Calculate the ph of the resulting solution if 25.0 ml of 0.250 m hcl (aq) is added to (b) 35.0 ml of 0.300 m naoh (aq).

In this question, the solution have 25.0 ml of 0.250 m hcl (aq). The amount of HCl in mole would be: 0.25 mol/1000ml * 25ml = 0.00625 mol

The solution then added by 35.0 ml of 0.300 m naoh (aq). The NaOH content would be: 0.3mol/1000ml * 35ml = 0.0105 mol

If you add them up, the number of NaOH remain after reaction would be: 0.0105 - 0.00625 = 0.00425mol

The total solution volume would be: 25ml+35ml = 60ml = 0.06liter

The concentration of the OH - ion would be: 0.00425mol / 0.06l = 0.07083 molar

The pH would be: 14 - log 0.07083 molar = 12.86