The reaction for HNO3 ad KOH is given by,
HNO3 + KOH -------> KNO3 + H2O
Therefore 1 mol of HNO3 will neutralize 1 mol of KOH.
The amount of KOH in 39 mL of 2.0 M KOH is,
= (2/1000) x 39 mol
= 0.078...
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The reaction for HNO3 ad KOH is given by,
HNO3 + KOH -------> KNO3 + H2O
Therefore 1 mol of HNO3 will neutralize 1 mol of KOH.
The amount of KOH in 39 mL of 2.0 M KOH is,
= (2/1000) x 39 mol
= 0.078 mol
Therefore mols of HNO3 needed is = 0.078 mol
Volume of 6.0 M HNO3 needed is, v then,
0.078 = (6/1000) x v
v = 13 mL.
Therefore, the volume of 6.0 M HNO3 needed to neutralize 39mL of 2.0 KOH is 13 mL.