Question

The question text is cut off in the image, so the full data for this question is not visible. Only the ending part
“\(\dots\) mole of \(CH_3COONa\). The \([H^+]\) in the resultant solution is”
and the given value
\[ K_a(CH_3COOH)=1.1\times 10^{-5} \] are visible. Therefore, the complete numerical information needed to solve this question is missing.

• A. \(1.47\times 10^{-5}\,M\)
• B. \(2\times 10^{-5}\,M\)
• C. \(2.5\times 10^{-5}\,M\)
• D. \(1.5\times 10^{-5}\,M\)
• E. \(0.9\times 10^{-5}\,M\)

Hint:

For acidic buffer solutions, always use \([H^+]=K_a\cdot \frac{[\text{acid}]}{[\text{salt}]}\). But this requires the full acid-to-salt ratio, so incomplete question data cannot give a reliable numerical answer.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The initial solution is a buffer solution containing a weak acid (acetic acid, CH\(_3\)COOH) and its conjugate base (acetate, from sodium acetate, CH\(_3\)COONa). A strong base (NaOH) is added to this buffer. The strong base will react with the weak acid component of the buffer. We need to calculate the new [H\(^+\)] after the reaction.
Step 2: Key Formula or Approach:
1. Calculate the moles of NaOH added.
2. Write the neutralization reaction between the added base (NaOH) and the buffer's acid component (CH\(_3\)COOH).
3. Calculate the new moles of the acid and its conjugate base after the reaction.
4. Use the Henderson-Hasselbalch equation or the acid dissociation constant expression to find the new [H\(^+\)].
The expression for K\(_a\) is:
\[ K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \implies [\text{H}^+] = K_a \frac{[\text{CH}_3\text{COOH}]}{[\text{CH}_3\text{COO}^-]} \] Step 3: Detailed Explanation:
1. Moles of NaOH added:
Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol.
Moles of NaOH = \(\frac{\text{mass}}{\text{molar mass}} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ mol}\).
2. Neutralization Reaction:
The strong base NaOH reacts with the weak acid CH\(_3\)COOH:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \] 3. Moles after reaction:
Initial moles in 1 L solution:
Moles of CH\(_3\)COOH = 1.0 mol
Moles of CH\(_3\)COONa (source of CH\(_3\)COO\(^-\)) = 1.0 mol
Moles of NaOH added = 0.1 mol
The 0.1 mol of NaOH will consume 0.1 mol of CH\(_3\)COOH and produce 0.1 mol of CH\(_3\)COONa.
New moles of CH\(_3\)COOH = 1.0 - 0.1 = 0.9 mol
New moles of CH\(_3\)COONa = 1.0 + 0.1 = 1.1 mol
Since the volume is 1 litre, these mole values are also the molar concentrations.
[CH\(_3\)COOH] = 0.9 M
[CH\(_3\)COO\(^-\)] = 1.1 M
4. Calculate [H\(^+\)]:
Using the rearranged K\(_a\) expression:
\[ [\text{H}^+] = K_a \times \frac{[\text{Acid}]}{[\text{Conjugate Base}]} \] \[ [\text{H}^+] = (1.1 \times 10^{-5}) \times \frac{0.9}{1.1} \] The '1.1' terms cancel out:
\[ [\text{H}^+] = 10^{-5} \times 0.9 = 0.9 \times 10^{-5} \text{ M} \] Step 4: Final Answer:
The hydrogen ion concentration in the resultant solution is 0.9 \(\times\) 10\(^{-5}\) M. This corresponds to option (E).