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A 25.0 mL aliquot of 0.10 M acetic acid is titrated with 0.10M sodium hydroxide solution.

A. What is the pH of the acid solution before NaOH is added?
B. What is the pH after 15.0mL of sodium hydroxide solution is added
C. According to your calculations [in(B)], which species is the limiting reagent and by what amount?
D. What is the pH of the solution after 25.0mL of 0.10 M NaOH is added?

E.Sketch the titration curve for 25.0mL of 0.10 M acetic acid with 0.10M sodium hydroxide according to your calculations

1 answer

To answer these questions, we will need to utilize the properties of acetic acid (a weak acid) and sodium hydroxide (a strong base), as well as some basic concepts from acid-base chemistry.

Given Data:

  • Volume of acetic acid = 25.0 mL
  • Concentration of acetic acid (\(CH_3COOH\)) = 0.10 M
  • Volume of sodium hydroxide = V mL
  • Concentration of sodium hydroxide (\(NaOH\)) = 0.10 M

A. pH of the Acid Solution Before NaOH is Added

To calculate the pH of a weak acid solution, we will use the dissociation constant (\(K_a\)) for acetic acid, which is approximately \(1.8 \times 10^{-5}\).

  1. Calculate moles of acetic acid: \[ \text{Moles of } CH_3COOH = \text{Volume} \times \text{Concentration} = 0.025,L \times 0.10,mol/L = 0.0025,mol \]

  2. Set up the equilibrium expression: \[ CH_3COOH \rightleftharpoons H^+ + CH_3COO^- \]

    We can let \(x\) be the concentration of \(H^+\) at equilibrium: \[ K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \] Which gives us: \[ K_a = \frac{x^2}{[0.10 - x]} \approx \frac{x^2}{0.10} \quad (since , K_a , is , small, , x , is , negligible) \]

  3. Solve for \(x\): \[ 1.8 \times 10^{-5} = \frac{x^2}{0.10} \implies x^2 = 1.8 \times 10^{-6} \implies x = \sqrt{1.8 \times 10^{-6}} \approx 0.00134 , M \]

  4. Calculate pH: \[ pH = -\log[H^+] = -\log(0.00134) \approx 2.87 \]

B. pH After 15.0 mL of Sodium Hydroxide Solution is Added

  1. Calculate moles of \(NaOH\) added: \[ \text{Moles of } NaOH = 0.015, L \times 0.10, mol/L = 0.0015, mol \]

  2. Determine the reaction between \(CH_3COOH\) and \(NaOH\): \[ CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \]

    Moles of \(CH_3COOH\) before reaction = 0.0025 mol
    Moles of \(NaOH\) = 0.0015 mol

    Thus, \(NaOH\) is the limiting reagent:

    • Moles of \(CH_3COOH\) remaining = \(0.0025 - 0.0015 = 0.0010,mol\)
    • Moles of \(CH_3COO^-\) formed = \(0.0015,mol\)

  3. Calculate concentrations after titration in new volume: Total volume = \(25.0 + 15.0 = 40.0, mL = 0.040, L\)

    • \([CH_3COOH] = \frac{0.0010, mol}{0.040, L} = 0.025, M\)
    • \([CH_3COO^-] = \frac{0.0015, mol}{0.040, L} = 0.0375, M\)

  4. Calculate pH using the Henderson-Hasselbalch equation: \[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \] \(pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74\)

    \[ pH = 4.74 + \log\left(\frac{0.0375}{0.025}\right) = 4.74 + \log(1.5) \approx 4.74 + 0.176 = 4.92 \]

C. Limiting Reagent and Amount

Limiting Reagent: \(NaOH\)

  • Amount of \(NaOH\) used: \(0.0015, mol\)
  • Amount of \(CH_3COOH\) remaining: \(0.0010, mol\)

D. pH After 25.0 mL of 0.10 M NaOH is Added

  1. Calculate moles of \(NaOH\) added: \[ \text{Moles of } NaOH = 0.025, L \times 0.10, mol/L = 0.0025, mol \]

  2. Stoichiometry of the reaction: \[ Moles of CH_3COOH = 0.0025, mol \]

    Since the moles are equal, they completely neutralize each other:

    • No acetic acid or sodium hydroxide remaining.
    • All is converted to acetate ion, \(CH_3COO^-\).

  3. Calculate the concentration of acetate ion in the new volume: \[ \text{Total volume} = 25.0, mL + 25.0, mL = 50.0, mL = 0.050, L \] Moles of acetate ion = 0.0025 mol
    Concentration of acetate ion = \(\frac{0.0025, mol}{0.050, L} = 0.050, M\)

  4. Calculate pH: The acetate ion will hydrolyze to give: \[ CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^- \]

    For acetate (\(CH_3COO^-\)), we will use its \(K_b\) and derived from \(K_w\): \[ K_b = \frac{K_w}{K_a} \quad (K_w = 1.0 \times 10^{-14}) \] \[ K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]

    Using \(K_b\) to find pOH: \[ K_b = \frac{x^2}{[0.050]} \implies x^2 = K_b \cdot 0.050 \implies x = \sqrt{5.56 \times 10^{-10} \cdot 0.050} \approx 1.73 \times 10^{-5} , M , (OH^- concentration) \]

    Calculate pOH: \[ pOH = -\log(1.73 \times 10^{-5}) \approx 4.76 \]

    Calculate pH: \[ pH + pOH = 14 \implies pH = 14 - 4.76 = 9.24 \]

Therefore, the pH after 25.0 mL of NaOH is added is approximately 9.24.

E. Titration Curve

To sketch the titration curve for 25.0 mL of 0.10 M acetic acid with 0.10 M sodium hydroxide:

  1. Start at a low pH around 2.87 (initial).
  2. As NaOH is added, the pH gradually rises.
  3. At 15.0 mL, the pH is around 4.92 (midway).
  4. Approaching 25.0 mL (the equivalence point), the pH starts to increase more rapidly, reaching around 9.24.
  5. Beyond the equivalence point, the curve will continue to rise steeply and approach a horizontal asymptote around 14.

The curve will generally have a gradual slope, a more pronounced rise around the equivalence point, and will taper off as it approaches the terminus of the titration.

Make sure to label axes and key points for clarity.