pH - MCAT Physical
Card 0 of 77
What is the resulting pH when 7.0g of HCl is dissolved in 3L of water?
What is the resulting pH when 7.0g of HCl is dissolved in 3L of water?
HCl is a strong acid; it will fully dissociate in water, meaning that the concentration of H+ is equal to the concentration of HCl (they are in a 1 : 1 ratio). 7.0g of HCl is equal to 0.2mol (MW of HCl is 35.3g/mol). 0.2mol HCl goes into 3L of water, resulting in a concentration of 0.067M, or 
.
Now we know that ![\dpi{100} \small \left [ H^{+} \right ]=6.7\times 10^{-2}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/18636/gif.latex)
and pH=-log\[H+\]. Using our trick for -log, we can see that 
. Since 
and 
, we know our answer is between 1 and 2.
is closer to 
, so we can pick the answer closer to 1. i.e. 1.2.
HCl is a strong acid; it will fully dissociate in water, meaning that the concentration of H+ is equal to the concentration of HCl (they are in a 1 : 1 ratio). 7.0g of HCl is equal to 0.2mol (MW of HCl is 35.3g/mol). 0.2mol HCl goes into 3L of water, resulting in a concentration of 0.067M, or
Now we know that
Compare your answer with the correct one above
2.0g of a monoprotic strong acid are dissolved in 1L of water. The resulting pH is 2.0. What is the molecular weight of the acid?
2.0g of a monoprotic strong acid are dissolved in 1L of water. The resulting pH is 2.0. What is the molecular weight of the acid?
Using the pH of 2.0, we can find that ![\dpi{100} \small \left [ H^{+} \right ]=10^{-2}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/18654/gif.latex)
, because 
.
Since the acid is strong (fully dissociates) and monoprotic (one H+ per molecule),
.
Our solution has 1L of water, meaning that we have 
of acid. We know that only 2.0g of acid were used to achieve this concentration, meaning that there is a ratio of 
. Simplifying this ratio gives 
.
Using the pH of 2.0, we can find that
Since the acid is strong (fully dissociates) and monoprotic (one H+ per molecule),
Our solution has 1L of water, meaning that we have
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47.0g of nitrous acid, HNO2, is added to 4L of water. What is the resulting pH? 
47.0g of nitrous acid, HNO2, is added to 4L of water. What is the resulting pH?
HNO2 is a weak acid; it will not fully dissociate, so we need to use the HA → H+ + A– reaction, with ![\dpi{100} \small K_{a}=\frac{\left [ products \right ]}{\left [ reactants \right ]}=\left \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=4.1\times 10^{-4}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/18682/gif.latex)
.
47.0g HNO2 is equal to 1mol. 1mol into 4L gives a concentration of 0.25M when the acid is first dissolved; however, we want the pH at equilibrium, not at the initial state. As the acid dissolves, we know \[HNO2\] will decrease to become ions, but we don't know by how much so we indicate the decrease as "x". As HNO2 dissolves by a factor of x, the ion concentrations will increase by x.
HNO2 → H+ + NO2–
Initial 0.25M 0 0
Equilibrium 0.25 – x x x
Now, we can fill in our equation: ![\dpi{100} \small K_{a}=\left \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=\frac{\left ( x \right )\left ( x \right )}{0.25-x}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/18683/gif.latex)
.
Since x is very small, we can ignore it in the denominator: 
(they expect you to do this on the MCAT; you will never have to solve with x in the denominator on the exam!)
Solve for x, and you find 
. Looking at our table, we know that ![\dpi{100} \small x=\left [ H^{+} \right ]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/18686/gif.latex)
Now we can solve for pH: 
HNO2 is a weak acid; it will not fully dissociate, so we need to use the HA → H+ + A– reaction, with
47.0g HNO2 is equal to 1mol. 1mol into 4L gives a concentration of 0.25M when the acid is first dissolved; however, we want the pH at equilibrium, not at the initial state. As the acid dissolves, we know \[HNO2\] will decrease to become ions, but we don't know by how much so we indicate the decrease as "x". As HNO2 dissolves by a factor of x, the ion concentrations will increase by x.
