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NaOH(s) was added to 1.0 L of HNO3(aq) 0.48 M. 

1,Calculate [H3O+] in the solution after the addition of 0.11 mol of NaOH(s).

2,Calculate [H3O+] in the solution after the addition of 0.89 mol of NaOH(s).

Apr 2nd, 2015

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.

[H3O+]=0.37 M

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

Apr 3rd, 2015

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