To answer your questions step by step, we will first review the key principles involved in titration and the dissociation of acetic acid and the hydroxide ion.
A. What is the pH of the acid solution before NaOH is added?
Given:
- Volume of acetic acid = 25.0 mL
- Concentration of acetic acid (\( \text{CH}_3\text{COOH} \)) = 0.10 M
The dissociation of acetic acid can be represented as follows:
\[ \text{CH}_3\text{COOH} \leftrightarrow \text{H}^+ + \text{CH}_3\text{COO}^- \]
Let \( K_a \) of acetic acid (\( \text{CH}_3\text{COOH} \)) be about \( 1.8 \times 10^{-5} \).
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Calculate the moles of acetic acid: \[ \text{Moles of CH}_3\text{COOH} = \text{Volume (L)} \times \text{Molarity} = 0.025 \times 0.10 = 0.0025 \text{ moles} \]
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Now set up the ICE table for the dissociation: \[ \begin{array}{c|c|c|c} & \text{CH}_3\text{COOH} & \text{H}^+ & \text{CH}_3\text{COO}^- \ \hline \text{Initial} & 0.10 & 0 & 0 \ \text{Change} & -x & +x & +x \ \text{Equilibrium} & 0.10-x & x & x \end{array} \]
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Use the \( K_a \) expression: \[ K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = \frac{x^2}{0.10 - x} \]
Assuming \( x \) is small compared to 0.10: \[ K_a \approx \frac{x^2}{0.10} \]
Substituting \( K_a \): \[ 1.8 \times 10^{-5} = \frac{x^2}{0.10} \] \[ x^2 = 1.8 \times 10^{-6} \] \[ x = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \text{ M} \]
- Therefore, \( \text{pH} \): \[ \text{pH} = -\log(1.34 \times 10^{-3}) \approx 2.87 \]
B. What is the pH after 15.0 mL of sodium hydroxide solution is added?
Volume of NaOH added = 15.0 mL
- Molarity of NaOH = 0.10 M
- Moles of NaOH added = \( 0.015 , \text{L} \times 0.10 , \text{M} = 0.0015 , \text{moles} \)
Total moles of acetic acid (in 25.0 mL):
- Moles of acetic acid = 0.0025 moles
Neutralization Reaction: \[ \text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \]
Initial amounts:
- Before reaction:
- Acetic acid: 0.0025 moles
- Sodium hydroxide: 0.0015 moles
After reaction:
- Remaining acetic acid = \( 0.0025 - 0.0015 = 0.0010 \) moles
- Acetate ions produced = \( 0.0015 \) moles
Now, the total volume after mixing is \( 25.0 + 15.0 = 40.0 \) mL = \( 0.040 \) L
Calculate concentrations after the reaction:
- [Acetic acid] = \( \frac{0.0010 \text{ moles}}{0.040 \text{ L}} = 0.025 \text{ M} \)
- [Acetate] = \( \frac{0.0015 \text{ moles}}{0.040 \text{ L}} = 0.0375 \text{ M} \)
Use the Henderson-Hasselbalch equation: \[ \text{pH} = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] Where \( pK_a \approx -\log(1.8 \times 10^{-5}) \approx 4.74 \).
\[ \text{pH} = 4.74 + \log\left(\frac{0.0375}{0.025}\right) \approx 4.74 + 0.176 \approx 4.92 \]
C. According to your calculations in B, which species is the limiting reagent and by what amount?
The limiting reagent in this scenario is sodium hydroxide. It completely reacts with some acetic acid, converting 0.0015 moles of acetic acid to acetate:
Limiting Reagent: NaOH
- Amount of acetic acid reacted = 0.0015 moles out of 0.0025 moles
- Remaining acetic acid = 0.0010 moles (after reaction)
- So, the amount of NaOH used is the limiting quantity: 0.0015 moles.
D. What is the pH of the solution after 25.0mL of 0.10M NaOH is added?
Total volume after adding 25.0 mL of NaOH:
- Initial volume = 25.0 mL (acetic acid) + 25.0 mL (NaOH) = 50.0 mL
Now, moles of NaOH added:
- Moles of NaOH = \( 0.0250 , \text{L} \times 0.10 , \text{M} = 0.0025 , \text{moles} \)
Total moles:
- Acetic acid before reaction = 0.0025 moles
- NaOH added = 0.0025 moles
At this point, they will completely neutralize each other: \[ \text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \]
After the neutralization:
- Both reactants are consumed, resulting in 0 moles of acetic acid and 0.0025 moles of acetate.
Since acetate ions (\( \text{CH}_3\text{COO}^- \)) will now be present, we will use hydrolysis to find the pH. In water, acetate ion acts as a weak base: \[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \leftrightarrow \text{CH}_3\text{COOH} + \text{OH}^- \] Let \( x \) = concentration of OH\(^-\) produced.
Using Kb: \[ K_b = \frac{K_w}{K_a} \] Where \( K_w = 1.0 \times 10^{-14} \) and \( K_a = 1.8 \times 10^{-5} \): \[ K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]
For the hydrolysis of acetate: \[ K_b = \frac{x^2}{[\text{CH}_3\text{COO}^-]} \quad \text{(initial concentration is about 0.05 M in 50 mL)} \] \[ 0.05 - x \approx 0.05 \] \[ 5.56 \times 10^{-10} = \frac{x^2}{0.05} \] \[ x^2 = 5.56 \times 10^{-10} \times 0.05 \approx 2.78 \times 10^{-11} \] \[ x = \sqrt{2.78 \times 10^{-11}} \approx 5.27 \times 10^{-6} \]
Now calculate \( \text{pOH} \): \[ \text{pOH} = -\log(5.27 \times 10^{-6}) \approx 5.28 \]
Finally, using \( \text{pH} + \text{pOH} = 14 \): \[ \text{pH} = 14 - 5.28 \approx 8.72 \]
E. Sketch the titration curve for 25.0 mL of 0.10 M NaOH according to your calculation.
In general, the sketch of a titration curve of a weak acid (acetic acid) with a strong base (NaOH) will show:
- A gradual increase in pH from the start (initial pH ~ 2.87).
- A steep increase around the equivalence point, which in this case occurs at 25.0 mL of NaOH added.
- After the equivalence point (beyond 25.0 mL), the pH rises sharply as there is excess NaOH.
The curve will begin at a low initial pH, rise gradually, steeply incline just before the equivalence point, and then level off at a higher pH.
In summary:
A) pH before NaOH: ~2.87
B) pH after 15.0 mL NaOH: ~4.92
C) Limiting Reagent: NaOH (0.0015 moles)
D) pH after 25.0 mL NaOH: ~8.72
For the visual representation, it's best to graph these points on graph paper or using software to have a clear picture of the titration curve.