Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:
a. 60 mL M10 HCl + 40 mL M10 NaOH
b. 55 mL M10 HCl + 45 mL M10 NaOH
c. 75 mL M5 HCl + 25 mL M5 NaOH
d. 100 mL M10 HCl + 100 mL M10 NaOH

pH of which one of them will be equal to 1?

The correct option is c.

The pH of 75 mL M5 HCl + 25 mL M5 NaOH will be equal to 1.

In option c, we have 75 mL of 5 M HCl and 25 mL of 5 M NaOH.

Millimoles of HCl = 75 mL × 5 M = 375 millimoles
Millimoles of NaOH = 25 mL × 5 M = 125 millimoles

The reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O

Since NaOH is a limiting reagent, it will react completely with HCl.

Millimoles of HCl remaining = 375 millimoles - 125 millimoles = 250 millimoles

Total volume = 75 mL + 25 mL = 100 mL

Concentration of HCl remaining = 250 millimoles / 100 mL = 2.5 M

pH = -log[H+]
Since HCl is a strong acid, [H+] = 2.5 M

Therefore, pH = -log(2.5) ≈ -0.398

However, there appears to be a mistake in the question or the provided solution. A pH of 1 requires a much lower concentration of H+ ions. Let's re-examine the options:

Option a: 60 mL of 10M HCl and 40mL of 10M NaOH
Millimoles HCl = 600; Millimoles NaOH = 400
Remaining millimoles HCl = 200
Concentration HCl = 200/100mL = 2M. pH = -log(2) ≈ 0.3

Option b: 55 mL of 10M HCl and 45 mL of 10M NaOH
Millimoles HCl = 550; Millimoles NaOH = 450
Remaining millimoles HCl = 100
Concentration HCl = 100/100mL = 1M. pH = -log(1) = 0

Option c: 75 mL of 5M HCl and 25 mL of 5M NaOH
Millimoles HCl = 375; Millimoles NaOH = 125
Remaining millimoles HCl = 250
Concentration HCl = 250/100mL = 2.5M. pH = -log(2.5) ≈ -0.398

Option d: 100 mL of 10M HCl and 100 mL of 10M NaOH
This is a complete neutralization, resulting in a pH of 7.

It seems there's an error in either the question or the provided solution. None of the options result in a pH of 1. The closest is option b, which yields a pH of 0.