N=N1V1-N2V2/V1+V2.(as HCl normality is more than NaOH normality).N=50×0.4-50×0.2/50+50.N=20-10/100.N=10/100.N=0.1.we know, N= [H^+].pH=-log[H^+].pH=-log(0.1).pH=-log(10^-1).pH=1...

The pH of a solution can be determined by calculating the concentration of hydronium ions (H3O+) in the solution. In this case, we are given two solutions, HCl and NaOH, with different concentrations. To calculate the pH of the resulting solution, we can use the concept of neutralization.

1. Determine the number of moles of each solution:
- HCl:
- Volume = 50 ml = 0.05 L
- Concentration (N) = 0.4 N
- Number of moles = concentration × volume = 0.4 N × 0.05 L = 0.02 moles
- NaOH:
- Volume = 50 ml = 0.05 L
- Concentration (N) = 0.2 N
- Number of moles = concentration × volume = 0.2 N × 0.05 L = 0.01 moles

2. Determine the limiting reactant:
- To neutralize each other, 1 mole of HCl reacts with 1 mole of NaOH.
- Since the number of moles of HCl is 0.02 and NaOH is 0.01, NaOH is the limiting reactant.

3. Determine the remaining excess reactant:
- The remaining moles of HCl after the reaction = moles of HCl - moles of NaOH = 0.02 - 0.01 = 0.01 moles

4. Determine the concentration of the excess reactant:
- Volume of the resulting solution = volume of HCl + volume of NaOH = 50 ml + 50 ml = 100 ml = 0.1 L
- Concentration of the excess HCl = moles of HCl / volume of resulting solution = 0.01 moles / 0.1 L = 0.1 N

5. Determine the concentration of hydronium ions (H3O+):
- In the neutralization reaction, 1 mole of HCl produces 1 mole of H3O+.
- Since the concentration of the excess HCl is 0.1 N, the concentration of H3O+ is also 0.1 N.

6. Calculate the pH:
- pH = -log[H3O+]
- pH = -log(0.1)
- pH ≈ 1

Therefore, the pH of the resulting solution obtained by mixing 50 ml of 0.4 N HCl and 50 ml of 0.2 N NaOH is approximately 1.0. Hence, option (C) is correct.