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MCQEasyJEE 2023Concentration Terms (Molarity, Molality, etc.)

JEE Chemistry 2023 Question with Solution

The volume of 0.02 M0.02 \, \text{M} aqueous HBr required to neutralize 10.0 mL10.0 \, \text{mL} of 0.01 M0.01 \, \text{M} aqueous Ba(OH)_2 is (Assume complete neutralization)

(1) 5.0 mL5.0 \, \text{mL}

(2) 10.0 mL10.0 \, \text{mL}

(3) 2.5 mL2.5 \, \text{mL}

(4) 7.5 mL7.5 \, \text{mL}_

  • A

    5.0 mL5.0 \, \text{mL}

  • B

    10.0 mL10.0 \, \text{mL}

  • C

    2.5 mL2.5 \, \text{mL}

  • D

    7.5 mL7.5 \, \text{mL}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Aqueous HBr has concentration 0.02 M0.02 \, \text{M} and aqueous Ba(OH)_2 has concentration 0.01 M0.01 \, \text{M} with volume 10.0 mL10.0 \, \text{mL}.

Find: The volume of HBr required for complete neutralization._

For neutralization, the milliequivalents of acid must equal the milliequivalents of base. Milliequivalents are calculated by:

Milliequivalents=M×n-factor×V\text{Milliequivalents} = M \times n\text{-factor} \times V

For HBr, the nn-factor is 11. For Ba(OH)_2, the nn-factor is 22. Let the required volume of HBr be VV._

Equating milliequivalents:

0.02×1×V=0.01×2×100.02 \times 1 \times V = 0.01 \times 2 \times 10 0.02V=0.20.02V = 0.2 V=0.20.02=10 mLV = \frac{0.2}{0.02} = 10 \, \text{mL}

Therefore, the volume of HBr required is 10 mL10 \, \text{mL}.

Equation Form

Given: M1=0.02 MM_1 = 0.02 \, \text{M} for HBr and M2=0.01 MM_2 = 0.01 \, \text{M} for Ba(OH)_2 with V2=10 mLV_2 = 10 \, \text{mL}.

Find: V1V_1, the volume of HBr required._

Using the equivalence relation:

m.e.q of HBr=m.e.q of Ba(OH)2\text{m.e.q of HBr} = \text{m.e.q of Ba(OH)}_2 M1×n1×V1=M2×n2×V2M_1 \times n_1 \times V_1 = M_2 \times n_2 \times V_2

Substitute the given values:

0.02×1×V1=0.01×2×100.02 \times 1 \times V_1 = 0.01 \times 2 \times 10 V1=0.01×100.02=10 mLV_1 = \frac{0.01 \times 10}{0.02} = 10 \, \text{mL}

Thus, the correct numerical result is 10.0 mL10.0 \, \text{mL}.

Common mistakes

  • Using the nn-factor of Ba(OH)_2 as 11 is incorrect because each mole of Ba(OH)_2 provides 2OH−2OH^- ions. Use n=2n = 2 for the base.

  • Equating only molarity × volume is wrong here because neutralization depends on equivalents, not just moles of solution. Multiply by the appropriate nn-factor before comparing acid and base.

  • Choosing the option directly from the displayed answer key without checking the calculation can cause an error here, because the working gives 10.0 mL10.0 \, \text{mL} while the marked option letter is inconsistent with the listed options. Match the computed value to the option position.

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