Reactions - College Chemistry
Card 0 of 20
Determine the oxidation number of each element in the compound 
Determine the oxidation number of each element in the compound
The oxidation number of each element in the compound 
is:
The oxidation number of chlorine is 
. There are 3 chlorine atoms present in the compound, so 
Because this is a neutral atom, the overall charge is 0. Therefore we can set 
Therefore, 
We can then solve for the oxidation number for 
The oxidation number of each element in the compound
The oxidation number of chlorine is
Because this is a neutral atom, the overall charge is 0. Therefore we can set
Therefore,
We can then solve for the oxidation number for
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Given the following equation, identify the reducing agent.
Given the following equation, identify the reducing agent.
Recall that a reducing agent is being oxidized and is losing electrons. Thus, when looking at the equation, look for which oxidation state is becoming more positive.
Start by assigning oxidation states to the elements.
For the reactants:
has an oxidation state of 
.
has an oxidation state of 
.
has an oxidation state of 
has an oxidation state of 
.
Now, assign the oxidation state of the products.
has an oxidation state of 
.
has an oxidation state of 
.
has an oxidation state of 
.
has an oxidation state of 
.
Only 
undergoes a loss of electrons. It must be the reducing agent.
Recall that a reducing agent is being oxidized and is losing electrons. Thus, when looking at the equation, look for which oxidation state is becoming more positive.
Start by assigning oxidation states to the elements.
For the reactants:
Now, assign the oxidation state of the products.
Only
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Which compound below has a nitrogen atom at a 
oxidation state?
Which compound below has a nitrogen atom at a
Let's start by going through each answer case-by-case applying the elementary rules of oxidation states we are given. The most applicable of these rules in this problem are the oxidation state of hydrogen in a compound is generally equal to 
, and the oxidation state of oxygen in a compound is generally equal to 
.
Let's start by looking at 
. We know this compound as a whole is neutrally charged (equal to 
overall charge), and applying our knowledge of the general oxidation state of hydrogen in a compound and knowing there are 
hydrogens we know that the 
hydrogens have a total charge of 
together, and nitrogen has an unknown oxidation state 
, so for our equation we have:
solving for 
gives 
, giving us the answer we need. Therefore the nitrogen atom in 
has an oxidation state of 
, which is the correct answer.
Let's still look at the other cases, however:
As this compound is uncharged we know it's net charge is equal to 
Using our knowledge of oxidation states of hydrogen and oxygen and counting the number of hydrogens and oxygen in this compound, we can determine that the total charge of all the hydrogens together is equal to 
, and that the total charge of the one oxygen is 
.
Just as we did above we can create an equation and solve for our unknown.
therefore 
, so the oxidation state of nitrogen in 
is 
. This means this isn't the right compound.
Next let's look at 
:
We can quickly apply our knowledge of oxidation states of oxygen knowing that the oxidation state of oxygen in compounds is generally 
. Since there are 
oxygens we must multiply 
to find out the total charge contributed by all the oxygens, which is 
.
So our equation for this becomes:
therefore 
and the oxidation state of nitrogen is 
, which isn't the answer we are looking for.
Finally the last compound is 
:
With this compound we can apply the rule we know about hydrogen's oxidation state in a compound being equal to 
. Since there are 
hydrogens, we must multiply 
which equals 
.
Since there are two nitrogens, our equation is:
and 
. Therefore the oxidation state of nitrogen is equal to 
.
Let's start by going through each answer case-by-case applying the elementary rules of oxidation states we are given. The most applicable of these rules in this problem are the oxidation state of hydrogen in a compound is generally equal to
Let's start by looking at
solving for
Let's still look at the other cases, however:
As this compound is uncharged we know it's net charge is equal to
Using our knowledge of oxidation states of hydrogen and oxygen and counting the number of hydrogens and oxygen in this compound, we can determine that the total charge of all the hydrogens together is equal to
Just as we did above we can create an equation and solve for our unknown.
therefore
Next let's look at
We can quickly apply our knowledge of oxidation states of oxygen knowing that the oxidation state of oxygen in compounds is generally
So our equation for this becomes:
therefore
Finally the last compound is
With this compound we can apply the rule we know about hydrogen's oxidation state in a compound being equal to
Since there are two nitrogens, our equation is:
and
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In the given reaction, which element(s) is/are oxidized?
In the given reaction, which element(s) is/are oxidized?
Let's start by looking at the equation and assigning oxidation numbers based off of the general rules we are given:
Let's start by looking at the reactants, namely 
. Knowing that an atom in its elemental form has an oxidation number of 
, 
has an oxidation number equal to 
.
