Study Guide

# Acids and Bases Questions

## Acids and Bases Questions

1. Which weak acid solution will have a greater buffering capacity: a 1 L solution of 0.02 M HCN or a 0.5 L solution of 0.04 M HCN?

2. Which antacid can provide the largest neutralizing effect on a per gram basis: Mg(OH)2 or CaCO3?

3. Why do you think there are multiple definitions of an acid?

4. What is the approximate ratio of H2O molecules to OH- molecules in a pH 12 solution?

5. Oxalic acid is a symmetrical molecule with two protons that readily dissociate. Which acid of oxalic acid is more acidic: H2C2O4 or HC2O4-? What is a reasonable physical explanation for why they might be the same or different?

6. The compound p-nitrophenol is more acidic than just plain old phenol (see structures below). Why do you think this might be? 7. Methane (CH4) is a very weak acid with a pKa of approximately 50. In a 1000 L solution of 0.001 M dissolved methane approximately how many molecules of methane are in the conjugate base form (CH3-)?

8. Explain, in words, the relationship between a buffer's pKa and a given pH of the buffer.

9. Determining the equilibrium concentrations of the products of a weak acid or base dissociation is made easier by making an assumption that the initial concentration of weak acid or base does not change significantly. Describe the necessary conditions that make this assumption valid.

10. Imagine a 1 L solution of a 1 M buffer with a pKa of 9 at a pH of 9. If this solution is diluted into 9 L of H2O (pKa = 15.7) how will the ratio of the buffer acid and its conjugate base change and what will the pH of the final 10 L solution be?

### Solutions

1. Which weak acid solution will have a greater buffering capacity: a 1 L solution of 0.02 M HCN or a 0.5 L solution of 0.04 M HCN?

The buffering capacity of a solution is determined by both the volume of the solution and the concentration of the buffer, in this case, HCN. The product of the buffer volume and concentration gives the number of moles of buffer. The number of moles of buffer determines the number of moles of titrant that can be absorbed before the buffer runs out. This is the solution's buffering capacity. In the question above each solution contains the same number of moles of HCN so the buffering capacities are the same.

2. Which antacid can provide the largest neutralizing effect on a per gram basis: Mg(OH)2 or CaCO3?

There are several ways to solve this problem but one of them is to start by imagining you have 1 g (a nice round number) each of Mg(OH)2 and CaCO3. By using the molecular weights of these antacids you can calculate that 1 g of each corresponds to 0.017 moles of Mg(OH)2 and 0.010 moles of CaCO3. When each antacid is dissolved in solution Mg(OH)2 will release 2 × 0.017 = 0.034 moles of OH- and CaCO3 will release 0.010 moles of the weak base CO32-. Since CO32- is a weak base (Kb = 1.8 × 10-4) less than 0.010 moles of OH- will form in solution from the reaction of CO32- with H2O. Therefore, Mg(OH)2, which can neutralize 0.034 moles of acid per gram, is the best antacid on a per gram basis.

3. Why do you think there are multiple definitions of an acid?

Ever since Arrhenius's day, scientists have wanted to be able to predict which molecules that are dissolved in water will result in giving the solution acidic or basic properties. Making this prediction based on the structure of the molecule is no easy task. Arrhenius thought proton and hydroxide groups were necessary. Later, Brønsted and Lowry noticed that molecules without hydroxide groups could also give solutions basic properties so Arrhenius's definition needed updates. Soon after, Lewis noted that certain metals have acidic properties and do so by accepting lone pair electrons. Again, the definition was modified to fit the context of the molecules of interest.

4. What is the approximate ratio of H2O molecules to OH- molecules in a pH 12 solution?

A pH 12 solution has a H+ concentration of 1 × 10-12 M (pH = -log([H+])). The OH- concentration can be computed using the relationship that [H+][OH-] = Kw. Plugging in 1 × 10-12 for [H+] and 1 × 10-14 for Kw gives a OH- concentration of 0.01 M. The concentration of H2O in a solution of H2O is 55 M. The approximate ratio of H2O to OH- molecules is therefore the ratio of 55 to 0.01 or 5500. Even at a very basic pH for every 1 OH- molecule there will be approximately 5500 molecules of H2O illustrating that the dissociation of water is a rare event indeed.

5. Oxalic acid is a symmetrical molecule with two protons that readily dissociate. Which acid of oxalic acid is more acidic: H2C2O4 or HC2O4-? What is a reasonable physical explanation for why they might be the same or different?

