trebuie calculata conc. molara a noii solutii
moli NaOH din 50ml
c=n/Vs
0,015= n/0,050---> n= 75x10⁻⁵mol
acelasi numar de moli se regasesc in 250ml(50mlsol+200mlapa)
c= 75x10⁻⁵mol/ 0,250l=3x10⁻³molNaOH = 3X10⁻³molOH⁻
pOH= - lg[OH⁻]= 3-lg3
pOH + pH=14---> pH= 14-3+lg3= 11+lg3=.............