HNO2 → H+ + NO2–
Initial 0.25M 0 0
Equilibrium 0.25 – x x x
Now, we can fill in our equation:
Since x is very small, we can ignore it in the denominator:
(they expect you to do this on the MCAT; you will never have to solve with x in the denominator on the exam!)
Solve for x, and you find
Now we can solve for pH:
Compare your answer with the correct one above
There are two containers, each containing different proton concentrations. In container A, 
. In container B, 
.
What is the difference in pH between these two containers?
There are two containers, each containing different proton concentrations. In container A,
What is the difference in pH between these two containers?
The MCAT will typically allow you to approximate the difference in pH between two solutions. Look at the problem this way: you should already know a proton concentration of 1 * 10-3M means a pH of 3. You do not need to know the exact pH of 5 * 10-6M, but you should recognize that it is in between 1 * 10-6M and 1 * 10-5M. This means it will have a pH between 5 and 6.
The difference in pH will therefore be between 5 minus 3 and 6 minus 3.
As a result, look for the answer that is between 2 and 3. The only answer that makes sense is 2.3.
The MCAT will typically allow you to approximate the difference in pH between two solutions. Look at the problem this way: you should already know a proton concentration of 1 * 10-3M means a pH of 3. You do not need to know the exact pH of 5 * 10-6M, but you should recognize that it is in between 1 * 10-6M and 1 * 10-5M. This means it will have a pH between 5 and 6.
The difference in pH will therefore be between 5 minus 3 and 6 minus 3.
As a result, look for the answer that is between 2 and 3. The only answer that makes sense is 2.3.
Compare your answer with the correct one above
HCN dissociates based on the following reaction.
The Ka for hydrogen cyanide is 
.
of 
is added to 
of water. What is the pH of the resulting solution?
HCN dissociates based on the following reaction.
The Ka for hydrogen cyanide is
Since HCN is a weak acid, we must use the equilibrium equation.
Because the HCN dissociates in solution, we expect the concentrations of protons and cyanide ions to increase, while the concentration of HCN will decrease. After determining the molarity of the solution, we can set up the equation below, using X as the amount of moles that dissociate.
Because X is small, we can neglect its impact in the denominator.
Since X is the concentration of protons in the solution, we can calculate the pH by using the equation 
.
Since HCN is a weak acid, we must use the equilibrium equation.
Because the HCN dissociates in solution, we expect the concentrations of protons and cyanide ions to increase, while the concentration of HCN will decrease. After determining the molarity of the solution, we can set up the equation below, using X as the amount of moles that dissociate.
Because X is small, we can neglect its impact in the denominator.
Since X is the concentration of protons in the solution, we can calculate the pH by using the equation
Compare your answer with the correct one above
A sample of gastric juice has a pH of 2.5. What is the hydrogen ion concentration in this secretion?
A sample of gastric juice has a pH of 2.5. What is the hydrogen ion concentration in this secretion?
The concentration of hydrogen ions must lie somewhere between 
and 
; alternatively stated, it is between 
and 
. The pH of a solution with hydrogen ion concentration of 
will be 3, and the pH of a solution with hydrogen ion concentration 
will be 2; thus, our concentration must lie between these two values, since our pH is 2.5
To find the exact concentration, you must be familiar with the logarithmic scale. A difference of 0.5 is equivalent to a log of 3.
Our answer must therefore be 
, or 
.
We can calculate the pH in reverse to check our answer.
The concentration of hydrogen ions must lie somewhere between
To find the exact concentration, you must be familiar with the logarithmic scale. A difference of 0.5 is equivalent to a log of 3.
Our answer must therefore be
We can calculate the pH in reverse to check our answer.
Compare your answer with the correct one above
Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.
All of the bases proceed in a similar fashion.
The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.
The pH of a solution of 
is lowered from 4 to 3, and then from 3 to 2. Which of the following is the most accurate description of what happens during these transitions?
Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.
All of the bases proceed in a similar fashion.
The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.
The pH of a solution of
The pH scale is logarithmic. Every pH unit drop corresponds to a tenfold increase in protons.
The pH scale is logarithmic. Every pH unit drop corresponds to a tenfold increase in protons.
Compare your answer with the correct one above
Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.
All of the bases proceed in a similar fashion.
The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.
