the energy required to break down the hydrogen bonding network.
• Polar molecules, such as acids, strongly hydrogen bond to water. This hydrogen bonding not only stabilizes the molecular dipoles, but also weakens the H—A covalent bond (A represents the acid molecule). As a result of this weakening, the H—A bond in these acids stretches (the bond length increases) and then fully breaks. The hydrogen that was hydrogen-bonded to the water molecule now becomes fully bonded to the oxygen, forming the species H₃O⁺ (the hydronium ion) and the acid now exists as an anion (A^–); this is the process of acid dissociation.
• The hydronium ion and the acid anion that are formed in an acid dissociation can react to re-form the original acid. This represents a set of forward- and back-reactions that occur together on a very fast time-scale; this type of a set of reactions is called an equilibrium and a double arrow is used in the chemical reaction to show this. This type of reaction is referred to as an acid dissociation equilibrium.

HA (aq) + H₂O ⇄ H₃O⁺ (aq) + A^– (aq)

• For any equilibrium, an equilibrium constant can be written that describes whether the products or the reactants will be the predominant species in solution. For the dissociation of the simple acid, HA, the equilibrium constant, K_a, is simply given by the ratio of the concentrations of the products and the reactants, excluding any solid or liquid reactants or products (such as water). Thus for the ionization of HA;

$K_{a}={\frac {\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[{\text{HA}}\right]}}$

.

• According to the Brønsted Acid-Base Theory, any substance that ionizes in water to form hydronium ions (a proton donor) is called an acid; any substance that accepts a proton from a hydronium ion is a base. In an acid-base equilibrium, the conjugate acid is defined as the species that donates a hydrogen (a proton) in the forward reaction, and the conjugate base is the species that accepts a hydrogen (a proton) the reverse reaction. Thus for the ionization of HCl, HCl is the conjugate acid and Cl^– is the conjugate base.

HCl (aq) + H₂O ⇄ H₃O⁺ (aq) + Cl^– (aq)

• Metal hydroxides, such as NaOH, dissolve in water to form metal cations and hydroxide anion. Hydroxide anion is a strong Brønsted base and, therefore, hydroxide anion accepts a proton from the hydronium ion to form two moles of water. The reaction of a Brønsted acid with a Brønsted base to form water is the process of neutralization.
• Just like water can promote the ionization of acids, water can also promote the ionization of itself. This equilibrium process occurs very rapidly in pure water and any sample of pure water will always contain a small concentration of hydronium and hydroxide ions. In pure water, at 25 ^oC, the concentration of hydronium ions ([H₃O⁺]) and hydroxide ions ([HO^–]) will both be equal to exactly 1 × 10^-7 M. This is referred to as the autoprotolysis of water.
• The equilibrium for the autoprotolysis of water is defined as' K_w', according to the equation shown below:

$K_{W}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{HO}}^{-}\right]$

, and at neutrality, [H₃O⁺] and [HO^–] are both 1 × 10^-7 M, making the value of K_w:

$K_{W}=\left[1\times 10^{-7}\right]\left[1\times 10^{-7}\right]=1\times 10^{-14}$

.

• Based on the autoprotolysis equilibrium, acidic, basic and neutral solutions can be defined as:
  1. A solution is acidic if [H₃O⁺] > 1 × 10^-7 M.
  2. A solution is basic if [H₃O⁺] < 1 × 10^-7 M.
  3. A solution is neutral if [H₃O⁺] = 1 × 10^-7 M.
• A pH value is simply the negative of the logarithm of the hydronium ion concentration (-log[H₃O⁺]).
• Remember that a logarithm consists of two sets of numbers; the digits to the left of the decimal point (the characteristic) reflect the integral power of 10, and are not included when you count significant figures. The numbers after the decimal (the mantissa) have the same significance as your experimental number, thus a logarithm of 4.15482 represents five significant figures.
• In an acid-base titration a solution of acid or base with a known concentration is slowly add to an acid or base solution with an unknown concentration, using a volumetric burette, to until neutrality has been achieved. Typically an indicator or a pH meter is used to signify neutrality.
• At neutrality, the volume and concentration of the reactant you have added is known, which means that you can calculate the number of moles that you added (remember, concentration × volume = moles). Based on the stoichiometry of your neutralization reaction, you then know how many moles of acid or base were in the unknown sample.

  Supplementary Problems

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