Reactions and Titrations - MCAT Physical
Card 0 of 280
Compound Base strength, K b 1 1012 2 105 3 101 4 10-8
Students in a chemistry class are given one of four unknown samples in a laboratory. A student is told that his compound is the strongest acid of the four compounds. Based on the information in the above table, which compound was the student given?
| Compound | Base strength, K b |
|---|---|
| 1 | 1012 |
| 2 | 105 |
| 3 | 101 |
| 4 | 10-8 |
Students in a chemistry class are given one of four unknown samples in a laboratory. A student is told that his compound is the strongest acid of the four compounds. Based on the information in the above table, which compound was the student given?
The strongest acid of the group will also have the smallest Kb value. As the weakest base (smallest Kb), compound 4 will only partially dissociate in solution because it has a fairly strong conjugate acid. Were this question asking which base was the strongest, compound 1 would be the answer, due to its large Kb value.
The strongest acid of the group will also have the smallest Kb value. As the weakest base (smallest Kb), compound 4 will only partially dissociate in solution because it has a fairly strong conjugate acid. Were this question asking which base was the strongest, compound 1 would be the answer, due to its large Kb value.
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HCN dissociates based on the following reaction.
The Ka for hydrogen cyanide is 
.
Suppose that a solution with a pH of 4.5 has 2M HCN added. Which of the following values will change?
HCN dissociates based on the following reaction.
The Ka for hydrogen cyanide is
Suppose that a solution with a pH of 4.5 has 2M HCN added. Which of the following values will change?
Remember that equilibrium constants are not affected by the concentrations of the reactants and products. Since an acid is being added to the solution, the pH of the solution will be affected. This means that the pOH will be affected as well.
Remember that equilibrium constants are not affected by the concentrations of the reactants and products. Since an acid is being added to the solution, the pH of the solution will be affected. This means that the pOH will be affected as well.
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A weak acid is slowly titrated with a strong base. Where on the titration curve would the solution be the most well-buffered?
A weak acid is slowly titrated with a strong base. Where on the titration curve would the solution be the most well-buffered?
At the half equivalence point, the concentration of acid in the solution is equal to the concentration of the conjugate base in solution. At this point, the graph shows a line that is near horizontal. This means that base or acid could be added and the pH of the solution would change very slowly.
Remember that pH is on the y-axis of the titration curve; thus a near-horizontal line will signify a region where pH is most stable. At the half equivalence point, pH = pKa.
At the half equivalence point, the concentration of acid in the solution is equal to the concentration of the conjugate base in solution. At this point, the graph shows a line that is near horizontal. This means that base or acid could be added and the pH of the solution would change very slowly.
Remember that pH is on the y-axis of the titration curve; thus a near-horizontal line will signify a region where pH is most stable. At the half equivalence point, pH = pKa.
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What is the pKa of acetic acid? (Ka = 1.8 * 10–5)
What is the pKa of acetic acid? (Ka = 1.8 * 10–5)
We know that pKa is equal to –log(Ka). Thus, pKa of acetic acid is –log(1.8 * 10–5). This is not an easy problem to solve in your head, but there is a trick.
We know that 1 * 10–4 > 1.8 * 10–5 > 1 * 10–5, and we know that –log(1 * 10–4) = 4 and –log(1 * 10–5) = 5. Now we can conclude that our pKa is somewhere between 4 and 5.
Two answer choices fall in this range: 4.2 and 4.7.
1.8 * 10–5 is closer to 1 * 10–5 than it is to 1 * 10–4, so we can pick the answer closer to 5 than to 4 : 4.7.
We know that pKa is equal to –log(Ka). Thus, pKa of acetic acid is –log(1.8 * 10–5). This is not an easy problem to solve in your head, but there is a trick.
We know that 1 * 10–4 > 1.8 * 10–5 > 1 * 10–5, and we know that –log(1 * 10–4) = 4 and –log(1 * 10–5) = 5. Now we can conclude that our pKa is somewhere between 4 and 5.
Two answer choices fall in this range: 4.2 and 4.7.