Next let's look at 
. This is a neutral compound so know the oxidation number of the whole compound must equal 
. Going off of our general rules, the oxidation number of fluorine is equal 
. Since we have 3 fluorines we multiply the oxidation number by 
to get the cumulative oxidation number of all the fluorines together to get 
. So now we must solve for the oxidation of Cl, which is unknown. This is done through a simple algebraic equation: 
so 
. Therefore the oxidation number of Chlorine is 
.
Now we have the oxidation numbers of all the elements present in this equation that are on the reactant side. I recommend writing down their oxidation numbers next to each other, so the elements that are oxidized and reduced can quickly be determined.
Oxidation numbers of elements in reactants:
Now let's look at the product side of the equation and determine the oxidation numbers of the elements there. First let's look at 
. Knowing the general rules of oxidation numbers we know that fluorine has an oxidation number of 
. Since there are 
fluorines we multiply 
which equals 
. We can now solve for the oxidation number of Uranium.
, therefore 
has an oxidation number of 
.
Finally, let's look 
. We can once again use the general rule of oxidation numbers that fluorine has a 
oxidation number to simplify this. Therefore 
, therefore Cl has an oxidation number of 
in the product.
Let's now write down the oxidation numbers of all the elements in the products.
Oxidation numbers of elements in products:
Now let's compare the oxidation numbers of the elements in the reactants with those in the products. 
goes from a 
oxidation number to a 
, therefore it is the element that's oxidized (oxidation represents a loss in electrons). 
goes from 
to 
, therefore it is reduced (reduction means gaining electrons). Finally 
's oxidation number is 
in both the products and reactants therefore it is neither oxidized nor reduced. This means that 
is the only element that's oxidized.
Let's start by looking at the equation and assigning oxidation numbers based off of the general rules we are given:
Let's start by looking at the reactants, namely
Next let's look at
so
Now we have the oxidation numbers of all the elements present in this equation that are on the reactant side. I recommend writing down their oxidation numbers next to each other, so the elements that are oxidized and reduced can quickly be determined.
Oxidation numbers of elements in reactants:
Now let's look at the product side of the equation and determine the oxidation numbers of the elements there. First let's look at
Finally, let's look
Let's now write down the oxidation numbers of all the elements in the products.
Oxidation numbers of elements in products:
Now let's compare the oxidation numbers of the elements in the reactants with those in the products.
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In which of the following compounds is the oxidation state of phosphorus the least?
In which of the following compounds is the oxidation state of phosphorus the least?
Let's start by examining each of the compounds we are given case-by-case and applying the general rules of oxidation numbers we know.
Let's start with 
:
We know that based off of the general rules of oxidation states, Oxygen generally has an oxidation number of 
, since we have 
oxygens we must multiply 
to get the total cumulative charge of all the oxygens. This equals 
.
Now we can solve for the oxidation state of phosphorus:
Since we have 
Phosphoruses we multiply our unknown 
by 
.
.
This is equal to 
because the compound has an overall 
net charge.
Solving for 
gives you 
therefore the oxidation state of Phosphorus in this compound is equal to 
.
Next let's looking at 
:
Knowing that an atom in its elemental form has an oxidation state of 
. This compound has an oxidation number of 
.
Now let's look at 
:
Knowing that the oxidation number of hydrogen is 
and that there are 
hydrogens, the total oxidation of all the hydrogens together is 
, therefore our equation is:
so 
, therefore the oxidation state of Phosphorus in this compound is 
.
Now let's consider 
:
Using our elementary rules for the oxidation numbers, we know oxygen has a 
oxidation number. We can also determine the oxidation number of chlorine, using the rule stating that the oxidation number of halogens is usually 
. Since there are three chlorines we must multiply 
, which gives us 
.
Therefore our equation is:
therefore 
, so the oxidation number of phosphorus in this case equals 
.
Finally let's look at 
:
Using the rules stated above for 
, we obtain the equation:
Therefore 
meaning the oxidation number of 
.
This means that the compound with the lowest oxidation state is 
, therefore it is the right answer.
Let's start by examining each of the compounds we are given case-by-case and applying the general rules of oxidation numbers we know.
Let's start with
We know that based off of the general rules of oxidation states, Oxygen generally has an oxidation number of
Now we can solve for the oxidation state of phosphorus:
Since we have
This is equal to
Solving for
Next let's looking at
Knowing that an atom in its elemental form has an oxidation state of
Now let's look at
Knowing that the oxidation number of hydrogen is
so
Now let's consider
Using our elementary rules for the oxidation numbers, we know oxygen has a
Therefore our equation is:
therefore
Finally let's look at
Using the rules stated above for
Therefore
This means that the compound with the lowest oxidation state is
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What is the oxidation state of gold in 
?