A scientist is studying an aqueous sample of 
, and finds that the hydroxide concentration is 
. Which of the following is true?
Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.
All of the bases proceed in a similar fashion.
The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.
A scientist is studying an aqueous sample of
Given the hydroxide ion concentration, we will need to work using pOH to find the pH. We know that the sum of pH and pOH is equal to 14.
Use our value for the concentration to find the pOH.
Now that we have the pOH, we can use it to solve for the pH.
Given the hydroxide ion concentration, we will need to work using pOH to find the pH. We know that the sum of pH and pOH is equal to 14.
Use our value for the concentration to find the pOH.
Now that we have the pOH, we can use it to solve for the pH.
Compare your answer with the correct one above
What is the pOH of a 
aqueous solution of 
?
What is the pOH of a
The first step for this problem is to find the pH. We can then derive the pOH from the pH value.
The pH is given by the equation 
. Since hydrochloric acid is monoprotic, the concentration of the solution is equal to the concentration of protons.
Using this value and the pH equation, we can calculate the pH.
Now we can find the pOH. The sum of the pH and the pOH is always 14.
The pOH of the solution is 7.8.
Alternatively, a shortcut can be used to estimate the pH. If 
is in the form 
, then pH is roughly 
.
For this question, this shortcut gets us a pH of 6.4, which produces a pOH of 7.6; very close to the real answer!
The first step for this problem is to find the pH. We can then derive the pOH from the pH value.
The pH is given by the equation
Using this value and the pH equation, we can calculate the pH.
Now we can find the pOH. The sum of the pH and the pOH is always 14.
The pOH of the solution is 7.8.
Alternatively, a shortcut can be used to estimate the pH. If
For this question, this shortcut gets us a pH of 6.4, which produces a pOH of 7.6; very close to the real answer!
Compare your answer with the correct one above
An arterial blood sample from a patient has a pH of 7.4. One day later, the same patient has an arterial blood pH of 7.15. How many times more acidic is the patient's blood on the second day?
An arterial blood sample from a patient has a pH of 7.4. One day later, the same patient has an arterial blood pH of 7.15. How many times more acidic is the patient's blood on the second day?
The equation to calculate pH is:
The normal pH of arterial blood is around 7.4. This reflects a concentration of hydrogen ions that can be found using the pH equation.
Using similar calculations for the second blood sample, we can find the hydrogen ion concentration again.
Now that we have both concentrations, can find the ratio of the acidity of the two samples.
You may know from biological sciences that this is approaching a lethal level of acidosis.
The equation to calculate pH is:
The normal pH of arterial blood is around 7.4. This reflects a concentration of hydrogen ions that can be found using the pH equation.
Using similar calculations for the second blood sample, we can find the hydrogen ion concentration again.
Now that we have both concentrations, can find the ratio of the acidity of the two samples.
You may know from biological sciences that this is approaching a lethal level of acidosis.
Compare your answer with the correct one above
You find a bottle in a lab that has a 
solution of acid. The acid has the following dissociative properties:
What is the pH of this solution?
You find a bottle in a lab that has a
What is the pH of this solution?
is a strong acid, meaning it will completely dissociate in solution. As such, the concentration of the acid will be equal to the proton concentration. Thus, to find pH, you should just plug the molar concentration of the acid solution into the pH formula.
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What is the resulting pH when 7.0g of HCl is dissolved in 3L of water?
What is the resulting pH when 7.0g of HCl is dissolved in 3L of water?
HCl is a strong acid; it will fully dissociate in water, meaning that the concentration of H+ is equal to the concentration of HCl (they are in a 1 : 1 ratio). 7.0g of HCl is equal to 0.2mol (MW of HCl is 35.3g/mol). 0.2mol HCl goes into 3L of water, resulting in a concentration of 0.067M, or .
Now we know that and pH=-log\[H+\]. Using our trick for -log, we can see that . Since and , we know our answer is between 1 and 2.
is closer to , so we can pick the answer closer to 1. i.e. 1.2.