1.8 * 10–5 is closer to 1 * 10–5 than it is to 1 * 10–4, so we can pick the answer closer to 5 than to 4 : 4.7.
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As the value of Ka increases, __________
As the value of Ka increases, __________
Ka represents the equilibrium constant for the acid dissociation in water, HA → H+ + A–, so it is a measure of the products divided by the reactants. As this value increases, the reaction favors the products (the ions) more, meaning that the acid dissociates more. The definition of a strong acid is one that fully dissociates in water, so as Ka increases the strength of the acid increases, and the strength of the conjugate base decreases. pKa is defined as the –log(Ka), so as Ka increases pKa decreases. All of these answer choices are correct.
Ka represents the equilibrium constant for the acid dissociation in water, HA → H+ + A–, so it is a measure of the products divided by the reactants. As this value increases, the reaction favors the products (the ions) more, meaning that the acid dissociates more. The definition of a strong acid is one that fully dissociates in water, so as Ka increases the strength of the acid increases, and the strength of the conjugate base decreases. pKa is defined as the –log(Ka), so as Ka increases pKa decreases. All of these answer choices are correct.
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Diffusion can be defined as the net transfer of molecules down a gradient of differing concentrations. This is a passive and spontaneous process and relies on the random movement of molecules and Brownian motion. Diffusion is an important biological process, especially in the respiratory system where oxygen diffuses from alveoli, the basic unit of lung mechanics, to red blood cells in the capillaries.
Figure 1 depicts this process, showing an alveoli separated from neighboring cells by a capillary with red blood cells. The partial pressures of oxygen and carbon dioxide are given. One such equation used in determining gas exchange is Fick's law, given by:
ΔV = (Area/Thickness) · Dgas · (P1 – P2)
Where ΔV is flow rate and area and thickness refer to the permeable membrane through which the gas passes, in this case, the wall of the avlveoli. P1 and P2 refer to the partial pressures upstream and downstream, respectively. Further, Dgas, the diffusion constant of the gas, is defined as:
Dgas = Solubility / (Molecular Weight)^(1/2)
Which following pair gives the correct equation, and change in pH, when carbon dioxide diffuses into blood?
Diffusion can be defined as the net transfer of molecules down a gradient of differing concentrations. This is a passive and spontaneous process and relies on the random movement of molecules and Brownian motion. Diffusion is an important biological process, especially in the respiratory system where oxygen diffuses from alveoli, the basic unit of lung mechanics, to red blood cells in the capillaries.
Figure 1 depicts this process, showing an alveoli separated from neighboring cells by a capillary with red blood cells. The partial pressures of oxygen and carbon dioxide are given. One such equation used in determining gas exchange is Fick's law, given by:
ΔV = (Area/Thickness) · Dgas · (P1 – P2)
Where ΔV is flow rate and area and thickness refer to the permeable membrane through which the gas passes, in this case, the wall of the avlveoli. P1 and P2 refer to the partial pressures upstream and downstream, respectively. Further, Dgas, the diffusion constant of the gas, is defined as:
Dgas = Solubility / (Molecular Weight)^(1/2)
Which following pair gives the correct equation, and change in pH, when carbon dioxide diffuses into blood?
This is essentially a two-part question. The first asks you to generate the production of carbonic acid; and the second, requires you to infer that an acid will lower blood pH. Only one answer choice is both balanced correctly and gives an accurate change in pH. Although this knowledge is not required for the MCAT, your body constantly monitors blood pH to determine if concentrations of CO2 are too high, or alternatively, if O2 concentrations are too low.
This is essentially a two-part question. The first asks you to generate the production of carbonic acid; and the second, requires you to infer that an acid will lower blood pH. Only one answer choice is both balanced correctly and gives an accurate change in pH. Although this knowledge is not required for the MCAT, your body constantly monitors blood pH to determine if concentrations of CO2 are too high, or alternatively, if O2 concentrations are too low.
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Given the above values of Ka, place the acids in order from strongest to weakest.
Given the above values of Ka, place the acids in order from strongest to weakest.
The acid dissociation constant, Ka, describes how strongly an acid tends to break apart into hydrogen ions (H+) and its conjugate base (A-). The higher the dissociation constant, the stronger the acid. HF has the largest Ka of these acids, making it the strongest, and H2CO3 has the smallest Ka, making it the weakest.
The acid dissociation constant, Ka, describes how strongly an acid tends to break apart into hydrogen ions (H+) and its conjugate base (A-). The higher the dissociation constant, the stronger the acid. HF has the largest Ka of these acids, making it the strongest, and H2CO3 has the smallest Ka, making it the weakest.
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What is the pH of a solution which has a hydroxide ion concentration of 5 * 10-4M?
What is the pH of a solution which has a hydroxide ion concentration of 5 * 10-4M?
First convert concentration of OH- into concentration of H+. Remember that Kw is 1*10-14.
Then, convert concentration of H+ into pH.
Alternatively, you can use the hydroxide concentration to solve for pOH and convert to pH.
First convert concentration of OH- into concentration of H+. Remember that Kw is 1*10-14.
Then, convert concentration of H+ into pH.
Alternatively, you can use the hydroxide concentration to solve for pOH and convert to pH.
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HCN dissociates based on the following reaction.
The Ka for hydrogen cyanide is 
.
What is the Kb for CN-?
HCN dissociates based on the following reaction.
The Ka for hydrogen cyanide is
What is the Kb for CN-?
Remember that the Ka for the acid and the Kb for the conjugate base, when multiplied will equal the autoionization of water constant (Kw).
Remember that the Ka for the acid and the Kb for the conjugate base, when multiplied will equal the autoionization of water constant (Kw).
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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.
In aqueous conditions the equilibrium constant for a Brønsted-Lowry base, 
, can be expressed as which of the following?
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.
In aqueous conditions the equilibrium constant for a Brønsted-Lowry base,
The base reaction will essentially be the reverse of the acid reaction. In the question, the aqueous conditions mean that the base, 
, reacts with water to give the following reaction:
Following normal equilibrium convention, we omit water from the equation because it is a pure liquid.
The base reaction will essentially be the reverse of the acid reaction. In the question, the aqueous conditions mean that the base,
Following normal equilibrium convention, we omit water from the equation because it is a pure liquid.
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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.
Which of the following expressions most closely approximates the equilibrium constant of pure water?
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.
Which of the following expressions most closely approximates the equilibrium constant of pure water?
The autionization of water proceeds as follows:
The equilibrium constant is calculated by the following formula.
The equilibrium expression is 
, with the pure water reactant excluded. Pure solids and liquids are not included in the equilibrium calculation.
The autionization of water proceeds as follows:
The equilibrium constant is calculated by the following formula.
The equilibrium expression is
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What is the hydroxide concentration in an aqueous solution with a pH of 2?
What is the hydroxide concentration in an aqueous solution with a pH of 2?
The hydroxide concentration can be determined by considering the autoionization of water in a solution. At 
, water has the equilibrium constant of 
. This value is based on the autoionization of water.
Since the reaction is in terms of proton and hydroxide ion concentrations, we can set the expression equal to this value in order to determine the concentration of hydroxide ions at the given pH. We start by determining the concentration of protons in the solution by using the following equation:
Now, we can solve for the hydroxide concentration.
This problem can also be solved by calculating the pOH of the solution and using this value to find the hydroxide ion concentration.
The hydroxide concentration can be determined by considering the autoionization of water in a solution. At
Since the reaction is in terms of proton and hydroxide ion concentrations, we can set the expression equal to this value in order to determine the concentration of hydroxide ions at the given pH. We start by determining the concentration of protons in the solution by using the following equation:
Now, we can solve for the hydroxide concentration.
This problem can also be solved by calculating the pOH of the solution and using this value to find the hydroxide ion concentration.
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Based on the above information, it is expected that __________.
Based on the above information, it is expected that __________.
