Acids and Bases - MCAT Chemical and Physical Foundations of Biological Systems
Card 1 of 231
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?
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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 6.7times 10^{-2}.
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 -logleft (1times 10^{-1} right )=1 and -logleft (1times 10^{-2} right )=2, we know our answer is between 1 and 2.
6.7times 10^{-2} is closer to 1times 10^{-1}, 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 6.7times 10^{-2}.
Now we know that
6.7times 10^{-2} is closer to 1times 10^{-1}, so we can pick the answer closer to 1. i.e. 1.2.
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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?
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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 $10^{-2}$mole of acid. We know that only 2.0g of acid were used to achieve this concentration, meaning that there is a ratio of $$\frac{2.0g}{10^{-2}$$mole}. Simplifying this ratio gives $\frac{200g}{mol}$.
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 $10^{-2}$mole of acid. We know that only 2.0g of acid were used to achieve this concentration, meaning that there is a ratio of $$\frac{2.0g}{10^{-2}$$mole}. Simplifying this ratio gives $\frac{200g}{mol}$.
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47.0g of nitrous acid, HNO2, is added to 4L of water. What is the resulting pH? left $(K_{a}$=4.1times $10^{-4}$ right )
47.0g of nitrous acid, HNO2, is added to 4L of water. What is the resulting pH? left $(K_{a}$=4.1times $10^{-4}$ right )
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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: K_{a}=left frac{left ( x right )left ( x right )}{left (0.25 right )}=4.1times 10^{-4}
(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: K_{a}=left frac{left ( x right )left ( x right )}{left (0.25 right )}=4.1times 10^{-4}
(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:
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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?
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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.
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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
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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
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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?
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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.
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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
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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.
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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
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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.
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What is the pOH of a 
aqueous solution of 
?
What is the pOH of a
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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!
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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?
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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.
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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?
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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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Which of the following would be most useful as a buffer?
Which of the following would be most useful as a buffer?
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A buffer must contain either a weak base and its salt or a weak acid and its salt. A mixture of ammonia and ammonium chloride is an example of the first case, since ammonia, NH3, is a weak base and ammonium chloride, NH4Cl, contains its salt.
Though autoionization of water produces small amounts of H3O+ and OH-, each conjugate salts of H2O, they exist in such small amounts as to make any buffering effects negligible.
A buffer must contain either a weak base and its salt or a weak acid and its salt. A mixture of ammonia and ammonium chloride is an example of the first case, since ammonia, NH3, is a weak base and ammonium chloride, NH4Cl, contains its salt.
Though autoionization of water produces small amounts of H3O+ and OH-, each conjugate salts of H2O, they exist in such small amounts as to make any buffering effects negligible.
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Carbonic anhydrase is an important enzyme that allows CO2 and H2O to be converted into H2CO3. In addition to allowing CO2 to be dissolved into the blood and transported to the lungs for exhalation, the products of the carbonic anhydrase reaction, H2CO3 and a related compound HCO3-, also serve to control the pH of the blood to prevent acidosis or alkalosis. The carbonic anhydrase reaction and acid-base reaction are presented below.
CO2 + H2O 
H2CO3
H2CO3 
HCO3- + H+
HCO3- may be considered a(n) .
Carbonic anhydrase is an important enzyme that allows CO2 and H2O to be converted into H2CO3. In addition to allowing CO2 to be dissolved into the blood and transported to the lungs for exhalation, the products of the carbonic anhydrase reaction, H2CO3 and a related compound HCO3-, also serve to control the pH of the blood to prevent acidosis or alkalosis. The carbonic anhydrase reaction and acid-base reaction are presented below.
CO2 + H2O
H2CO3
HCO3- may be considered a(n) .
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First, we need to determine from both the information presented in the paragraph and the chemical equation what HCO3- is doing. The paragraph tells us “HCO3- also \[serves\] to control the pH of the blood.” This is the definition of a buffer. A buffer is able to mitigate the addition of acid or base to a solution, or in this case, blood, by being deprotonated or protonated. For example, if the blood becomes too acidic, H+ will recombine with HCO3- to form H2CO3, thus stabilizing the pH. Alternatively, if the pH becomes too basic, H2CO3 will dissociate, releasing H+ into solution.
