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

Use the Henderson-Hasselbalch equation:

Assuming that these ions occupy the same volume,

We know we have 30g of acetic acid, which is equal to 0.5mol (you should memorize the formula for acetic acid).

Plugging in our values gives us $5.75=4.75+log$\frac{A^{-}$$}{0.5}

Solving for A– gives us 5mol.

To solve this question you need to think about the chemical reaction occurring.

We can ignore water and sodium ions for the sake of this question. The reactants exist in a 1:1 ratio, so that for every mol of NaOH we add, we lose one mol of acetic acid and gain one mol of acetate. We can determine the moles of acetic acid by using M = mol/L, which gives us mol = ML = (2M) * (0.5L) = 1mol acetic acid. If we use the Hendersen Hasselbach equation we can see that the pH equals the pKa when the concentration of conjugate base (acetate) equals the concentration of acid.

If we have 1mol of acetic acid and add 0.5mol of NaOH, we will lose 0.5mol of acetic acid and gain 0.5mol of acetate. We will then be at a point where acetic acid equals acetate. This is summarized in the ICE table below. Now we know the moles of NaOH (0.5 moles) and the concentration (2M) so we can find the volume by doing M = mol/L.

L = mol/M = (0.5mol)/(2M) = 0.25L

| | Acetic acid | NaOH | Acetate | | | -------------- | --------- | --------- | -------- | | I | 1 mol | 0.5 mol | 0 mol | | C | -0.5 mol | -0.5 mol | +0.5 mol | | E | 0.5 mol | 0 mol | 0.5 mol |

To answer this question, we need to be able to compare the concentrations of acid and conjugate base in the solution with the pH. The Henderson-Hasselbach equation is used to compare these values, and is written as:. In this question, this equation can be written with the given values of Ka and pH.

Since we know the Ka of HCN, we can derive the pKa, which turns out to be 9.2.

As a result, we want to see to it that the amount of conjugate base is equal to the concentration of acid, so that . Because log(1) = 0, we want to see to it that the concentrations of the acid and the conjugate base are equal to one another. We know from the question that \[HCN\] = 2M.

As a result, a concentration of 2M NaCN will allow the pH of the solution to be 9.2.

If we use the Henderson-Hasselbalch equation, we do not need to worry about using the molar amounts of both the acid and the base. At the half equivalence point, the conjugate base concentration is equal to that of the weak acid. This means that the equation can be simplified.

Henderson-Hasselbalch equation:

Simplified equation for half equivalence point:

Because we know the acid dissociation constant for hydrofluoric acid, the pH is calculated as:

To answer this question you need to use the Henderson-Hasselbalch equation:

The ratio given in the question is , or .

To use the correct ratio for the Henderson-Hasselbalch equation, we need to convert this ratio to its reciprocal:

Plugging the given values into the equation gives us:

The question is asking for the concentration of hydrogen ions. To solve for this we have to use the definition of pH.

Solving for the concentration of hydrogen ions gives us:

The Henderson-Hasselbalch equation is a tool that allows us to calculate the pH of an acid solution using the pKa of the acid and the relative concentrations of the acid and its conjugate base. It is defined as:

By looking at the equation we can determine that if the ratio inside the logarithm is greater than 1, then the pH of the solution will be greater than the pKa; however, if the ratio is less than 1 (meaning, if the concentration of the acid is greater than the concentration of conjugate base), then the pH will be less than the pKa. Statement I is false.

Increasing the ratio of to will increase the logarithm, and subsequently the pH of the solution. This makes sense because you will have more conjugate base than acid, thereby making the solution more alkaline and increasing the pH. Statement II is true.

pH and pKa are defined as follows:

If we have the same concentration of hydrogen ions as the acid dissociation constant (), then the pH will equal the pKa. According to the Henderson-Hasselbalch equation, the pH equals the pKa if the concentration of the conjugate base equals the concentration of acid; therefore, it is possible for the hydrogen ion concentration to equal the acid dissociation constant. Statement III is false.

The definition of pH is as follows:

The pH of a solution heavily depends on the concentration of hydrogen ions. Recall that the concentration is in molarity (M), which is defined as:

Increasing the volume of the solution will decrease the concentration (molarity) of the hydrogen ions which will, subsequently, increase the pH; therefore, increasing volume will increase the pH.

Recall that pKa of an acid can never be altered. pKa is a reflection of the strength of the acid, which stays constant under all circumstances.

The Henderson-Hasselbalch equation states that:

The question gives us information regarding the ratio of conjugate base to acid AND the pH for each acidic solution. Using this information, we can solve for the pKa values of both solutions.

The pKa values of both solutions are the same. This means that both solution contains the same acid; therefore, the identity of HA is the same as the identity of HB.

The hydrogen ion concentration of solution A is lower than that of solution B because the pH of solution A is greater. Acidity, or strength, of an acid is determined by the pKa. Since we have the same pKa for both acids, HA and HB will have the same acidity. Acid dissociation constant, Ka, is defined as:

Acid dissociation constant only depends on the pKa; therefore, the Ka for both acids is the same.

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.

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.

Each of these compounds requires one equivalent of H+ is added to 100mL of 5N NH3 mol, so for HClO4, 2N = 2M, and for NH3, 5N = 5M.

Using the concentrations and volumes, we can find the moles, finding 0.25times 2=0.5mol HClO4 and 0.10times 5=0.5mol NH3.

In this case, equivalents of acid are equal to the equivalents of base, meaning that we are at the equivalence point in a titration. HClO4 is a strong acid, and NH3 is a weak base.

Thus, the acid will fully dissociate, while the base will not, resulting in a greater concentration of H+ than OH– in the solution. This means the resulting solution will be acidic. We know that the indicator changes from yellow to green at 8.3, which is a basic pH. Our initial solution is basic, and we must pass through the pH of 8.3 to reach our final acidic solution, with pH < 8.3, meaning that the indicator must change from yellow to green during the reaction. This gives out final answer that the solution will be acidic and green.

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.