Answer: 127.5ml
Step-by-step explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is

are the n-factor, molarity and volume of base which is KOH.
We are given:

Putting values in above equation, we get:

Thus 127.5 ml of 0.5M of HNO3 would be needed to react with 85ml of 0.75M of KOH