First, we need to determine from both the information presented in the paragraph and the chemical equation what HCO3- is doing. The paragraph tells us “HCO3- also \[serves\] to control the pH of the blood.” This is the definition of a buffer. A buffer is able to mitigate the addition of acid or base to a solution, or in this case, blood, by being deprotonated or protonated. For example, if the blood becomes too acidic, H+ will recombine with HCO3- to form H2CO3, thus stabilizing the pH. Alternatively, if the pH becomes too basic, H2CO3 will dissociate, releasing H+ into solution.
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HCN dissociates based on the following reaction:
The Ka for hydrogen cyanide is 
.
Two moles of HCN are mixed with one mole of NaCN, resulting in 1L of solution. What is the pH of the solution?
HCN dissociates based on the following reaction:
The Ka for hydrogen cyanide is
Two moles of HCN are mixed with one mole of NaCN, resulting in 1L of solution. What is the pH of the solution?
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Whenever a concentration of the conjugate base is initially present in the solution, we say that the solution has been buffered. NaCN will completely dissolve in solution, meaning that there will be 1M of CN- ions in the solution initially (one mole of NaCN in 1L of soultion equals 1M). With this in mind, the new equilibrium equation for the acid's dissociation is as follows:
The initial concentrations of CN- and HCN are given in the question as 1M and 2M, respectively. The initial concentration of H+ is zero. As HCN begine to dissociate, H+ and CN- will each increase by X amount, while HCN will decrease by X amount.
Since 1M and 2M are much larger numbers than the value for X, the X values next to these numbers can be omitted from the equation.
By plugging this value into the equation 
, we determine that the pH of the solution is 8.9. (Remember that X is equal to the amount of increase in H+ ions).
This may seem wrong, because HCN is an acid, and we are used to acids resulting in solutions with pH values less than 7; however, HCN is a weak acid, and a large concentration of its conjugate base can result in a basic solution.
Whenever a concentration of the conjugate base is initially present in the solution, we say that the solution has been buffered. NaCN will completely dissolve in solution, meaning that there will be 1M of CN- ions in the solution initially (one mole of NaCN in 1L of soultion equals 1M). With this in mind, the new equilibrium equation for the acid's dissociation is as follows:
The initial concentrations of CN- and HCN are given in the question as 1M and 2M, respectively. The initial concentration of H+ is zero. As HCN begine to dissociate, H+ and CN- will each increase by X amount, while HCN will decrease by X amount.
Since 1M and 2M are much larger numbers than the value for X, the X values next to these numbers can be omitted from the equation.
By plugging this value into the equation
This may seem wrong, because HCN is an acid, and we are used to acids resulting in solutions with pH values less than 7; however, HCN is a weak acid, and a large concentration of its conjugate base can result in a basic solution.
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The pH of a buffered solution is 
. What is the approximate ratio of the concentration of acid to conjugate base if the 
of the acid is 
?
The pH of a buffered solution is
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In a buffered solution, when the concentrations of acid and conjugate base are equal, we know the 
.
This is derived from the Henderson-Hasselbalch equation: 
.
When concentrations are equal, the log of 
is 
, and 
.
In our question, the pH is approximately one unit greater than the 
.
Because of this, we know that the log of the two concentrations must be equal to one.
The log of 
is 
, and therefore the conjugate base must be ten times greater than the acid; therefore the ratio of acid to base is approximately 
.
We can quickly narrow this question to two possible answers, since the buffered solution is more basic than the 
, and thus we would expect there to be a greater concentration of base than acid in the solution.
In a buffered solution, when the concentrations of acid and conjugate base are equal, we know the
This is derived from the Henderson-Hasselbalch equation:
When concentrations are equal, the log of
In our question, the pH is approximately one unit greater than the
Because of this, we know that the log of the two concentrations must be equal to one.
The log of
We can quickly narrow this question to two possible answers, since the buffered solution is more basic than the
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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 making a buffer to maintain the pH of a solution at about 6. Which of the following describes the best composition of the solution?
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 making a buffer to maintain the pH of a solution at about 6. Which of the following describes the best composition of the solution?
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A buffer is best made of equal and copious amounts of an acid and its conjugate base. The acid must have a pKa as close as possible to the pH desired. For this question, the desired pH is 6. The best option will have a pKa of about 5.9.
Adding larger amounts of the acid than the base will skew the equilibrium in favor of the acid, increasing the overall proton concentration, and limiting the buffer ability to absorb more proton addition without changing pH significantly.
A buffer is best made of equal and copious amounts of an acid and its conjugate base. The acid must have a pKa as close as possible to the pH desired. For this question, the desired pH is 6. The best option will have a pKa of about 5.9.
