Question

A volume of 80.0 mL of a 0.690 M HNO3 solution is titrated with 0.790 M KOH. Calculate the volume of KOH required to reach the equivalence point. Express your answer to three significant figures and include the appropriate units.

Answer 1

Given :

A volume of 80.0 mL of a 0.690 M
`HNO_3` solution is titrated with 0.790 M KOH.

To Find :

The volume of KOH required to reach the equivalence point.

Solution :

We know, at equivalent point :

moles of
`HNO_3` = moles of KOH

`M_(HNO_3)V_(HNO_3)=M_(KOH)V_(KOH)\\\\0.690* 80 = 0.790* V_(KOH)\\\\V_(KOH)=(0.690* 80 )/( 0.790)\ ml\\\\V_(KOH)=69.87\ ml`

Therefore, volume of KOH required is 69.87 ml.

Hence, this is the required solution.