1.76g of sodium hydroxide was added to 50cm^3 of 0.025mol.l^-1 nitric acid. Calculate the pH of the resulting solution.

Answer 1

NaOH+HNO3>>NaNO3 + water
moles acid= .050dm^3*.025mol/dm^3
= 1.25mMole (check that).
Moles NaOH=1.76/40= 44mMole (check that).
moles of NaOH in excess:44-1.25=you do it.

This represents the OH concentration...

OH=(molesOHabove/.05)

H= 1E-14/OHabove

pH= -log(H)