Adding larger amounts of the acid than the base will skew the equilibrium in favor of the acid, increasing the overall proton concentration, and limiting the buffer ability to absorb more proton addition without changing pH significantly.
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Which of the following would be a good buffer solution?
Which of the following would be a good buffer solution?
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A good buffer solution consists of a weak acid with its conjugate base, or a weak base and its conjugate acid. A strong acid or base will never be a good buffer. We will go through each answer choice in detail.
Sulfuric acid is a strong acid and cannot be used in a buffer.
Neither sodium chloride, nor calcium chloride constitute a weak acid or weak base. Both of these are simple salts and will not be able to create a buffer solution.
Hydrochloric acid is a strong acid and cannot be used in a buffer.
In solution, these salts will produce sulfate ions and hydrogen sulfate ions. Hydrogen sulfate is a weak acid and sulfate is its conjugate base. These ions will form a buffer solution.
A good buffer solution consists of a weak acid with its conjugate base, or a weak base and its conjugate acid. A strong acid or base will never be a good buffer. We will go through each answer choice in detail.
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Which of the following pairs cannot constitute a buffer system?
Which of the following pairs cannot constitute a buffer system?
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A buffer system is composed of one of the following scenarios:
1. A weak base with its conjugate acid.
2. A weak acid with its conjugate base.
The only option that does not qualify is the nitrate, nitrite pair. A difference of one oxygen does not make a buffer system.
A buffer system is composed of one of the following scenarios:
1. A weak base with its conjugate acid.
2. A weak acid with its conjugate base.
The only option that does not qualify is the nitrate, nitrite pair. A difference of one oxygen does not make a buffer system.
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None of the questions in this set require the use of a calculator. Math problems are intended to mimic the level of math intensity that you will see on the MCAT exam.
A Bronsted-Lowry acid will be able to .
None of the questions in this set require the use of a calculator. Math problems are intended to mimic the level of math intensity that you will see on the MCAT exam.
A Bronsted-Lowry acid will be able to .
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The Bronsted-Lowry definition of an acid is one of three acid definitions that can be used on the MCAT. The three definitions are as follows:
Arrhenius acid: increases H+ concentration in water (releases an H+)
Bronsted-Lowry acid: donates a proton to a Bronsted base (Bronsted bases accept protons)
Lewis acid: accepts a pair of electrons from a Lewis base (Lewis bases donate electrons)
An easy way to remember the difference between Bronsted-Lowry and Lewis is that the "e" in Lewis comes before the "e" in Bronsted; therefore the Lewis definition has to do with electrons.
The Bronsted-Lowry definition of an acid is one of three acid definitions that can be used on the MCAT. The three definitions are as follows:
Arrhenius acid: increases H+ concentration in water (releases an H+)
Bronsted-Lowry acid: donates a proton to a Bronsted base (Bronsted bases accept protons)
Lewis acid: accepts a pair of electrons from a Lewis base (Lewis bases donate electrons)
An easy way to remember the difference between Bronsted-Lowry and Lewis is that the "e" in Lewis comes before the "e" in Bronsted; therefore the Lewis definition has to do with electrons.
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Which of the following is not a Lewis acid?
Which of the following is not a Lewis acid?
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The definition of a Lewis acid is an electron acceptor. Try drawing the Lewis dot structures for these compounds, and you will find that NH4+, Na+, and Al3+ are missing electrons. They each have an empty orbital, allowing them to accept electrons.
BF3 is not an ion; however, we know that boron only has 3 valence electrons, which means that even when they are all bound to fluorine the molecule does not satisfy the octet rule. BF3 only has 6 valence electrons around boron, and can accept another electron pair to get to the octet state.
CaO is a Lewis base. In the Lewis dot structure, we can see that the oxygen molecule has two lone pairs that it can donate to other molecules, making it an electron donor.
The definition of a Lewis acid is an electron acceptor. Try drawing the Lewis dot structures for these compounds, and you will find that NH4+, Na+, and Al3+ are missing electrons. They each have an empty orbital, allowing them to accept electrons.
BF3 is not an ion; however, we know that boron only has 3 valence electrons, which means that even when they are all bound to fluorine the molecule does not satisfy the octet rule. BF3 only has 6 valence electrons around boron, and can accept another electron pair to get to the octet state.
CaO is a Lewis base. In the Lewis dot structure, we can see that the oxygen molecule has two lone pairs that it can donate to other molecules, making it an electron donor.
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