What is the concentration of H+ or OH- in excess after the reaction of 28 ml HNO3 , concentration 0.250 + 53 ml KOH 0.320 concentration?

We'l write the chemical reaction
first:

KOH + HNO3 -> KNO3 +
H2O

To calculate the concentration of the excess ions we'll
apply the following formula:

Concentration of excess of
ions = excess moles/total volume of solution

We'll
determine the total volume of solution:

Total volume = 28
mL + 53mL = 81mL = 0.081L

We'll determine the initial moles
of H+;

Initial moles of H+ = volume * concentration of HNO3
= 0.028*0.250

Initial moles of H+ = 0.007
mol

Initial moles of OH- = volume * concentration of KOH =
0.053*0.320

Initial moles of OH- = 0.01696
mol

We'll calculate the excess moles of OH = (Initial moles
of OH-) - (Initial moles of H+)

excess moles of OH =
0.01696 - 0.007 = 0.00996 mol

Concentration of excess of
OH- = excess moles OH-/total volume of
solution

Concentration of excess of OH- = 0.00996
mol/0.081L = 0.123

The concentration of the
excess of OH- ions is of 0.123.