HCl is a strong acid; it will fully dissociate in water, meaning that the concentration of H+ is equal to the concentration of HCl (they are in a 1 : 1 ratio). 7.0g of HCl is equal to 0.2mol (MW of HCl is 35.3g/mol). 0.2mol HCl goes into 3L of water, resulting in a concentration of 0.067M, or
Now we know that
Compare your answer with the correct one above
2.0g of a monoprotic strong acid are dissolved in 1L of water. The resulting pH is 2.0. What is the molecular weight of the acid?
2.0g of a monoprotic strong acid are dissolved in 1L of water. The resulting pH is 2.0. What is the molecular weight of the acid?
Using the pH of 2.0, we can find that , because .
Since the acid is strong (fully dissociates) and monoprotic (one H+ per molecule),.
Our solution has 1L of water, meaning that we have of acid. We know that only 2.0g of acid were used to achieve this concentration, meaning that there is a ratio of . Simplifying this ratio gives .
Using the pH of 2.0, we can find that
Since the acid is strong (fully dissociates) and monoprotic (one H+ per molecule),
Our solution has 1L of water, meaning that we have
Compare your answer with the correct one above
47.0g of nitrous acid, HNO2, is added to 4L of water. What is the resulting pH?
47.0g of nitrous acid, HNO2, is added to 4L of water. What is the resulting pH?
HNO2 is a weak acid; it will not fully dissociate, so we need to use the HA → H+ + A– reaction, with .
47.0g HNO2 is equal to 1mol. 1mol into 4L gives a concentration of 0.25M when the acid is first dissolved; however, we want the pH at equilibrium, not at the initial state. As the acid dissolves, we know \[HNO2\] will decrease to become ions, but we don't know by how much so we indicate the decrease as "x". As HNO2 dissolves by a factor of x, the ion concentrations will increase by x.
HNO2 → H+ + NO2–
Initial 0.25M 0 0
Equilibrium 0.25 – x x x
Now, we can fill in our equation: .
Since x is very small, we can ignore it in the denominator:
(they expect you to do this on the MCAT; you will never have to solve with x in the denominator on the exam!)
Solve for x, and you find . Looking at our table, we know that
Now we can solve for pH:
HNO2 is a weak acid; it will not fully dissociate, so we need to use the HA → H+ + A– reaction, with
47.0g HNO2 is equal to 1mol. 1mol into 4L gives a concentration of 0.25M when the acid is first dissolved; however, we want the pH at equilibrium, not at the initial state. As the acid dissolves, we know \[HNO2\] will decrease to become ions, but we don't know by how much so we indicate the decrease as "x". As HNO2 dissolves by a factor of x, the ion concentrations will increase by x.
HNO2 → H+ + NO2–
Initial 0.25M 0 0
Equilibrium 0.25 – x x x
Now, we can fill in our equation:
Since x is very small, we can ignore it in the denominator:
(they expect you to do this on the MCAT; you will never have to solve with x in the denominator on the exam!)
Solve for x, and you find
Now we can solve for pH:
Compare your answer with the correct one above
There are two containers, each containing different proton concentrations. In container A, . In container B, .
What is the difference in pH between these two containers?
There are two containers, each containing different proton concentrations. In container A,
What is the difference in pH between these two containers?
The MCAT will typically allow you to approximate the difference in pH between two solutions. Look at the problem this way: you should already know a proton concentration of 1 * 10-3M means a pH of 3. You do not need to know the exact pH of 5 * 10-6M, but you should recognize that it is in between 1 * 10-6M and 1 * 10-5M. This means it will have a pH between 5 and 6.
The difference in pH will therefore be between 5 minus 3 and 6 minus 3.
As a result, look for the answer that is between 2 and 3. The only answer that makes sense is 2.3.
The MCAT will typically allow you to approximate the difference in pH between two solutions. Look at the problem this way: you should already know a proton concentration of 1 * 10-3M means a pH of 3. You do not need to know the exact pH of 5 * 10-6M, but you should recognize that it is in between 1 * 10-6M and 1 * 10-5M. This means it will have a pH between 5 and 6.
The difference in pH will therefore be between 5 minus 3 and 6 minus 3.
As a result, look for the answer that is between 2 and 3. The only answer that makes sense is 2.3.
Compare your answer with the correct one above
HCN dissociates based on the following reaction.
The Ka for hydrogen cyanide is .
of is added to of water. What is the pH of the resulting solution?
