find the ph of the solution that results from the addition of 25ml of 0.1mol/l NaOH solution to 50ml of 0.1mol/l HCL solution?

To find the pH of the resulting solution from the addition of NaOH and HCl, you need to determine the concentration of the resulting solution.

First, let's calculate the number of moles of NaOH and HCl in their respective solutions:

Moles of NaOH = 0.1 mol/L * 0.025 L (volume in liters) = 0.0025 mol
Moles of HCl = 0.1 mol/L * 0.05 L = 0.005 mol

Since NaOH and HCl react in a 1:1 ratio, the number of moles of NaOH is equal to the number of moles of HCl. Therefore, the resulting solution will have no excess of either NaOH or HCl.

Next, let's calculate the total volume of the resulting solution:

Total volume = 25 mL (NaOH) + 50 mL (HCl) = 75 mL = 0.075 L

Now, we can determine the concentration of the resulting solution:

Concentration of the resulting solution = (moles of NaOH + moles of HCl) / total volume
Concentration = (0.0025 mol + 0.005 mol) / 0.075 L = 0.1 mol/L

Since the resulting solution has a concentration of 0.1 mol/L, it is equivalent to the original concentrations of NaOH and HCl.

To find the pH of the solution, we need to know if it is a strong acid or a strong base. In this case, since the concentrations of NaOH and HCl are the same and they react in a 1:1 ratio, the resulting solution will be neutral.

Therefore, the pH of the resulting solution is 7.