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MCQEasyJEE 2024Buffer Solutions

JEE Chemistry 2024 Question with Solution

The pH of an aqueous solution containing 1 M1 \, \text{M} benzoic acid (pKa=4.20pK_a = 4.20) and 1 M1 \, \text{M} sodium benzoate is 4.54.5. The volume of benzoic acid solution in 300 mL300 \, \text{mL} of this buffer solution is _____ mL.

  • A

    9090

  • B

    100100

  • C

    110110

  • D

    120120

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Buffer contains 1 M1 \, \text{M} benzoic acid and 1 M1 \, \text{M} sodium benzoate. The pH is 4.54.5, pKa=4.20pK_a = 4.20, and total volume is 300 mL300 \, \text{mL}.

Find: Volume of the benzoic acid solution used.

Use the Henderson-Hasselbalch equation:

pH=pKa+log⁡([A−][HA])\text{pH} = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)

Substitute the given values:

4.5=4.20+log⁡([A−][HA])4.5 = 4.20 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) 0.30=log⁡([A−][HA])0.30 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) [A−][HA]=100.30≈2.0\frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.30} \approx 2.0

Let the volume of 1 M1 \, \text{M} benzoic acid be VacidV_{\text{acid}} mL and the volume of 1 M1 \, \text{M} sodium benzoate be VsaltV_{\text{salt}} mL. Since both stock solutions have the same molarity, the concentration ratio in the final buffer equals the volume ratio:

[A−][HA]=VsaltVacid=2\frac{[\text{A}^-]}{[\text{HA}]} = \frac{V_{\text{salt}}}{V_{\text{acid}}} = 2

So,

Vsalt=2VacidV_{\text{salt}} = 2V_{\text{acid}}

Now use the total volume condition:

Vacid+Vsalt=300V_{\text{acid}} + V_{\text{salt}} = 300 Vacid+2Vacid=300V_{\text{acid}} + 2V_{\text{acid}} = 300 3Vacid=3003V_{\text{acid}} = 300 Vacid=100 mLV_{\text{acid}} = 100 \, \text{mL}

Therefore, the volume of benzoic acid solution is 100 mL100 \, \text{mL}. The correct option is B.

Ratio of Acid and Salt

Given: An acidic buffer is prepared from benzoic acid and sodium benzoate, each of concentration 1 M1 \, \text{M}. The pH is 4.54.5 and pKa=4.20pK_a = 4.20.

Find: The volume of benzoic acid present in 300 mL300 \, \text{mL} buffer.

For a weak acid buffer,

pH=pKa+log⁡([salt][acid])\text{pH} = pK_a + \log\left(\frac{[\text{salt}]}{[\text{acid}] }\right)

Hence,

4.5−4.20=log⁡([A−][HA])4.5 - 4.20 = \log\left(\frac{[\text{A}^-]}{[\text{HA}] }\right) 0.30=log⁡([A−][HA])0.30 = \log\left(\frac{[\text{A}^-]}{[\text{HA}] }\right) [A−][HA]=100.30≈2\frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.30} \approx 2

So the final solution has sodium benzoate and benzoic acid in the ratio

[A−]:[HA]=2:1[\text{A}^-] : [\text{HA}] = 2 : 1

Since both starting solutions are 1 M1 \, \text{M}, the mole ratio is the same as the volume ratio. Therefore,

Vsalt:Vacid=2:1V_{\text{salt}} : V_{\text{acid}} = 2 : 1

Let the benzoic acid volume be xx mL. Then sodium benzoate volume is 2x2x mL. Total volume is 300 mL300 \, \text{mL}, so

x+2x=300x + 2x = 300 3x=3003x = 300 x=100x = 100

Therefore, the required volume of benzoic acid solution is 100 mL100 \, \text{mL}.

Common mistakes

  • Using the Henderson-Hasselbalch ratio as [HA][A−]\frac{[\text{HA}]}{[\text{A}^-]} instead of [A−][HA]\frac{[\text{A}^-]}{[\text{HA}]} gives the inverse result. For an acidic buffer, use pH=pKa+log⁡([salt][acid])\text{pH} = pK_a + \log\left(\frac{[\text{salt}]}{[\text{acid}] }\right).

  • Assuming equal volumes because both solutions are 1 M1 \, \text{M} is incorrect. Equal molarity does not imply equal volume; the required ratio must first be obtained from the pH equation.

  • Taking 100.3010^{0.30} as 0.50.5 instead of about 22 reverses the acid-salt ratio. Convert the logarithmic relation carefully before applying the volume condition.

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