HCN dissociates based on the following reaction.
The Ka for hydrogen cyanide is
Since HCN is a weak acid, we must use the equilibrium equation.
Because the HCN dissociates in solution, we expect the concentrations of protons and cyanide ions to increase, while the concentration of HCN will decrease. After determining the molarity of the solution, we can set up the equation below, using X as the amount of moles that dissociate.
Because X is small, we can neglect its impact in the denominator.
Since X is the concentration of protons in the solution, we can calculate the pH by using the equation .
Since HCN is a weak acid, we must use the equilibrium equation.
Because the HCN dissociates in solution, we expect the concentrations of protons and cyanide ions to increase, while the concentration of HCN will decrease. After determining the molarity of the solution, we can set up the equation below, using X as the amount of moles that dissociate.
Because X is small, we can neglect its impact in the denominator.
Since X is the concentration of protons in the solution, we can calculate the pH by using the equation
Compare your answer with the correct one above
A sample of gastric juice has a pH of 2.5. What is the hydrogen ion concentration in this secretion?
A sample of gastric juice has a pH of 2.5. What is the hydrogen ion concentration in this secretion?
The concentration of hydrogen ions must lie somewhere between and ; alternatively stated, it is between and . The pH of a solution with hydrogen ion concentration of will be 3, and the pH of a solution with hydrogen ion concentration will be 2; thus, our concentration must lie between these two values, since our pH is 2.5
To find the exact concentration, you must be familiar with the logarithmic scale. A difference of 0.5 is equivalent to a log of 3.
Our answer must therefore be , or .
We can calculate the pH in reverse to check our answer.
The concentration of hydrogen ions must lie somewhere between
To find the exact concentration, you must be familiar with the logarithmic scale. A difference of 0.5 is equivalent to a log of 3.
Our answer must therefore be
We can calculate the pH in reverse to check our answer.
Compare your answer with the correct one above
Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.
All of the bases proceed in a similar fashion.
The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.
The pH of a solution of is lowered from 4 to 3, and then from 3 to 2. Which of the following is the most accurate description of what happens during these transitions?
Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.
All of the bases proceed in a similar fashion.
The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.
The pH of a solution of
The pH scale is logarithmic. Every pH unit drop corresponds to a tenfold increase in protons.
The pH scale is logarithmic. Every pH unit drop corresponds to a tenfold increase in protons.
Compare your answer with the correct one above
Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.
All of the bases proceed in a similar fashion.
The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.
A scientist is studying an aqueous sample of , and finds that the hydroxide concentration is . Which of the following is true?
Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.
All of the bases proceed in a similar fashion.
The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.
A scientist is studying an aqueous sample of
Given the hydroxide ion concentration, we will need to work using pOH to find the pH. We know that the sum of pH and pOH is equal to 14.
Use our value for the concentration to find the pOH.
Now that we have the pOH, we can use it to solve for the pH.
Given the hydroxide ion concentration, we will need to work using pOH to find the pH. We know that the sum of pH and pOH is equal to 14.
Use our value for the concentration to find the pOH.
Now that we have the pOH, we can use it to solve for the pH.
Compare your answer with the correct one above
What is the pOH of a aqueous solution of ?
What is the pOH of a
The first step for this problem is to find the pH. We can then derive the pOH from the pH value.
The pH is given by the equation . Since hydrochloric acid is monoprotic, the concentration of the solution is equal to the concentration of protons.
Using this value and the pH equation, we can calculate the pH.
Now we can find the pOH. The sum of the pH and the pOH is always 14.
The pOH of the solution is 7.8.
Alternatively, a shortcut can be used to estimate the pH. If is in the form , then pH is roughly .
For this question, this shortcut gets us a pH of 6.4, which produces a pOH of 7.6; very close to the real answer!
The first step for this problem is to find the pH. We can then derive the pOH from the pH value.
The pH is given by the equation
Using this value and the pH equation, we can calculate the pH.
Now we can find the pOH. The sum of the pH and the pOH is always 14.
The pOH of the solution is 7.8.
Alternatively, a shortcut can be used to estimate the pH. If
For this question, this shortcut gets us a pH of 6.4, which produces a pOH of 7.6; very close to the real answer!
Compare your answer with the correct one above