100 ml of each HCl solution having pH=5 and NaOH having pH=8 is mixture. How much volume of `(N)/(100)` NaOH is required to neutralise to 20 ml of this mixture ?

Options

A. 0.009ml
B. 9 ml
C. 20 ml
D. 15 ml

Answer

To solve the problem, we need to determine how much volume of \( \frac{N}{100} \) NaOH is required to neutralize 20 ml of the mixture of HCl and NaOH solutions. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the concentration of HCl
Given that the pH of the HCl solution is 5, we can calculate the concentration of hydrogen ions \([H^+]\) using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-5} \, \text{mol/L} \]

### Step 2: Calculate the number of moles of HCl in 100 ml
Using the concentration from Step 1, we can find the number of moles of HCl in 100 ml (0.1 L):
\[ \text{Moles of HCl} = [H^+] \times \text{Volume} = 10^{-5} \, \text{mol/L} \times 0.1 \, \text{L} = 10^{-6} \, \text{mol} \]

### Step 3: Calculate the concentration of NaOH
The pH of the NaOH solution is given as 8. We can calculate the concentration of hydroxide ions \([OH^-]\) using:
\[ pOH = 14 - pH = 14 - 8 = 6 \]
\[ [OH^-] = 10^{-pOH} = 10^{-6} \, \text{mol/L} \]

### Step 4: Calculate the number of moles of NaOH in 100 ml
Now, we find the number of moles of NaOH in 100 ml (0.1 L):
\[ \text{Moles of NaOH} = [OH^-] \times \text{Volume} = 10^{-6} \, \text{mol/L} \times 0.1 \, \text{L} = 10^{-7} \, \text{mol} \]

### Step 5: Determine the remaining moles of H+ after reaction
Since HCl is a strong acid and NaOH is a strong base, they will react in a 1:1 ratio:
\[ \text{Remaining moles of } H^+ = \text{Moles of HCl} - \text{Moles of NaOH} = 10^{-6} \, \text{mol} - 10^{-7} \, \text{mol} = 9 \times 10^{-7} \, \text{mol} \]

### Step 6: Calculate the concentration of H+ in the total mixture
The total volume of the mixture is 200 ml (100 ml of HCl + 100 ml of NaOH). Therefore, the concentration of \( H^+ \) in the mixture is:
\[ \text{Concentration of } H^+ = \frac{\text{Remaining moles of } H^+}{\text{Total Volume}} = \frac{9 \times 10^{-7} \, \text{mol}}{0.2 \, \text{L}} = 4.5 \times 10^{-6} \, \text{mol/L} \]

### Step 7: Calculate the moles of H+ in 20 ml of the mixture
Now, we need to find the moles of \( H^+ \) in 20 ml (0.02 L) of this mixture:
\[ \text{Moles of } H^+ \text{ in 20 ml} = 4.5 \times 10^{-6} \, \text{mol/L} \times 0.02 \, \text{L} = 9 \times 10^{-8} \, \text{mol} \]

### Step 8: Calculate the volume of NaOH required to neutralize the remaining H+
Using the neutralization reaction, we know that:
\[ \text{Moles of NaOH required} = \text{Moles of } H^+ = 9 \times 10^{-8} \, \text{mol} \]
Since the NaOH solution is \( \frac{N}{100} \) (which is equivalent to \( 0.1 \, \text{mol/L} \)):
\[ \text{Volume of NaOH required} = \frac{\text{Moles of NaOH}}{\text{Concentration of NaOH}} = \frac{9 \times 10^{-8} \, \text{mol}}{0.1 \, \text{mol/L}} = 9 \times 10^{-7} \, \text{L} = 0.0009 \, \text{L} = 0.9 \, \text{ml} \]

### Final Answer
The volume of \( \frac{N}{100} \) NaOH required to neutralize 20 ml of the mixture is **0.9 ml**.

---