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
