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.

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 .

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.