Acid-Base Reactions - MCAT Chemical and Physical Foundations of Biological Systems

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 0.5Mleft ( $\frac{0.05}{0.1}$=0.5 right ).

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.

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.

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.

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.

The acidity of the solution results from the amount of hydrogen ions present in the solution. We can increase or decrease the pH of the solution by decreasing or increasing the amount of hydrogen ions present, respectively. pKa is a measure of the strength of an acid (meaning how easily it can dissociate into hydrogen ions and its conjugate base). Altering the pH of the solution will have no effect on the strength, and subsequently pKa, of the acid.

To answer this question you need to use Le Chatelier’s principle. Adding sodium acetate to the solution will cause it to dissociate as follows:

The dissociation reaction will produce more acetate ions. According to Le Chatelier’s principle, the increase in acetate ions will shift the equilibrium of the reaction (given in the question) to the left. This means that and will be utilized to form . This will cause a decrease in the amount of hydronium ions in solution. Recall that pH is increased when the concentration of hydrogen ions (or hydronium ions, ) is decreased; therefore, adding sodium acetate will increase the pH of the solution.

Increasing acetic acid concentration will shift the equilibrium to the right and produce more hydronium ions, thereby decreasing the pH. Recall that you can never change the pKa of an acid. The pKa of acetic acid is around 4.75, and it cannot be altered. Le Chatelier’s principle only applies when there is a change in amount of aqueous or gaseous substances; liquid and solid substances will not shift the equilibrium. Changing the volume of liquid water will not change the concentration of hydronium ions.

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 0.5Mleft ( $\frac{0.05}{0.1}$=0.5 right ).

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.

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.

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.

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.

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.

The acidity of the solution results from the amount of hydrogen ions present in the solution. We can increase or decrease the pH of the solution by decreasing or increasing the amount of hydrogen ions present, respectively. pKa is a measure of the strength of an acid (meaning how easily it can dissociate into hydrogen ions and its conjugate base). Altering the pH of the solution will have no effect on the strength, and subsequently pKa, of the acid.

To answer this question you need to use Le Chatelier’s principle. Adding sodium acetate to the solution will cause it to dissociate as follows:

The dissociation reaction will produce more acetate ions. According to Le Chatelier’s principle, the increase in acetate ions will shift the equilibrium of the reaction (given in the question) to the left. This means that and will be utilized to form . This will cause a decrease in the amount of hydronium ions in solution. Recall that pH is increased when the concentration of hydrogen ions (or hydronium ions, ) is decreased; therefore, adding sodium acetate will increase the pH of the solution.

Increasing acetic acid concentration will shift the equilibrium to the right and produce more hydronium ions, thereby decreasing the pH. Recall that you can never change the pKa of an acid. The pKa of acetic acid is around 4.75, and it cannot be altered. Le Chatelier’s principle only applies when there is a change in amount of aqueous or gaseous substances; liquid and solid substances will not shift the equilibrium. Changing the volume of liquid water will not change the concentration of hydronium ions.

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 0.5Mleft ( $\frac{0.05}{0.1}$=0.5 right ).

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.

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.

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.