### All College Chemistry Resources

## Example Questions

### Example Question #151 : College Chemistry

What is the of a solution that contains hydrofluoric acid and fluoride?

Note: for hydrofluoric acid.

**Possible Answers:**

**Correct answer:**

In this question, we're given the concentrations of both a weak acid and its conjugate base in solution. We're also provided with the acid-dissociation constant for this acid, and we're asked to find the pH of the solution.

The easiest way to go about solving this problem is to first convert the acid-dissociation constant given into pKa. Then, we can use the Henderson-Hasselbalch equation to solve for the pH of this solution.

Now that we have the pKa, we can use it, along with the concentrations of the weak acid and its conjugate base, in order to solve for the pH of the solution.

### Example Question #1 : Henderson Hasselbalch Equation

Ibuprofen has a pKa of . What is the ratio of to in the blood ( )?

**Possible Answers:**

**Correct answer:**

The Henderson-Hasselbach equation can be used to find the answer.

represents the concentration of conjugate base, while represents the concentration of weak acid.

When we plug the given numbers into this equation, we get:

### Example Question #1 : Henderson Hasselbalch Equation

A buffer contains ammonia and ammonium chloride. What is its pH?

**Possible Answers:**

**Correct answer:**

To find the answer to this question, we must use the Henderson-Hasselbach equation:

refers to the concentration of base (in this case, ammonium chloride). refers to the concentration of weak acid (in this case, ammonia). We must then plug the given numbers into the equation and solve for pH.

### Example Question #1 : Henderson Hasselbalch Equation

Suppose that a solution containing equal amounts of acetic acid and acetate is combined with a solution containing hydrochloric acid. What is the pH of the resulting solution? The pKa of acetic acid is .

**Possible Answers:**

**Correct answer:**

In this question, we're given two different solutions. For each, we're told the volume as well as the concentration of acid within that solution. Then, we're told that the two solutions are mixed. Upon combining them, we're asked to determine what the pH of the resultant mixture will be.

To answer this question, the first thing we have to realize is that acetic acid and its conjugate base, acetate, can act as a buffer system. This means that it acts to resist dramatic changes in pH. Because we are told that acetic acid and its conjugate base are in equal amounts, we know that the pH of the starting solution is equal to the pKa of acetic acid.

When is added to the solution, we know that it will completely dissociate because it is a strong acid. This will, in turn, make the solution more acidic. The buffer system, however, will help to ensure that the pH does not drop too significantly, although we can still expect the pH to fall slightly below the pKa.

It's important to remember that when the two solutions are combined, the volume changes as well. Since each solution begins at , combining them together will result in a final solution with twice the volume, or . The significance of this is that the molarity of acetic acid and acetate will be cut in half, each to a value of . And upon reacting with the extra protons produced by the , this will also change the concentration slightly.

In this new mixture, of the protons produced by will combine with an equivalent of acetate to produce an additional of acetic acid. Hence, we can calculate the final molarity of the buffer components. Acetic acid will have a concentration of and acetate will have a concentration of .

We now have what we need to calculate the pH of the mixture. Plugging these values into the Henderson-Hasselbalch equation will give us the answer.

### Example Question #1 : Henderson Hasselbalch Equation

Determine the pH of an aqueous solution of 10 mL of 0.03 M acetic acid, and 15 mL of 0.025 M acetate, . pKa of acetic acid is 4.75.

**Possible Answers:**

4.85

4.67

3.80

4.83

**Correct answer:**

4.85

To answer this question, we can use the Henderson-Hasselbalch equation since we have a buffer system, a weak acid and its conjugate base.

Acetic acid is our weak acid and acetate is the conjugate base.

Since we are given volumes and concentration we need to figure out the molar ratio between acetate and acetic acid. We cannot just plug in the given concentrations.

This pH makes sense because the number should be more basic. We have more moles of the base than we do of the acid. If we had more moles of acid we would expect the pH to be lower.

### Example Question #71 : Reactions

You are in chemistry lab performing a titration. You were given an aqueous solution with an unknown concentration of acetate, to solve through titration with concentrated hydrochloric acid. You know that the pKa of acetic acid is 4.75 and that your titrant is 0.1 M hydrochloric acid.

After you have added 1 mL of 0.1 M hydrochloric acid, what is the dominant species in the solution? Note: The predominant species is the pH-determining species.

**Possible Answers:**

Acetate,

Water,

Acetic acid,

Hydrochloric acid, hydrochloric acid

**Correct answer:**

Acetate,

The answer is the acetate ion because at this point, there has not been enough hydrochloric acid added to make a lot of acetic acid. And the hydrochloric acid is immediately used up upon addition into the solution. The water is at a constant concentration so is not considered the predominant species.

### Example Question #26 : Acid Base Reactions

Determine the pH of the following combination of solutions:

8 mL of 0.04 M HCl

5 mL of 0.8 M ammonia

*pKa of ammonium = 9.26*

**Possible Answers:**

10.56

7.95

8.2

10.32

**Correct answer:**

8.2

The first reaction that has to happen is the creation of , ammonium.

We need to determine how many moles of hydrogen ions that we have from the HCl, that is equal to the amount of that is formed.

We need to determine the total amount of moles of ammonia, and then how many are remaining.

Now we can use the Henderson-Hasselbalch equation to solve for the pH

### Example Question #161 : College Chemistry

Determine the pH of the following combination of solutions:

6 mL of 0.25 M ammonium

5 mL of 0.34 M ammonia

*pKa of ammonium = 9.26*

**Possible Answers:**

9.39

9.20

9.12

9.31

**Correct answer:**

9.31

The relevant chemical reaction here is:

First we need to determine how many moles we have of ^{ }and _{. }

Now we can use the Henderson-Hasselbalch equation to solve for the pH.

### Example Question #28 : Acid Base Reactions

Determine the pH after combining these two solutions:

10 mL of 0.5 M NaOH

25 mL of 0.5 M acetic acid

pKa of acetic acid = 4.75

**Possible Answers:**

4.57

4.92

4.35

5.15

**Correct answer:**

4.57

The first reaction that will happen is the formation of acetate from the reaction of the strong base with acetic acid:

How many moles of OH^{-} are reacting? Using the volume:

How many moles of CH_{3}COOH are reacting? Doing the same as above:

So the amount of acetate, formed is equal to the number of moles of added.

And to figure out how much acetic acid is leftover:

moles of acetic acid - moles of moles of acetic acid leftover

**So now that we know how many moles we have of the weak acid and its conjugate base, we can use the Henderson-Hasselbalch equation: **

### Example Question #29 : Acid Base Reactions

Determine what volume of 0.1 M ammonia, in mL, should be added to 45 mL of 0.025 M ammonium to create a buffer solution with a pH of 7.34.

*pKa ammonia = 9.26*

**Possible Answers:**

0.001 mL

1.35 mL

3 mL

0.135 mL

**Correct answer:**

0.135 mL

We need to use the Henderson-Hasselbalch equation to determine how many moles of ammonia we need.

First we need to determine the number of moles of ammonium we have:

When we plug that into the Henderson-Hasselbalch equation:

In this case, we have the acid, and we are looking for the base

Raise both sides to the 10

Cross multiply

moles of ammonia

Now we need to determine the volume of 0.1 M ammonia that gives us that number of moles: