Logo
Jeeify
MCQEasyJEE 2023Stoichiometry & Calculations

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):_

  • 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:B

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 = Molarity ×\times n-factor ×\times Volume.

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

Using equality of 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}. The correct option is B.

Equivalent Formula Method

Given:

  • M1=0.02 MM_1 = 0.02 \, \text{M} for HBr
  • n1=1n_1 = 1 for HBr
  • M2=0.01 MM_2 = 0.01 \, \text{M} for Ba(OH)_2
  • n2=2n_2 = 2 for Ba(OH)_2
  • V2=10 mLV_2 = 10 \, \text{mL}

Find: V1V_1, the volume of HBr.

Apply the relation:

M1×n1×V1=M2×n2×V2M_1 \times n_1 \times V_1 = M_2 \times n_2 \times V_2

Substituting the given values:

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

Hence, the required volume is 10.0 mL10.0 \, \text{mL}, so the correct option is B.

Common mistakes

  • Using molarity ×\times volume directly without considering n-factor is incorrect here because Ba(OH)_2 provides two OH⁻ ions per mole. Use milliequivalents or the balanced stoichiometric ratio instead._

  • Taking the n-factor of Ba(OH)_2 as 11 is wrong. It must be 22 because one mole of Ba(OH)_2 neutralizes two moles of HBr.

  • Ignoring unit consistency can lead to error. Since the relation is applied in milliequivalent form, keep the base volume in mL consistently and solve for the acid volume in mL as well.

Practice more Stoichiometry & Calculations questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Stoichiometry & Calculations

Related questions