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A 100 ml solution of 0.1 N HCl was titra...

A `100 ml` solution of `0.1 N HCl` was titrated with `0.2``N NaOH` solution. The titration was discontinued after adding `30 ml NaOH` solution. The remaining titration was completed by adding `0.25 N KOH` solution. The volume of `KOH` required for completing the titration is

A

16 ml

B

32 ml

C

35 ml

D

70 ml

Text Solution

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The correct Answer is:
To solve the problem step by step, we need to find the volume of KOH solution required to complete the titration after adding 30 ml of NaOH to the HCl solution. ### Step 1: Calculate the number of equivalents of HCl The number of equivalents of HCl can be calculated using the formula: \[ \text{Number of equivalents} = \text{Volume (L)} \times \text{Normality (N)} \] Given: - Volume of HCl = 100 ml = 0.1 L - Normality of HCl = 0.1 N Calculating the number of equivalents: \[ \text{Number of equivalents of HCl} = 0.1 \, \text{L} \times 0.1 \, \text{N} = 0.01 \, \text{equivalents} \] ### Step 2: Calculate the number of equivalents of NaOH added Next, we calculate the number of equivalents of NaOH that were added during the titration. Given: - Volume of NaOH = 30 ml = 0.03 L - Normality of NaOH = 0.2 N Calculating the number of equivalents: \[ \text{Number of equivalents of NaOH} = 0.03 \, \text{L} \times 0.2 \, \text{N} = 0.006 \, \text{equivalents} \] ### Step 3: Calculate the remaining equivalents of HCl To find the remaining equivalents of HCl after adding NaOH: \[ \text{Remaining equivalents of HCl} = \text{Total equivalents of HCl} - \text{Equivalents of NaOH added} \] \[ \text{Remaining equivalents of HCl} = 0.01 - 0.006 = 0.004 \, \text{equivalents} \] ### Step 4: Calculate the volume of KOH required to neutralize the remaining HCl Now, we need to find out how much KOH is required to neutralize the remaining 0.004 equivalents of HCl. Given: - Normality of KOH = 0.25 N Using the formula for equivalents: \[ \text{Number of equivalents} = \text{Volume (L)} \times \text{Normality (N)} \] Let \( V \) be the volume of KOH in liters. Therefore: \[ 0.004 = V \times 0.25 \] Solving for \( V \): \[ V = \frac{0.004}{0.25} = 0.016 \, \text{L} = 16 \, \text{ml} \] ### Conclusion The volume of KOH required to complete the titration is **16 ml**. ---
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