Avolume of 90.0 ml of a 0.550 m hno3 solution is titrated with 0.890 m koh. calculate the volume of koh required to reach the equivalence point.

55.61 mL

Explanation:

Given;

Volume of HNO₃ solution, V₁ = 90.0 mL

Molarity of the solution, M₁ = 0.550 M

Molarity of KOH, M₂ = 0.890 M

the reaction between the two is given as:

HNO₃ + KOH ⇒ KNO₃ + H₂O

Here,

1 mol of HNO₃ reacts with 1 mol of KOH

thus, we have

M₁V₁ = M₂V₂

here,

V₂ is the volume of KOH

on substituting the values, we get

0.550 × 90.0 = 0.890 × V₂

or

V₂ = 55.61 mL

Hence,

The volume of KOH required to reach the equivalence point = 55.61 mL

the answer is b. dilution