For a general chemical equation , where A, B, C, and D are elements and the Greek letters are their coefficients, we have the reaction quotient equation:

We can find the reaction quotient equation for our reaction by substituting the variables.

Notice that the concentration of  is in the denominator and is squared, so doubling the concentration of  changes the reaction quotient by a factor of one-fourth.

For any titration curve the equivalence point corresponds to the steepest part of the curve. In this case, the question indicates that the pH at equivalence was 8.9, which would correspond to a volume between 20.0 and 24.0 on the x-axis of the graph. This leads to our approximate answer of 21.8mL.

The pK_a corresponds to the pH when half of the acid has been neutralized. In this case, that would correspond to when half of the 0.5M NaOH has been added. We can see in the curve that the equivalence point is at around 22mL of NaOH, as this is the steepest change in pH. The half equivalence point will be found at approximately 11mL of NaOH addition. On the graph, 11mL of NaOH results in a pH of approximately 5.0, giving us our pK_a value.

H₂CO₃ will have two equivalence points and two half equivalence points, one set corresponding to each of its two protons.

After the first equivalence point, all of the acid is in the form HCO₃^-. When an acid is at a half equivalence point, the acid's concentration will be equal to the concentration of the conjugate base. For the second half equivalence point, the acid is in the  HCO₃^- form, and the conjugate base is CO₃^2-. As a result, at the second half equivalence point, the concentrations of HCO₃^- and CO₃^2- will be equal.

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 pK_a 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

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 K_a and pH.

Since we know the K_a of HCN, we can derive the pK_a, 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.

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 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. 

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: