Acid-Base Reactions - College Chemistry
Card 0 of 20
Determine the pH of an aqueous solution of 0.01 M acetic acid, 
. The pKa of acetic acid is 4.75.
Determine the pH of an aqueous solution of 0.01 M acetic acid,
Since acetic acid is a weak acid, it has a Ka that is rather small, we have to do a RICE table to determine the equilibrium amount of hydronium, H3O+ to then determine the pH.
R 
I 0.1 M - 0 0
C -x +x +x
E 0.1 -x x x
So first we need to change our pKa to a Ka
where 
therefore
= 
= 
If we assume that x is very small compared to 0.1...
Where 
(note: when solving using the quadratic we come up with the same answer)
So if 
Since acetic acid is a weak acid, it has a Ka that is rather small, we have to do a RICE table to determine the equilibrium amount of hydronium, H3O+ to then determine the pH.
R
I 0.1 M - 0 0
C -x +x +x
E 0.1 -x x x
So first we need to change our pKa to a Ka
where
If we assume that x is very small compared to 0.1...
Where
(note: when solving using the quadratic we come up with the same answer)
So if
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Which combination(s) would create a buffer solution?
I. Weak acid
II. Weak acid's conjugate base
III. Strong acid
IV. Strong base
V. Weak base
VI. Weak base's conjugate acid
Which combination(s) would create a buffer solution?
I. Weak acid
II. Weak acid's conjugate base
III. Strong acid
IV. Strong base
V. Weak base
VI. Weak base's conjugate acid
A buffer solution is formed from the equilibrium of a weak acid and its conjugate base, or from a weak base and its conjugate acid. It's ability to "buffer" the pH or keep it from changing in large amounts in from the switching between these two forms weak and its conjugate.
A buffer solution is formed from the equilibrium of a weak acid and its conjugate base, or from a weak base and its conjugate acid. It's ability to "buffer" the pH or keep it from changing in large amounts in from the switching between these two forms weak and its conjugate.
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You are in chemistry lab performing a titration. You were given 15 mL of an aqueous solution with an unknown concentration of acetic acid, 
to solve through titration with concentrated sodium hydroxide, 
. You know that the pKa of acetic acid is 4.75 and that your titrant is 0.1 M sodium hydroxide, 
.
The endpoint was determined at 10 mL of sodium hydroxide, 
. What is the pH after 5 mL of 
was added?
You are in chemistry lab performing a titration. You were given 15 mL of an aqueous solution with an unknown concentration of acetic acid,
The endpoint was determined at 10 mL of sodium hydroxide,
At the half end point, the 
. This can be determined by the Henderson-Hasselbalch equation if it is not clear.
Since the endpoint of the titration is that there are 10 mL of 0.1 M NaOH added, that means that there are 0.001 moles of acetic acid.
When 5 mL of NaOH is added, there are 0.0005 moles of acetic acid and 0.0005 moles of acetate formed.
Therefore pH= pKa
At the half end point, the
Since the endpoint of the titration is that there are 10 mL of 0.1 M NaOH added, that means that there are 0.001 moles of acetic acid.
When 5 mL of NaOH is added, there are 0.0005 moles of acetic acid and 0.0005 moles of acetate formed.
Therefore pH= pKa
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Determine which of these solution combinations form a buffer.
Determine which of these solution combinations form a buffer.
First to go through why the other ones are wrong:
Strong base + strong acid neutralizes and does not form a buffer solution
Strong base + weak base does not form a buffer - would need an acid
Strong base + weak acid = all weak acid converted to conjugate base
The correct answer is:
Strong base + weak acid = half converted to conjugate base with half leftover as weak acid, with all the components for a buffer
First to go through why the other ones are wrong:
Strong base + strong acid neutralizes and does not form a buffer solution
Strong base + weak base does not form a buffer - would need an acid
Strong base + weak acid = all weak acid converted to conjugate base
The correct answer is:
Strong base + weak acid = half converted to conjugate base with half leftover as weak acid, with all the components for a buffer
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Determine which combination of solutions would create a buffer solution.
Determine which combination of solutions would create a buffer solution.
For all the other options there is no ammonium leftover with which to serve as the weak acid in the buffer system, the ammonium is all used up and converted to ammonia. However in the correct answer choice, there is enough ammonium leftover after the reaction with the sodium hydroxide.
For all the other options there is no ammonium leftover with which to serve as the weak acid in the buffer system, the ammonium is all used up and converted to ammonia. However in the correct answer choice, there is enough ammonium leftover after the reaction with the sodium hydroxide.
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Determine which solution(s) will yield a buffer solution.
I. 10 mL of 0.5 M HCl + 20 mL of 0.5 M acetate
II. 10 mL of 0.5 M HCl + 10 mL of 0.5 M acetate
III. 10 mL of 0.5 M HCl + 10 mL of 1.0 M acetate
IV. 10 mL of 0.5 M HCl + 10 mL of 1.5 M acetate
Determine which solution(s) will yield a buffer solution.
