Question

Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. $$ \begin{array}{l}{\text { a. } 100.0 \mathrm{mL} \text { of } 0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{HsO}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right) \text { titrated }} \\ {\text { by } 0.10 \mathrm{M} \mathrm{NaOH}}\end{array} $$ $$ \begin{array}{l}{\text { b. } 100.0 \mathrm{mL} \text { of } 0.10 M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right) \text { titrated }} \\ {\text { by } 0.20 \mathrm{M} \mathrm{HNO}_{3}}\end{array} $$ $$ 100.0 \mathrm{mL} \text { of } 0.50 \mathrm{M} \mathrm{HCl} \text { titrated by } 0.25 \mathrm{M} \mathrm{NaOH} $$

Short answer

At the halfway and equivalence points, a) (Weak acid and strong base): Halfway point pH: 4.19 Equivalence point pH: 8.10 b) (Weak base and strong acid): Perform similar calculations as in scenario a, considering weak base and conjugate acid. Convert calculated pOH to pH. c) (Strong acid and strong base): Halfway point pH: 0.90 Equivalence point pH: 7.00

Step-by-step solution

Calculate the initial moles of HC7H5O2 (HA)

To calculate the initial moles of the weak acid, use the formula: moles = molarity × volume, which in this case is (0.10 mol/L)(0.100 L) = 0.01 moles.

Calculate the moles of NaOH at the halfway point and equivalence point.

At the halfway point, we have added half the moles of NaOH required to reach the equivalence point. Thus, the number of moles of NaOH is 0.005 moles. At the equivalence point, the moles of added NaOH are equal to the initial moles of the weak acid, which is 0.01 moles.

Calculate the pH at the halfway point.

At the halfway point, we treat the solution as a buffer. This is because the number of moles of the weak acid equals the moles of its conjugate base. We can calculate the pH using the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]). In this case, pKa is calculated as -log(Ka), which is -log(6.4 × 10^(-5)) = 4.19. Since [A-] = [HA] at the halfway point, the pH = pKa = 4.19.

Calculate the pH at the equivalence point.

At the equivalence point, we have the presence of the conjugate base C7H5O2^- (A^-) in the Solution. We will ignore any effects of H3O^+ contribution from water, as the effect is minimal. To determine the pH, we need to treat the solution as if it is a solution of a weak base. We use Kb for the dissociation of the weak base: Kb = Kw / Ka, where Kw = 1×10^(-14). Kb = (1×10^(-14))/(6.4×10^(-5)) = 1.56×10^(-10). Setting up an ICE table and solving for the hydroxide ion [OH^-], we get: [OH^-] = sqrt((1.56×10^(-10))(0.10 M)) = 1.25×10^(-6) M. To find the pH, first, we calculate pOH = -log[OH^-] = 5.90, and then we calculate pH by using the formula pH = 14 - pOH = 14 - 5.90 = 8.10. #For scenario b (weak base and strong acid):# Follow a similar process as in scenario a, but this time, consider the weak base and its conjugate acid. Calculate the values required for the Kb to obtain the pOH and convert it into pH at the halfway point and the equivalence point. #For scenario c (strong acid and strong base):#

Calculate the initial moles of HCl.

To calculate the initial moles of the strong acid, use the formula: moles = molarity × volume, which in this case is (0.50 mol/L)(0.100 L) = 0.050 moles of HCl.

Calculate the moles of NaOH required for the halfway and equivalence points.

At the halfway point, we have added half the moles of NaOH required to reach the equivalence point. Thus, the number of moles of NaOH is 0.025 moles. At the equivalence point, the moles of added NaOH are equal to the initial moles of the strong acid, which is 0.050 moles.

Calculate the pH at the halfway point.

At the halfway point, use the concept of a buffer solution, as the number of moles of the strong acid equals the moles of its conjugate base. First, we calculate [H3O^+] = (initial moles of HCl)/(total volume in liters) = 0.025 moles / (0.100 L + 0.100 L) = 0.125 mol/L. Then, we calculate pH = -log[H3O^+] = -log(0.125) = 0.90.

Calculate the pH at the equivalence point.

At the equivalence point, there is only water present, because strong acids and strong bases neutralize each other completely. So, the pH of the solution will be 7, as it is a neutral solution.

Key concepts

Halfway Point

The halfway point in a titration is a crucial juncture, especially for weak acids or bases. At this stage, exactly half of the initial amount of the acidic or basic solution has reacted with the titrant. This means, for a weak acid that we are titrating with a strong base, half of the acid will be neutralized, forming its conjugate base.

So, why is the halfway point particularly significant for pH calculations? At this point, the concentration of the weak acid \([HA]\) is equal to the concentration of its conjugate base \([A^-]\). As a result, this is when the solution acts as a buffer.

Using the Henderson-Hasselbalch equation, calculating pH becomes straightforward because the log term \(\log([A^-]/[HA])\) equals zero. Therefore, \( ext{pH} = ext{p}K_a\). This tells us that at the halfway point, the pH of the solution is equal to the pKa value of the weak acid, offering a simple yet powerful estimation of the acid’s strength.

Equivalence Point

The equivalence point in a titration signifies the moment when the moles of titrant added equals the moles of the substance initially present in the solution. For both strong and weak acids and bases, this point is essential for determining how much acid or base is needed to completely neutralize the other.

For strong acid-strong base titrations, the equivalence point results in a neutral solution, leading to a pH of 7.
In weak acid with strong base titrations, for example, at the equivalence point, all of the weak acid has been converted to its conjugate base. Since the conjugate base is weakly basic, the solution will have a pH greater than 7.

Conversely, when titrating a weak base with a strong acid, the pH at the equivalence point will be less than 7, as the resulting solution contains the conjugate acid of the base. Thus, understanding the chemical nature of the species present at the equivalence point is key to predicting the resultant pH.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a critical equation in chemistry that simplifies the calculation of pH in buffered solutions. It is expressed as \( ext{pH} = ext{p}K_a + ext{log}rac{[A^-]}{[HA]}\).

This equation describes the relationship between pH, the pKa of the acid, and the concentrations of the acid \[HA\] and its conjugate base \[A^-\]. It assumes that the acid and base are in equilibrium, a useful condition for calculating pH in buffer systems created during titration.

Because it takes the ratio of base to acid into account, it provides a straightforward way to calculate pH changes in systems that resist drastic pH shifts, such as biological systems or chemical reactions requiring precise conditions.

• It helps predict the pH of a buffer solution.
• It assists in determining how much acid or base needs to be added to achieve a desired pH change.
• At the halfway point of a titration, where [A^-] equals [HA], it reduces to \( ext{pH} = ext{p}K_a\), making it extremely practical for quick calculations.

With its simplicity and practical applications, the Henderson-Hasselbalch equation is an invaluable tool for anyone studying acid-base chemistry or working in related fields.