What is the oxidation state of gold in
Let's first consider the overall compound 
. Since it is negatively charged overall it is equal to 
. Applying our elementary rules of oxidation states we know that halogens generally have an oxidation state of 
, so this means chlorine has an oxidation state of 
. Since there are 
chlorines we must multiply 
to get the charge of all the chlorines together. This is equal to 
.
We can now solve for oxidation state of gold through creating an equation as shown below:
Simplifying this we get that 
, therefore the oxidation state of gold is equal to 
Let's first consider the overall compound
We can now solve for oxidation state of gold through creating an equation as shown below:
Simplifying this we get that
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What species is reduced in the following chemical reaction?
What species is reduced in the following chemical reaction?
Reduction is defined as the gain of electrons so we want to find the element that gains electrons/becomes negatively charged.
Once again our equation is 
Let's start on the reactant side with 
:
Following our general rules of oxidation states, we know that an atom in its elemental form has an oxidation state equal to zero, therefore 
has an oxidation state equal to 
.
Next, consider 
: Based off of our general oxidation state rules we known the oxidation number of oxygen is 
, since we have two oxygens we must multiply 
. This means that the overall combined charge of the oxygens in this compound is equal to 
.
We can now solve for the oxidation state of 
using this equation, which set equal to zero because the net charge of the compound is 
.
, therefore 
Now let's consider 
:
Applying the rule for the oxidation number of oxygen being 
and the rule of hydrogen being equal to +1, we obtain the equation:
where 
is the oxidation number of sulfur.
This simplifies so that 
, therefore the oxidation number of 
.
Moving on to the products, let's consider 
:
We know that the oxidation state of Oxygen is 
based off the rules mentioned above. We can also determine that the oxidation number of 
is equal to 
because lead is most stable when it loses two electrons.
Now we must solve for sulfur's oxidation state:
therefore 
, therefore the oxidation state of sulfur 
.
As for 
, we can just use the general oxidation rules mentioned above to determine 
Looking at the changes of oxidation numbers between elements in the products and reactants we see that there is no change in the oxidation state of sulfur, hydrogen, and oxygen between the products and reactants, but lead goes from 
to 
, so therefore it is reduced.
Reduction is defined as the gain of electrons so we want to find the element that gains electrons/becomes negatively charged.
Once again our equation is
Let's start on the reactant side with
Following our general rules of oxidation states, we know that an atom in its elemental form has an oxidation state equal to zero, therefore
Next, consider
We can now solve for the oxidation state of
Now let's consider
Applying the rule for the oxidation number of oxygen being
where
This simplifies so that
Moving on to the products, let's consider
We know that the oxidation state of Oxygen is
Now we must solve for sulfur's oxidation state:
therefore
As for
Looking at the changes of oxidation numbers between elements in the products and reactants we see that there is no change in the oxidation state of sulfur, hydrogen, and oxygen between the products and reactants, but lead goes from
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What is the oxidation state of copper in 
?
What is the oxidation state of copper in
In order to determine the oxidation state of copper in this compound we first must note the fact that compound has an overall charge of 
.
We can also apply the general rules of oxidation numbers we know to determine the oxidation numbers of hydrogen and oxygen. Since hydrogen is a group 
element, it has an oxidation number of 
. Since there are two hydrogens we must multiple 
which equals 
to get the total charge of both hydrogens.
As for oxygen, based off of our general rules of oxidation numbers, we know that oxygen normally has an oxidation number of 
, so knowing the oxidation numbers of oxygen and hydrogen, and net the charge of the compound, we can setup an equation to solve for the oxidation number of copper as shown below:
therefore 
, therefore copper has an oxidation number equal to 
.
In order to determine the oxidation state of copper in this compound we first must note the fact that compound has an overall charge of
We can also apply the general rules of oxidation numbers we know to determine the oxidation numbers of hydrogen and oxygen. Since hydrogen is a group
As for oxygen, based off of our general rules of oxidation numbers, we know that oxygen normally has an oxidation number of
therefore
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What is the oxidation number of sulfur in 
?
What is the oxidation number of sulfur in
In order to determine the oxidation number we must apply the general rules of oxidation numbers we know. One of these rules states the oxidation number of fluorine is always 
. Since there are 
fluorines that means the fluorines together contribute a 
charge to the compound. Since the overall charge of the compound is 
, we can create an equation with an unknown (charge of sulfur) and solve for it to find the oxidation number of sulfur.