Since the product of the cation and anion is 
, the only true statement is that the concentration of the cation 
is the square root of this number:
Since the product of the cation and anion is
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A solution of hydrofluoric acid has a concentration of 
The 
for 
is 
.
What is the pH of the solution?
A solution of hydrofluoric acid has a concentration of
The
What is the pH of the solution?
Since hydrofluoric acid is a weak acid, an ICE table needs to be set up in order to determine the hydronium ion concentration. Since both fluoride ion and hydronium ion concentrations will increase by 
, while the acid concentration will decrease by 
, the equilibrium expression comes out to be:
Note that the 
in the denominator will have a negligible effect and can be ignored.
Since 
is equal to the hydronium ion concentration, we can calculate the pH by taking the negative log of the concentration:
Since hydrofluoric acid is a weak acid, an ICE table needs to be set up in order to determine the hydronium ion concentration. Since both fluoride ion and hydronium ion concentrations will increase by
Note that the
Since
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A solution of hydrofluoric acid has a concentration of .
The 
for 
is 
.
If sodium hydroxide is slowly added to the acid solution, what will the pH be at the equivalence point?
Assume the concentration of the acid is not changed with the addition.
A solution of hydrofluoric acid has a concentration of
The
If sodium hydroxide is slowly added to the acid solution, what will the pH be at the equivalence point?
Assume the concentration of the acid is not changed with the addition.
At the equivalence point, there are equimolar amounts of acid and base. This means that all weak acid has been neutralized, and only the conjugate base remains. Since the conjugate base of a weak acid will affect the pH, we need to use an ICE table in order to find the pH. First, we start by finding the base dissociation constant of the conjugate base, using the equation:
The balanced equation for the conjugate base dissociation is:
As the hydroxide ion and acid concentrations increase by 
, the fluoride ion concentration will decrease by 
. This makes the equilibrium expression:
Since this is the hydroxide concentration, we can find the pH by taking the negative log of this value, then subtracting from 14:
At the equivalence point, there are equimolar amounts of acid and base. This means that all weak acid has been neutralized, and only the conjugate base remains. Since the conjugate base of a weak acid will affect the pH, we need to use an ICE table in order to find the pH. First, we start by finding the base dissociation constant of the conjugate base, using the equation:
The balanced equation for the conjugate base dissociation is:
As the hydroxide ion and acid concentrations increase by
Since this is the hydroxide concentration, we can find the pH by taking the negative log of this value, then subtracting from 14:
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100mL of an unknown solution of NaOH is titrated with 3M HCl until neutralized. The resulting solution is evaporated, and 3.0g of white crystal are recovered. What was the concentration of the NaOH solution?
100mL of an unknown solution of NaOH is titrated with 3M HCl until neutralized. The resulting solution is evaporated, and 3.0g of white crystal are recovered. What was the concentration of the NaOH solution?
In the neutralization reaction between NaOH and HCl, NaCl salt is formed. When the solution is evaporated, this salt is left behind.
3.0g of NaCl is equivalent to 0.05mol NaCl. Since the titration is between a strong acid and a strong base, all of the NaOH in the original solution is converted to NaCl in a one-to-one ratio, meaning that mol NaCl = mol NaOH.
We now know that there was 0.05mol NaOH in the 100mL solution, so the concentration must have been 
.
In the neutralization reaction between NaOH and HCl, NaCl salt is formed. When the solution is evaporated, this salt is left behind.
3.0g of NaCl is equivalent to 0.05mol NaCl. Since the titration is between a strong acid and a strong base, all of the NaOH in the original solution is converted to NaCl in a one-to-one ratio, meaning that mol NaCl = mol NaOH.
We now know that there was 0.05mol NaOH in the 100mL solution, so the concentration must have been
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How many milliliters of 0.05M HCl are required to neutralize 200mL of 0.025M 
?
How many milliliters of 0.05M HCl are required to neutralize 200mL of 0.025M
First note that there are two moles of 
for each mole of 
.
Then calculate the number of moles of 
in the given volume of solution.
To neutralize, we need 
.
