Question 22 of 34intermediate🔧 ApplyMCQ1 mark

10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a given solution of HCl. If we take 20 mL of the same solution of NaOH, the amount HCl solution (the same solution as before) required to neutralise it will be

Exercise: EXERCISES | Q: 3 | (Page Page 18)
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Explanation

This question is based on the concept of neutralisation reaction where acids and bases neutralise each other in a fixed ratio. The reaction given in the context is: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l). Since the concentrations of both solutions remain the same, the ratio of volumes required for neutralisation remains constant. If 10 mL of NaOH requires 8 mL of HCl for complete neutralisation, then doubling the volume of NaOH to 20 mL will require double the volume of HCl as well. Therefore, 20 mL NaOH will require 16 mL HCl for neutralisation.

Solution Steps

  1. Step 1: Identify the given ratio - 10 mL NaOH is neutralised by 8 mL HCl.

  2. Step 2: Since the same solutions are used, the ratio of volumes remains constant.

  3. Step 3: Calculate the required HCl volume: (20 mL NaOH × 8 mL HCl) ÷ 10 mL NaOH = 16 mL HCl.

  4. Step 4: The correct answer is option (D) 16 mL.

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