This equation is 
The reason why it is 
instead of just 
is because there are 
sulfurs.
therefore the oxidation number of sulfur is 
.
In order to determine the oxidation number we must apply the general rules of oxidation numbers we know. One of these rules states the oxidation number of fluorine is always
This equation is
therefore the oxidation number of sulfur is
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When balanced, what is the value of 
in the following chemical equation?
When balanced, what is the value of
Recall that a balanced chemical equation has the same number of each element on one side.
Start by counting the number of each element on each side.
There are the following numbers of moles of each reactant:
There are the following numbers of moles of each product:
Add coefficients in front of the molecular compounds in the equation until there are the same numbers of sulfur, oxygen, lithium, and selenium on each side.
The balanced chemical equation is the following:
Both products and reactants now have the following number of moles:
Since the coefficient in front of 
is 
, 
must equal to 
.
Recall that a balanced chemical equation has the same number of each element on one side.
Start by counting the number of each element on each side.
There are the following numbers of moles of each reactant:
There are the following numbers of moles of each product:
Add coefficients in front of the molecular compounds in the equation until there are the same numbers of sulfur, oxygen, lithium, and selenium on each side.
The balanced chemical equation is the following:
Both products and reactants now have the following number of moles:
Since the coefficient in front of
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Balance the following chemical equation:
Balance the following chemical equation:
A balanced chemical equation will have the same number of each atom on both sides of the reaction.
Start by balancing the number of nitrate.
Since we have 
nitrate on the left, we must also have 
nitrate on the right.
This equation now gives 
potassium on the left, and 
potassium on the right. Balance the potassium.
This equation is balanced because there are equal numbers of each atom on both sides.
A balanced chemical equation will have the same number of each atom on both sides of the reaction.
Start by balancing the number of nitrate.
Since we have
This equation now gives
This equation is balanced because there are equal numbers of each atom on both sides.
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Balance the following redox reaction in acidic conditions:
Balance the following redox reaction in acidic conditions:
Start by writing the half reactions for the equation
All atoms except for O and H are already balanced, so we can go straight to balancing O and H atoms. First balance O atoms by adding H2O
Now balance H atoms with H+
Balance each half reaction's electrical charge with electrons
Now combine both half reactions to get the overall reaction. Be sure to cancel out electrons by multiplying the oxidation reaction by 5 and the reduction reaction by 2. This will give 10 e- on each side of the overall reaction so that they cancel out.
Overall Reaction:
Notice that we have H2O and H+ on both sides of the equation, so we need to simplify. Subtract the 5 H2O on the reactant side from the 8 H2O on the product side which leaves 3 H2O on the reactant side. Then subtract the 10 H+ on the product side from the 16 H+ on the reactant side which leave 6 H+ on the reactant side. Now we have the simplified overall reaction
Start by writing the half reactions for the equation
All atoms except for O and H are already balanced, so we can go straight to balancing O and H atoms. First balance O atoms by adding H2O
Now balance H atoms with H+
Balance each half reaction's electrical charge with electrons
Now combine both half reactions to get the overall reaction. Be sure to cancel out electrons by multiplying the oxidation reaction by 5 and the reduction reaction by 2. This will give 10 e- on each side of the overall reaction so that they cancel out.
Overall Reaction:
Notice that we have H2O and H+ on both sides of the equation, so we need to simplify. Subtract the 5 H2O on the reactant side from the 8 H2O on the product side which leaves 3 H2O on the reactant side. Then subtract the 10 H+ on the product side from the 16 H+ on the reactant side which leave 6 H+ on the reactant side. Now we have the simplified overall reaction
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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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Consider the following reaction:
Give the expression for the equilibrium constant for this reaction.
Consider the following reaction:
Give the expression for the equilibrium constant for this reaction.
Recall how to find the expression of the equilibrium constant for the simplified equation:
Since the given equation has gases, we will only consider the partial pressures of each gas in the expression for the equilibrium constant. Remember that only molecules in aqueous and gas forms are included in this expression. Pure solids and pure liquids are excluded.
Thus, we can then write the following equilibrium constant for the given equation:
Recall how to find the expression of the equilibrium constant for the simplified equation:
Since the given equation has gases, we will only consider the partial pressures of each gas in the expression for the equilibrium constant. Remember that only molecules in aqueous and gas forms are included in this expression. Pure solids and pure liquids are excluded.
Thus, we can then write the following equilibrium constant for the given equation:
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