To find the pH of the solution obtained by mixing 25 ml of 0.2 M HCl with 50 ml of 0.25 M NaOH, we need to determine the concentration of the resulting solution.

First, let's find the number of moles of HCl and NaOH used:

moles of HCl = volume (in L) x concentration
= 0.025 L x 0.2 M
= 0.005 mol

moles of NaOH = volume (in L) x concentration
= 0.050 L x 0.25 M
= 0.0125 mol

Since HCl and NaOH react in a 1:1 ratio, the number of moles of HCl used is equal to the number of moles of NaOH used. Therefore, all the HCl is neutralized by NaOH, and we are left with an excess of NaOH.

The total volume of the resulting solution is the sum of the volumes of HCl and NaOH used:

total volume = 25 ml + 50 ml
= 75 ml
= 0.075 L

To find the concentration of the resulting solution, we divide the moles of NaOH by the total volume:

concentration = moles of NaOH / total volume
= 0.0125 mol / 0.075 L
= 0.1667 M

Now, we can use the concentration to find the pOH of the solution. The pOH is calculated using the formula:

pOH = -log10(concentration)

pOH = -log10(0.1667)
≈ 0.778

Finally, we can find the pH of the solution using the relationship:

pH + pOH = 14

pH = 14 - pOH
= 14 - 0.778
≈ 13.222

Therefore, the pH of the solution obtained by mixing 25 ml of 0.2 M HCl with 50 ml of 0.25 M NaOH is approximately 13.222.