# What is molarity of an HCl sloution if 45.0mL of 0.250 mol/L KOH are needed to neutalize 12.3mL of HCl?

## volumeacid*molarity acid=volumebase*molarity base

MolarityHCl= 45*.250M/12.3

## To find the molarity (M) of the HCl solution, you need to use the concept of stoichiometry and the formula:

M1V1 = M2V2

where M1 and V1 are the molarity and volume of the potassium hydroxide (KOH) solution, and M2 and V2 are the molarity and volume of the hydrochloric acid (HCl) solution.

In this case, you have the following information:

- M1 (Molarity of KOH) = 0.250 mol/L

- V1 (Volume of KOH) = 45.0 mL

- V2 (Volume of HCl) = 12.3 mL

To find M2 (Molarity of HCl), you need to rearrange the formula as follows:

M2 = (M1V1) / V2

Now substitute the given values into the equation:

M2 = (0.250 mol/L * 45.0 mL) / 12.3 mL

Calculate the result:

M2 = 1.147 mol/L

Therefore, the molarity of the HCl solution is 1.147 mol/L.