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If 100 ml of 0.1 M CH3COOH and 200 ml of...

If 100 ml of 0.1 M `CH_3COOH` and 200 ml of 0.03 M NaOH solutions are mixed together, then the pH of resulting mixture will be given `[pk_a (CH_3COOH) = 4.74 and log 1.5 = 0.18)]`

A

3.97

B

4.92

C

4.35

D

5.52

Text Solution

AI Generated Solution

The correct Answer is:
To find the pH of the resulting mixture when 100 ml of 0.1 M acetic acid (CH₃COOH) is mixed with 200 ml of 0.03 M sodium hydroxide (NaOH), we can follow these steps: ### Step 1: Calculate the number of moles of CH₃COOH - Molarity (M) = moles/volume (L) - Volume of CH₃COOH = 100 ml = 0.1 L - Molarity of CH₃COOH = 0.1 M \[ \text{Moles of CH₃COOH} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol} = 10 \, \text{mmol} \] ### Step 2: Calculate the number of moles of NaOH - Volume of NaOH = 200 ml = 0.2 L - Molarity of NaOH = 0.03 M \[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.03 \, \text{mol/L} \times 0.2 \, \text{L} = 0.006 \, \text{mol} = 6 \, \text{mmol} \] ### Step 3: Determine the limiting reagent - We have 10 mmol of CH₃COOH and 6 mmol of NaOH. - Since NaOH is the limiting reagent, it will completely react with CH₃COOH. ### Step 4: Calculate the remaining moles of CH₃COOH after reaction \[ \text{Remaining CH₃COOH} = \text{Initial CH₃COOH} - \text{NaOH used} = 10 \, \text{mmol} - 6 \, \text{mmol} = 4 \, \text{mmol} \] ### Step 5: Calculate the moles of the salt formed (sodium acetate, CH₃COONa) - The moles of salt formed will be equal to the moles of NaOH reacted, which is 6 mmol. ### Step 6: Calculate the total volume of the solution \[ \text{Total Volume} = 100 \, \text{ml} + 200 \, \text{ml} = 300 \, \text{ml} = 0.3 \, \text{L} \] ### Step 7: Calculate the concentrations of the remaining acetic acid and the formed salt - Concentration of CH₃COOH: \[ [\text{CH₃COOH}] = \frac{4 \, \text{mmol}}{0.3 \, \text{L}} = \frac{4 \times 10^{-3} \, \text{mol}}{0.3 \, \text{L}} = 0.01333 \, \text{M} \] - Concentration of CH₃COONa: \[ [\text{CH₃COONa}] = \frac{6 \, \text{mmol}}{0.3 \, \text{L}} = \frac{6 \times 10^{-3} \, \text{mol}}{0.3 \, \text{L}} = 0.02 \, \text{M} \] ### Step 8: Use the Henderson-Hasselbalch equation to find the pH \[ \text{pH} = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \] Given \( pK_a = 4.74 \): \[ \text{pH} = 4.74 + \log\left(\frac{0.02}{0.01333}\right) \] Calculating the ratio: \[ \frac{0.02}{0.01333} \approx 1.5 \] Thus, \[ \text{pH} = 4.74 + \log(1.5) \approx 4.74 + 0.18 = 4.92 \] ### Final Answer: The pH of the resulting mixture is **4.92**. ---
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