I. 10 mL of 0.5 M HCl + 20 mL of 0.5 M acetate
II. 10 mL of 0.5 M HCl + 10 mL of 0.5 M acetate
III. 10 mL of 0.5 M HCl + 10 mL of 1.0 M acetate
IV. 10 mL of 0.5 M HCl + 10 mL of 1.5 M acetate
These answers are correct because the two components needed to create a buffer solution are a weak acid and its conjugate base, or a weak base and its conjugate acid. In these cases, the first reaction to occur upon addition of the strong acid is the formation of the conjugate acid, acetic acid.
If the amount of initial 
is greater than HCl, then we will have some 
left over to act as a buffer with the created conjugate acid. This can be through a greater volume, or through a higher concentration as shown in the correct answers.
These answers are correct because the two components needed to create a buffer solution are a weak acid and its conjugate base, or a weak base and its conjugate acid. In these cases, the first reaction to occur upon addition of the strong acid is the formation of the conjugate acid, acetic acid.
If the amount of initial
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Which of the following acid and base pairs are capable of acting as a buffer?
Which of the following acid and base pairs are capable of acting as a buffer?
In this question, we're presented with a variety of acid/base pairs and we're asked to identify which one could act as a buffer.
Remember that a buffer is a pair of acid and its conjugate base that acts to resist substantial changes in pH. In order for a buffer to work, the acid base pair needs to exist in equilibrium. This way, when the pH of the solution changes, the equilibrium of the acid/base reaction will shift, such that the pH will not change drastically.
To have an acid/base pair in equilibrium, we'll need to look for a pair that contains a weak acid. Acids like 
and 
are so strong that they will dissociate completely. Of the answer choices shown, only the carbonic acid/bicarbonate system (
and 
) exists in equilibrium. Thus, this is the correct answer.
In this question, we're presented with a variety of acid/base pairs and we're asked to identify which one could act as a buffer.
Remember that a buffer is a pair of acid and its conjugate base that acts to resist substantial changes in pH. In order for a buffer to work, the acid base pair needs to exist in equilibrium. This way, when the pH of the solution changes, the equilibrium of the acid/base reaction will shift, such that the pH will not change drastically.
To have an acid/base pair in equilibrium, we'll need to look for a pair that contains a weak acid. Acids like
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What is the 
of a solution that contains 
hydrofluoric acid and 
fluoride?
Note: 
for hydrofluoric acid.
What is the
Note:
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.
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.
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Ibuprofen has a pKa of 
. What is the ratio of 
to 
in the blood (
)?
Ibuprofen has a pKa of
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:
The Henderson-Hasselbach equation can be used to find the answer.
When we plug the given numbers into this equation, we get:
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A buffer contains 
ammonia and 
ammonium chloride. What is its pH?
A buffer contains
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.
To find the answer to this question, we must use the Henderson-Hasselbach equation:
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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 
.
Suppose that a
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.
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
It's important to remember that when the two solutions are combined, the volume changes as well. Since each solution begins at
In this new mixture,
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.
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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.
Determine the pH of an aqueous solution of 10 mL of 0.03 M acetic acid,
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.
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.
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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.
You are in chemistry lab performing a titration. You were given an aqueous solution with an unknown concentration of acetate,
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.
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.
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.
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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
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
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 CH3COOH 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:
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 CH3COOH are reacting? Doing the same as above:
So the amount of acetate,
And to figure out how much acetic acid is leftover:
moles of acetic acid - moles of
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:
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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
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
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.
The relevant chemical reaction here is:
First we need to determine how many moles we have of
Now we can use the Henderson-Hasselbalch equation to solve for the pH.
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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
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
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
The first reaction that has to happen is the creation of
We need to determine how many moles of hydrogen ions that we have from the HCl, that is equal to the amount of
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
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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
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
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:
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
Now we need to determine the volume of 0.1 M ammonia that gives us that number of moles:
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Identify the Bronsted-Lowry acid in the following equation:
Identify the Bronsted-Lowry acid in the following equation:
Recall that a Bronsted-Lowry acid donates a proton. From the equation, we can see that 
becomes 
, which it lost an 
, or a proton. Thus, water must be the Bronsted-Lowry acid.
Recall that a Bronsted-Lowry acid donates a proton. From the equation, we can see that
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Identify the Lewis Base in the following reaction:
Identify the Lewis Base in the following reaction:
Recall that a Lewis base accepts a proton, while a Lewis acid donates a proton. Looking at the equation, we can see that 
becomes 
. Since 
accepted a proton, it must be the Lewis base.
Recall that a Lewis base accepts a proton, while a Lewis acid donates a proton. Looking at the equation, we can see that
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What is the pH of a 
solution?
What is the pH of a
PH is defined as 
, where 
is hydronium concentration). In order to do this we simply take 
:
=
PH is defined as
=
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