Because oxalic acid is symmetrical it may be tempting to think that the acidity, or propensity of each proton to dissociate would be the same. This is not true, however, because once one proton dissociates the likelihood of the other proton dissociating goes down. One potential reason the second proton dissociation is less likely is because it would require leaving an additional negative charge on the molecule that would be repelled from the other negatively charged oxygen. This scenario is unfavorable from an electrostatic perspective and therefore, the acidity of the second proton is weaker than the first proton. The pKa of H2C2O4 is 5.9 × 10-2 while the pKa of HC2O4- is 6.4 × 10-5.

6. The compound p-nitrophenol is more acidic than just plain old phenol (see structures below). Why do you think this might be? The proton on the OH group of the p-nitrophenol compound is about 1000 times more likely to dissociate than the proton on the OH group of the phenol compound. This makes the p-nitrophenol compound about 1000 times more acidic than the phenol compound. Remarkably, the structure of these two molecules, particularly around the OH groups, is quite similar. However, at the tail end of the molecule, p-nitrophenol has a nitro group, while the phenol just has a hydrogen atom (not drawn). One reason p-nitrophenol could be more acidic is because the nitro group, even at the tail end of the molecule, is able to stabilize the negative charge left after the proton on the OH group dissociates. Notice that the nitrogen atom of the nitro group is positively charged. This nitro group is therefore able to accommodate, through a process known to organic chemists as resonance stabilization, the electrons left on the oxygen atom after dissociation. The hydrogen at the tail end of the phenol, on the other hand, cannot stabilize negative charge like the nitro group and thus, the phenol is not very acidic.

7. Methane (CH4) is a very weak acid with a pKa of approximately 50. In a 1000 L solution of 0.001 M dissolved methane approximately how many molecules of methane are in the conjugate base form (CH3-)?

The pKa of methane can be used to compute the Ka, 10-50, for the dissociation equilibrium of methane forming CH3-. Using the method discussed in Section 4 on the equilibrium of weak acids enables the final equilibrium concentration of CH3- to be calculated. It is 3.16 × 10-27 M. In 1000 L of 3.16 × 10-27 M CH3- there are 3.16 × 10-24 moles of CH3-. Converting moles to molecules using Avogadro's number reveals that there would be approximately 2 molecules of CH3- at any given time in this 1000 L solution.

8. Explain, in words, the relationship between a buffer's pKa and a given pH of the buffer.

The Henderson-Hasselbalch equation is the mathematical description of this relationship. The equation, which is simply derived from the definition of the equilibrium constant for a weak acid equilibrium, tells us that if the pH is below the buffer's pKa there will be more buffer in the acid (HA) form than in the conjugate base (A-) form. If the pH of the solution is the same as the buffer's pKa value, half of the weak acid is dissociated and the ratio of HA to A- is one. If the pH of the solution is above the buffer's pKa value, there will be more buffer in the conjugate base form (A-) than in the acid (HA) form.

9. Determining the equilibrium concentrations of the products of a weak acid or base dissociation is made easier by making an assumption that the initial concentration of weak acid or base does not change significantly. Describe the necessary conditions that make this assumption valid.

The Ka or Kb value of the weak acid or base must be small relative to the initial concentration of the acid or base in order for the assumption to hold. When this is the case, the change in the initial concentration of acid or base is negligible. Because the assumption is based on the relative values of equilibrium constant and initial concentration there is no hard and fast rule for what these values must be for the assumption to hold. However, if the initial concentration is more than 10-fold greater than the equilibrium constant you can bet on the assumption that the initial concentration of reactant will not change significantly.

10. Imagine a 1 L solution of a 1 M buffer with a pKa of 9 at a pH of 9. If this solution is diluted into 9 L of H2O (pKa = 15.7) how will the ratio of the buffer acid and its conjugate base change and what will the pH of the final 10 L solution be?

Witness the magic of a buffer. They hold their pH even after dilution. The Henderson-Hasselbalch equation tells us that when the pH of a buffer solution is at its pKa the ratio of the buffer acid and its conjugate base will be one. In other words there are equal concentrations of the buffer acid and conjugate base when the buffer is in a solution whose pH equals the buffer's pKa. Upon dilution the concentration of buffer goes down (by 10-fold in the example above) but the ratio of buffer acid and its conjugate base remains the same. The contribution of H+ and OH- from water is still miniscule compared to the overall equilibrium concentrations of H+ and OH- from the buffer because the acid dissociation constant of the buffer is about 105-fold larger than Kw (Kw = 10-14; Ka of buffer, 10-9) and the concentration of buffer is still quite significant at 0.1 M after the 10-fold dilution. Of course, after a much larger dilution, the equilibrium of water dissociation will start to dominate and the pH will begin to drift to neutral.