Verified Solution • 2025–26 Session moderate

QJEEM19Ch93eb: JEE Main 2019 Chemistry - Class 11 JEEM19Ch93eb Solution: 25 ml of the given HCl solution requires 30 mL of 0.1 M sodium…

25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solutions ?

Multiple Choice Options

  • A 50 mL
  • B 12.5 mL
  • C 25 mL
  • D 75 mL

Step-by-Step Verified Solution

2HCl(aq) + Na2CO3(aq) →\to H2CO3 + NaCl

moles of HCl
2
 =
moles of Na2CO3
1

⇒\Rightarrow M×2510002=0.1×3010001{{M \times {{25} \over {1000}}} \over 2} = {{0.1 \times {{30} \over {1000}}} \over 1}

⇒\Rightarrow Molarity of HCl (M) = 625M{6 \over {25}}M

HCl(aq) + NaOH(aq) →\to NaCl + H2O

moles of HCl
1
 =
moles of NaOH
1

625×V10001=0.2×3010001{{{6 \over {25}} \times {V \over {1000}}} \over 1} = {{0.2 \times {{30} \over {1000}}} \over 1}

⇒\Rightarrow V = 25 ml

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