Given :
A volume of 80.0 mL of a 0.690 M [tex]HNO_3[/tex] 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 [tex]HNO_3[/tex] = moles of KOH
[tex]M_{HNO_3}V_{HNO_3}=M_{KOH}V_{KOH}\\\\0.690\times 80 = 0.790\times V_{KOH}\\\\V_{KOH}=\dfrac{0.690\times 80 }{ 0.790}\ ml\\\\V_{KOH}=69.87\ ml[/tex]
Therefore, volume of KOH required is 69.87 ml.
Hence, this is the required solution.
