Reaction Types - AP Chemistry

The definition of a buffer solution is that it contains a weak acid and its conjugate base, or a weak base and its conjugate acid. Since we are starting with a weak acid in this case, we need its conjugate base.

For a buffer solution, you need a weak acid and its conjugate base, or a weak base and its conjugate acid. HCO3 from the NaHCO3 and CO3– from K2CO3 are this pair.

The Ka is the acid dissociation constant, and thus it is what determines how strong the acid is. Stronger acids dissociate to a greater extent and produce lower pH values.

The conjugate base has one less H atoms and one unit greater negative charge because of this. Thus, the correct answer is HC2O4–

A weak acid/base best buffers about 1 pH point above and below its pKa. The pKA closest to the middle of 4 and 6 (so want as close to 5) is acetic acid at 4.7.

The pH at half the equivalence point is equal to the pKa of the acid.

To answer this question we need to look at the reaction below:

An increase in the pH will result in a decrease in the concentration of hydrogen ions (). Using Le Chatelier’s principle we can find out which answer choices will decrease .

Removing carbonic acid will decrease the concentration of . To maintain equilibrium, the reaction will shift to the left and make more reactants from products; therefore, there will be a decrease in the and an increase in pH.

Recall that salts like sodium bicarbonate, or , will dissociate in water and form ions. Sodium bicarbonate will form sodium () and bicarbonate () ions. This side reaction will result in an increase in the bicarbonate ion concentration. Le Chatelier’s principle will shift the equilibrium of the given reaction to the left and, therefore, decrease the . Adding sodium bicarbonate will increase the pH.

Buffer solutions can be made via two methods. The first method involves adding equal amounts of a weak acid and a salt of its weak conjugate base (or vice versa). The second methods involves adding a weak acid and a half equivalent of a strong base (or vice versa).

is a weak acid and is a salt of its weak conjugate base; therefore, this can form a buffer.

is a weak base and is a salt of its weak conjugate acid; this can also form a buffer. Note that this is the converse of the first method (weak base with salt of weak acid), but it can still form a buffer solution.

is a strong acid and is a weak base; therefore, adding and a half equivalent of will create a buffer solution. This is the converse of the second method (adding a weak base to a half equivalent of strong acid).

and are both strong reagents (acid and base, respectively); therefore, they cannot form a buffer solution.

The question states that the patient has low levels of carbon dioxide. If we look at the given reaction, we will notice that the reaction will compensate for this by shifting the reaction equilibrium to the left. This phenomenon is called Le Chatelier’s principle and occurs to maintain the equilibrium of the reaction; therefore, the reaction will create more carbon dioxide by utilizing bicarbonate and hydrogen ions in the blood. A decrease in hydrogen ion concentration in blood will increase the pH and cause alkalosis (basicity in the blood). Since carbon dioxide is the cause of alkalosis, this patient will experience respiratory alkalosis. If he experienced alkalosis due to a change in bicarbonate ion concentration, the patient will have metabolic alkalosis.

The ratio of carbonic acid to bicarbonate will stay the same because both will be used in equal amounts (1:1 ratio) to produce carbon dioxide. Increasing respiratory rate, or hyperventilation, will result in an increase in the amount of carbon dioxide expelled by the patient; this will decrease the carbon dioxide concentration in the blood and will worsen the respiratory alkalosis. Recall that we are utilizing the bicarbonate ion (in conjunction with hydrogen ions) to create carbonic acid. The carbonic acid will be further broken down to replenish the carbon dioxide. A decrease in the bicarbonate concentration will slow down this process.

One way to make a buffer is by adding equal amounts of a weak acid to its weak conjugate base. For example, you can add 1M acetic acid to 1M acetate to create a buffer solution (note that both acetic acid and its conjugate base (acetate) are weak). However, when using this method you have to remember that the desired pH of the buffer solution has to equal the pKa of the weak acid. The question states that the pKa of the acid is 5.9 and the desired pH of the buffer is 3.5; therefore, it is not possible to make the buffer with the given acid. The researcher would have to find another acid that has a pKa near 3.5.

Acidic solutions have a pH level lower than 7, which is the neutral pH. Basic solutions have a pH level greater than 7.

When performing acid/base calculations, the Henderson-Hasselbalch equation is useful:

To use this equation we need to convert the to the , and can use the following definition to do so:

Since an individual will begin experiencing acidosis when the blood falls below 7.35 we can use 7.35 as the pH in the Henderson-Hasselbalch equation. Adding in the of 6.10 calculated from the gives the equation below:

Plug in known values and solve:

Correct Answer:

The answer is unitless because all units cancel out.

Larger halogen size leads to greater acidity because of weaker H-X interactions.

Acid strength increases from left to right across a period and increases going down a group.

Acid strength of an oxy acid increases with increasing electronegativity on the halogen.

For an oxyacid, the acid strength increases as the number of oxygens increase.

Acid strength increases going across a period.

pOH would be 4 (use –log \[OH–\]) and pH would be 14–pOH = 14 – 4 = 10

\[OH-\] = 4 X 10-6

pOH = 5.4 — use –log \[OH–\] to find pOH

pH = 14– pOH = 8.6