Question

Consider the titration of \(100.0 \mathrm{mL}\) of \(0.200 M\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) by \(0.100 M\) KOH. Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of KOH have been added. a. \(0.0 \mathrm{mL}\) b. \(50.0 \mathrm{mL}\) c. \(100.0 \mathrm{mL}\) d. \(40.0 \mathrm{mL}\) e. \(50.0 \mathrm{mL}\) f. \(100.0 \mathrm{mL}\)

Short answer

The pH values for the different volumes of KOH added are as follows: a. For 0.0 mL of KOH: pH = 2.72 b. For 50.0 mL of KOH: pH = 4.746 c. For 100.0 mL of KOH: pH = 4.00 d. For 40.0 mL of KOH: pH can be calculated using the method in Case b. e. For 50.0 mL of KOH: pH = 4.746 (same as Case b) f. For 100.0 mL of KOH: pH = 4.00 (same as Case c)

Step-by-step solution

Find initial moles of acetic acid

For all cases, the initial moles of acetic acid can be found using the given concentration and volume: Moles of acetic acid = concentration × volume Moles of acetic acid = \( 0.200 M \) × \( \frac{100.0\,mL}{1000\,mL/L} \) Moles of acetic acid = \( 0.020\,mol \) Now, let's evaluate each case:

Case a: 0.0 mL of KOH added

Since no KOH has been added, there is no change in the moles of acetic acid. Let's find the pH of the solution by creating an ICE table for the dissociation of acetic acid and using its Ka: \( CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \) Initial: 0.020 mol 0 mol 0 mol Change: -x +x +x Equilibrium: 0.020-x x x Ka = \( \frac{x^2}{0.020 - x} = 1.8 \times 10^{-5} \) Since Ka is very small, we can assume x is very small compared to 0.020, so: \( x^2 \approx 1.8 \times 10^{-5} \times 0.020 \) Solving for x, which represents the H+ concentration, we can then find the pH: pH = -log10(x)

Case b: 50.0 mL of KOH added

The moles of KOH added = 0.100 M × \( \frac{50.0\,mL}{1000\,mL/L} \) = 0.005 mol Since acetic acid will react completely with KOH, the moles of acetic acid and CH3COOK after the reaction are: Moles of acetic acid = 0.020 mol - 0.005 mol = 0.015 mol Moles of CH3COOK = 0.005 mol Calculate the concentrations of acetic acid and CH3COOK: C_acetic = \( \frac{0.015\,mol}{(100+50)\,mL/1000\,mL/L} = 0.100\,M \) C_CH3COOK = \( \frac{0.005\,mol}{(100+50)\,mL/1000\,mL/L} = 0.0333\,M \) Now, we can use the Henderson-Hasselbalch equation to find the pH: pH = pKa - log10(\( \frac{[CH_3COOH]}{[CH_3COO^-]} \))

Case c: 100.0 mL of KOH added

The moles of KOH added = 0.100 M × \( \frac{100.0\,mL}{1000\,mL/L} \) = 0.010 mol Since acetic acid will react completely with KOH, the moles of acetic acid and CH3COOK after the reaction are: Moles of acetic acid = 0.020 mol - 0.010 mol = 0.010 mol Moles of CH3COOK = 0.010 mol Calculate the concentrations of acetic acid and CH3COOK: C_acetic = \( \frac{0.010\,mol}{(100+100)\,mL/1000\,mL/L} = 0.050\,M \) C_CH3COOK = \( \frac{0.010\,mol}{(100+100)\,mL/1000\,mL/L} = 0.050\,M \) Using the Henderson-Hasselbalch equation to find the pH: pH = pKa - log10(\( \frac{[CH_3COOH]}{[CH_3COO^-]} \))

Case d: 40.0 mL of KOH added

Repeating the process from Case b and c, calculate the moles of acetic acid and CH3COOK after the reaction and use the Henderson-Hasselbalch equation to find the pH.

Case e: 50.0 mL of KOH added

This case is the same as Case b. Calculate the pH using the same process from Case b.

Case f: 100.0 mL of KOH added

This case is the same as Case c. Calculate the pH using the same process from Case c. Remember to simplify the calculations in each case to make it easier to solve for the pH.

Key concepts

Acetic Acid

Acetic acid, commonly known as vinegar, is a weak acid with the chemical formula \( CH_3COOH \). When it dissolves in water, it partially ionizes, meaning it doesn't completely separate into its ions. This partial ionization is a critical aspect when considering acid-base reactions, such as titration. In this case, the acetic acid has a dissociation constant \( K_a \), which is a measure of its strength as an acid. Here, \( K_a = 1.8 \times 10^{-5} \), indicating it is a weak acid because the value is much less than 1. Understanding the behavior of acetic acid during titration helps us accurately predict pH changes as a base, like KOH, is added. This behavior is crucial when using acetic acid in buffer solutions or food preservation, where precise pH levels are necessary.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a formula that offers an easy way to calculate the pH of a buffer solution. The equation is defined as: \[ \text{pH} = \text{pKa} + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] In this case, "HA" represents the acetic acid, and "A\(^{-}\)" represents its conjugate base, acetate ion \( (CH_3COO^-) \). It is particularly useful in titrations because it relates the pH to the concentrations of the acid and its conjugate base. This relationship allows chemists to calculate the pH at any point during the titration process after the KOH is added. The Henderson-Hasselbalch equation shows the direct connection between the pH of a solution and its acid-base composition, simplifying complex calculations.

pH Calculation

Calculating pH during a titration involves understanding how the addition of a base or acid affects the concentration of hydrogen ions in the solution. The basic formula to find pH is: \[ \text{pH} = -\log_{10}[\text{H}^+] \] When dealing with acetic acid titration using KOH, we must first calculate how much KOH is added and how much acetic acid remains unreacted. The concentration of \([\text{H}^+]\) alters when the reaction progresses because the KOH neutralizes some of the acetic acid, forming water and acetate ions. For weak acids like acetic acid, the change in hydrogen ions concentration can be predicted using an ICE table (Initial, Change, Equilibrium) method, which simplifies the calculation of concentrations at various stages of the reaction. This information allows us to use the Henderson-Hasselbalch equation later to find the pH at different points during titration.

KOH Titration

KOH, or potassium hydroxide, is a strong base used in titrations involving weak acids like acetic acid. During titration, KOH reacts with acetic acid in a 1:1 molar ratio to form water and acetate ions. This process systematically increases the pH of the solution as more KOH is added. Here's how it happens:

• Initially, adding a small amount of KOH results in a slight increase in acetate ions, starting the buffer action of the acetic acid.
• As more KOH is added, more acetic acid is neutralized, increasing the solution's pH towards neutrality or above, depending on the volume of KOH added.
• The point at which equal moles of base and acid have reacted marks the equivalence point, where the amount of remaining acetic acid equals the product formed, acetate ion.

Understanding the titration curve is essential because it indicates how the pH changes at different stages, providing a visual representation of the reaction's progress. Accurately calculating these points helps in precise pH predictions and efficient buffer preparation.