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Since we are looking at an acid-base reaction, the first thing to do is to write a balanced chemical equation.
KOH + HNO3 --> KNO3 + H2O
Once we have the balanced equation, we can see that we have a 1:1 ratio between the base and the acid. Therefore, at the equivalence point, the moles of KOH will be equal to the moles of HNO3. Knowing that we can set this up like we set up stoichiometry problems.
55 mL HNO3 (1 L / 1000 mL) (0.25 mol HNO3 / L)(1 mol KOH / 1 mol HNO3)
= 0.0138 mol KOH
Solving this will give us the mol of KOH needed at the equivalence point, which is 0.0138 mol KOH. In order to find the volume of KOH, we would need to know the molarity of KOH which was not given in the question.
If we assume a concentration of 0.100 M KOH, we can show how to set it up
0.014 mol (1 L / 0.10 M KOH) (1000 mL / 1 L) = 138 mL KOH
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