NaOH(s) was added to 1.0 L of HNO3(aq) 0.48 M.
2,Calculate [H3O+] in the solution after the addition of 0.89 mol of NaOH(s).
1. Moles of HNO3 present= 0.48 M* 1 L= 0.48 moles
After addition of 0.11 moles of NaOH equal number of moles of the acid has reacted. So
Moles of HNO3 remaining = 0.48-0.11= 0.37 moles
Total volume. Of solution remains 1 L because solid NaOH was added.
pH = -log (0.37)= 0.432
2. After addition of 0.89 moles of the base all the acid has reacted.
Moles of NaOH remaining = 0.89-0.48 = 0.41
[OH-]= 0.41moles/1L = 0.41 M
pOH = -log (0.41)=0.387
pH= 14-0.387= 13.613
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