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Some Basic Concepts of Chemistry question

2023 · 6 Apr · Shift 2 · Q15
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  5. /2023 · 6 Apr · Shift 2 · Q15

Some Basic Concepts of Chemistry question

2023 · 6 Apr · Shift 2 · Q15

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
The volume of 0.02 M0.02 ~\mathrm{M}0.02 M aqueous HBr\mathrm{HBr}HBr required to neutralize 10.0 mL10.0 \mathrm{~mL}10.0 mL of 0.01 M0.01 ~\mathrm{M}0.01 M aqueous Ba(OH)2\mathrm{Ba}(\mathrm{OH})_{2}Ba(OH)2​ is (Assume complete neutralization)
  1. A
    7.5 mL
  2. B
    5.0 mL
  3. C
    10.0 mL
  4. D
    2.5 mL
View written solutionFree

Correct answer: C

  1. Write the balanced neutralization reaction
Ba(OH)2+2HBr→BaBr2+2H2O\mathrm{Ba(OH)_2 + 2HBr \rightarrow BaBr_2 + 2H_2O}Ba(OH)2​+2HBr→BaBr2​+2H2​O

This shows that:

  • 111 mole of Ba(OH)2\mathrm{Ba(OH)_2}Ba(OH)2​ reacts with 222 moles of HBr\mathrm{HBr}HBr.
  1. Calculate moles of Ba(OH)2\mathrm{Ba(OH)_2}Ba(OH)2​ present

Given:

  • Concentration of Ba(OH)2=0.01 M\mathrm{Ba(OH)_2} = 0.01\,\mathrm{M}Ba(OH)2​=0.01M
  • Volume =10.0 mL=0.0100 L= 10.0\,\mathrm{mL} = 0.0100\,\mathrm{L}=10.0mL=0.0100L

So,

moles of Ba(OH)2=M×V=0.01×0.0100=1.0×10−4 mol\text{moles of } \mathrm{Ba(OH)_2} = M \times V = 0.01 \times 0.0100 = 1.0 \times 10^{-4}\,\mathrm{mol}moles of Ba(OH)2​=M×V=0.01×0.0100=1.0×10−4mol
  1. Use stoichiometry to find moles of HBr\mathrm{HBr}HBr required

From the balanced equation,

1 mol Ba(OH)2 requires 2 mol HBr1\,\text{mol } \mathrm{Ba(OH)_2} \text{ requires } 2\,\text{mol } \mathrm{HBr}1mol Ba(OH)2​ requires 2mol HBr

Therefore,

moles of HBr=2×1.0×10−4=2.0×10−4 mol\text{moles of } \mathrm{HBr} = 2 \times 1.0 \times 10^{-4} = 2.0 \times 10^{-4}\,\mathrm{mol}moles of HBr=2×1.0×10−4=2.0×10−4mol
  1. Calculate volume of 0.02 M0.02\,\mathrm{M}0.02M HBr\mathrm{HBr}HBr needed

Given:

  • Concentration of HBr=0.02 M\mathrm{HBr} = 0.02\,\mathrm{M}HBr=0.02M

Using,

V=nM=2.0×10−40.02=1.0×10−2 LV = \frac{n}{M} = \frac{2.0 \times 10^{-4}}{0.02} = 1.0 \times 10^{-2}\,\mathrm{L}V=Mn​=0.022.0×10−4​=1.0×10−2L

Convert to mL:

1.0×10−2 L=10.0 mL1.0 \times 10^{-2}\,\mathrm{L} = 10.0\,\mathrm{mL}1.0×10−2L=10.0mL
  1. Check options

The correct option is:

  • C: 10.0 mL10.0\,\mathrm{mL}10.0mL
  1. Comparison with stored correct answer

Stored correct answer: C

My derived answer is also C, so they agree.

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