Calculate the volume of 0.6M HNO3 necessary to neutralize 28.55 cm3 of 0.45M KOH.

please show work and detail.

Volumebase*concentrationKOH=Volumeacid*ConcentrationAcid

solve for volume acid. This works because the moles of acid are the same as the moles of base in the balanced neutralization equation.

To calculate the volume of 0.6M HNO3 (nitric acid) necessary to neutralize 28.55 cm3 of 0.45M KOH (potassium hydroxide), we need to use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between HNO3 and KOH.

The balanced chemical equation for the reaction is:

HNO3 + KOH → KNO3 + H2O

From the balanced equation, we can see that the mole ratio between HNO3 and KOH is 1:1. This means that it takes one mole of HNO3 to react with one mole of KOH.

First, let's calculate the number of moles of KOH using its concentration and volume:

Moles of KOH = concentration × volume
= 0.45 M × 28.55 cm3

Now, we have the number of moles of KOH. Since the mole ratio between HNO3 and KOH is 1:1, the number of moles of HNO3 needed is also equal to the number of moles of KOH.

Therefore, the volume of 0.6M HNO3 required to neutralize 28.55 cm3 of 0.45M KOH is given by:

Volume of HNO3 = Moles of HNO3 / Concentration of HNO3
= Moles of KOH / 0.6 M

Let's calculate the moles of KOH using the provided concentration and volume:

Moles of KOH = 0.45 M × 28.55 cm3 / 1000 cm3/L
= 0.0128475 moles

Now, using the moles of KOH, we can calculate the volume of 0.6M HNO3 required:

Volume of HNO3 = 0.0128475 moles / 0.6 M
= 0.0214125 L

To convert the volume from liters to milliliters:

Volume of HNO3 = 0.0214125 L × 1000 mL/L
= 21.4125 mL

Therefore, the volume of 0.6M HNO3 necessary to neutralize 28.55 cm3 of 0.45M KOH is approximately 21.4125 mL.