chem question! Volume and equivalence point

Question

28.6 mL of 0.207 M KOH is titrated with 0.143 M of HNO_3. What volume of HNO_3 (in mL) is needed to reach the equivalence point?

Answer 1

KOH + HNO3 ==> H2O + KNO3

(28.6 ml)(0.207 M) = (x ml)(0.143 M)

x = 41.4 ml

Done another way;

moles KOH present = 0.0286 L x 0.207 mol/L = 0.00592 moles. This is also moles of HNO3 needed

(x L)(0.143 mol/L) = 0.00592 moles HNO3

x = 0.0414 L = 41.4 mls