JEE Challenger
More from Equilibrium
ChemMedAdv 2015 • Paper 2

Calculate pH of Solution After Mixing Acetic Acid and NaOH

Comprehension Passage

PARAGRAPH 2

When 100 mL100\text{ mL} of 1.0 M HCl1.0\text{ M HCl} solution was mixed with 100 mL100\text{ mL} of 1.0 M NaOH1.0\text{ M NaOH} inside an insulated beaker under constant pressure, a temperature rise of 5.7∘C5.7^\circ\text{C} was recorded for the beaker along with its contents (Expt. 1). Since the enthalpy of neutralization for a strong acid with a strong base remains a constant (−57.0 kJ mol−1-57.0\text{ kJ mol}^{-1}), this setup was utilized to determine the calorimeter constant. In a subsequent trial (Expt. 2), 100 mL100\text{ mL} of 2.0 M2.0\text{ M} acetic acid (Ka=2.0×10−5K_a = 2.0 \times 10^{-5}) was mixed with 100 mL100\text{ mL} of 1.0 M NaOH1.0\text{ M NaOH} under identical conditions as Expt. 1, resulting in a temperature increase of 5.6∘C5.6^\circ\text{C}. (Assume heat capacity for all solutions to be 4.2 J g−1 K−14.2\text{ J g}^{-1}\text{ K}^{-1} and density for all solutions to be 1.0 g mL−11.0\text{ g mL}^{-1})

Find the pH\text{pH} value of the resulting solution following Expt. 2.

Options

A

2.8

B

4.7

Correct
C

5.0

D

7.0

Topics & Concepts

EquilibriumAcid-Base Equilibrium

Step-by-Step Solution

To find the pH\text{pH} value of the resulting solution after Experiment 2, we analyze the reaction between acetic acid (CH3COOH\text{CH}_3\text{COOH}) and sodium hydroxide (NaOH\text{NaOH}).

Step 1: Calculate the initial millimoles of the reactants

  • Millimoles of CH3COOH=Volume (in mL)×Molarity=100 mL×2.0 M=200 mmol\text{CH}_3\text{COOH} = \text{Volume (in mL)} \times \text{Molarity} = 100 \text{ mL} \times 2.0 \text{ M} = 200 \text{ mmol}
  • Millimoles of NaOH=Volume (in mL)×Molarity=100 mL×1.0 M=100 mmol\text{NaOH} = \text{Volume (in mL)} \times \text{Molarity} = 100 \text{ mL} \times 1.0 \text{ M} = 100 \text{ mmol}

Step 2: Neutralization Reaction

Acetic acid reacts with sodium hydroxide according to the following equation: CH3COOH+NaOH→CH3COONa+H2O\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}

  • Initial millimoles:

    • CH3COOH=200 mmol\text{CH}_3\text{COOH} = 200 \text{ mmol}
    • NaOH=100 mmol\text{NaOH} = 100 \text{ mmol}
    • CH3COONa=0 mmol\text{CH}_3\text{COONa} = 0 \text{ mmol}

  • Millimoles after reaction: Since NaOH\text{NaOH} is the limiting reagent, 100 mmol100 \text{ mmol} of NaOH\text{NaOH} will completely react with 100 mmol100 \text{ mmol} of CH3COOH\text{CH}_3\text{COOH}.

    • Remaining CH3COOH=200−100=100 mmol\text{CH}_3\text{COOH} = 200 - 100 = 100 \text{ mmol}
    • Remaining NaOH=0 mmol\text{NaOH} = 0 \text{ mmol}
    • Formed CH3COONa=100 mmol\text{CH}_3\text{COONa} = 100 \text{ mmol}

Step 3: Determine the pH of the Buffer Solution

The resulting solution contains a weak acid (CH3COOH\text{CH}_3\text{COOH}) and its salt with a strong base (CH3COONa\text{CH}_3\text{COONa}), forming an acidic buffer solution.

Using the Henderson-Hasselbalch equation: pH=pKa+log⁡10([Salt][Acid])\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right)

First, calculate pKa\text{p}K_a: pKa=−log⁡10(Ka)=−log⁡10(2.0×10−5)=5−log⁡10(2)≈5−0.301=4.699≈4.7\text{p}K_a = -\log_{10}(K_a) = -\log_{10}(2.0 \times 10^{-5}) = 5 - \log_{10}(2) \approx 5 - 0.301 = 4.699 \approx 4.7

Now, substitute the values into the Henderson-Hasselbalch equation: pH=4.7+log⁡10(100 mmol100 mmol)\text{pH} = 4.7 + \log_{10}\left(\frac{100 \text{ mmol}}{100 \text{ mmol}}\right) pH=4.7+log⁡10(1)\text{pH} = 4.7 + \log_{10}(1) pH=4.7+0=4.7\text{pH} = 4.7 + 0 = 4.7

Conclusion

The pH\text{pH} value of the resulting solution is 4.74.7.

Correct Option: (B)

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