We can plug in our value of 0.01mol and the given concrentration of 0.05M, and solve for the required volume.
First note that there are two moles of
Then calculate the number of moles of
To neutralize, we need
We can plug in our value of 0.01mol and the given concrentration of 0.05M, and solve for the required volume.
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What volume of 0.375M H2SO4 is needed to fully neutralize 0.5L of 0.125M NaOH?
What volume of 0.375M H2SO4 is needed to fully neutralize 0.5L of 0.125M NaOH?
This question requires use of the simple titration equation M1V1 = M2V2. The key is to identify that sulfuric acid has two equivalents of acidic hydrogens while NaOH has only one hydroxide equivalent. All wrong answer choices result from making this mistake or other calculation errors.
This question requires use of the simple titration equation M1V1 = M2V2. The key is to identify that sulfuric acid has two equivalents of acidic hydrogens while NaOH has only one hydroxide equivalent. All wrong answer choices result from making this mistake or other calculation errors.
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Which of the following is true regarding an acid and its pKa?
I. One can increase the strength of an acid by decreasing its pKa value
II. pKa increases as the acid dissociation constant decreases
III. pKa of an acid cannot be changed by altering the concentration of the acid
Which of the following is true regarding an acid and its pKa?
I. One can increase the strength of an acid by decreasing its pKa value
II. pKa increases as the acid dissociation constant decreases
III. pKa of an acid cannot be changed by altering the concentration of the acid
Recall that pKa is defined as follows:
Here, 
is the acid dissociation constant. 
is a measure of the equilibrium strength of an acid and is unique for each acid. The higher the value of 
, the stronger the acid; however, a particular acid’s 
value, and subsequently its strength, can never be changed. The only way you can change the 
of an acid is by changing the identity of the acid itself. This means that the pKa value of an acid is also always constant; therefore, you cannot decrease an acid’s pKa.
Using the definition of pKa, we can see that the pKa of an acid increases as you decrease the acid dissociation constant (
). A strong acid will have a high 
and a low pKa.
The pKa of an acid can never be altered; therefore, changing the concentration of the acid will not alter the pKa of the acid. It might change the amount of hydrogen ions produced and alter the pH; however, the pKa of the acid will stay constant.
Recall that pKa is defined as follows:
Here,
Using the definition of pKa, we can see that the pKa of an acid increases as you decrease the acid dissociation constant (
The pKa of an acid can never be altered; therefore, changing the concentration of the acid will not alter the pKa of the acid. It might change the amount of hydrogen ions produced and alter the pH; however, the pKa of the acid will stay constant.
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Consider two solutions: solution A and solution B. Solution A is a 0.1M hydrogen iodide solution and solution B is a 0.1M hydrochloric acid solution. What can you conclude about these two solutions?
Consider two solutions: solution A and solution B. Solution A is a 0.1M hydrogen iodide solution and solution B is a 0.1M hydrochloric acid solution. What can you conclude about these two solutions?
To answer this question you need to write out the acid dissociation reaction for hydrogen iodide and hydrochloric acid.
The acid dissociation reaction for hydrogen iodide is:
The acid dissociation reaction for hydrochloric acid is:
Recall that both 
and 
are very strong acids; therefore, they will dissociate completely in solution and produce their respective products. Since the ratio of acid to hydrogen ions is 1:1 for both acids AND the concentration of both acids is the same (0.1M), the amount of hydrogen ions produced will be the same for both solutions.
Recall that pKa decreases as the Ka (acid dissociation) increases. As mentioned, both acids are very strong; therefore, they will have very high Ka values and, subsequently, very low pKa values.
To answer this question you need to write out the acid dissociation reaction for hydrogen iodide and hydrochloric acid.
The acid dissociation reaction for hydrogen iodide is:
The acid dissociation reaction for hydrochloric acid is:
Recall that both
Recall that pKa decreases as the Ka (acid dissociation) increases. As mentioned, both acids are very strong; therefore, they will have very high Ka values and, subsequently, very low pKa values.
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