# Molarity of Concentrated Hydrochloric Acid (HCl) and Neutralization Calculation

## What is the molarity of concentrated HCl?

## What volume of it would you need to prepare 985 mL of 1.6 M HCl?

## What mass of sodium bicarbonate would be needed to neutralize the spill if a bottle containing 1.75 L of concentrated HCl dropped on a lab floor and broke open?

## Answer:

1) 11.64 mol/L is the molarity of concentrated HCl.

2) 135.40 mL of volume of 11.64 M will need to prepare 985 mL of 1.6 M HCl.

3) 1,711.08 grams of sodium bicarbonate would be needed to neutralize the spill HCl solution.

According to a label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density of 1.18 g/mL. This information allows us to calculate the molarity of the concentrated HCl and determine the volume required for dilution, as well as the mass of sodium bicarbonate needed for neutralization.

1) To determine the molarity of concentrated HCl, we first need to calculate the moles of HCl present in the solution. Given that the solution is 36.0% HCl by mass, we can calculate the moles of HCl present in 100 g of solution, which is approximately 0.9863 mol. With the density of the solution being 1.18 g/mL, the volume of the solution is 84.75 mL or 0.08475 L. By dividing the moles of HCl by the volume of the solution, we find that the molarity of concentrated HCl is 11.64 mol/L.

2) Using the dilution equation (M₁V₁ = M₂V₂), we can calculate the volume of concentrated HCl needed to prepare 985 mL of 1.6 M HCl. By substituting the given values into the equation, we find that 135.40 mL of 11.64 M HCl is required to prepare 985 mL of 1.6 M HCl.

3) The neutralization reaction between sodium bicarbonate and hydrochloric acid can be used to calculate the mass of sodium bicarbonate needed to neutralize the spill of concentrated HCl. With the concentration of HCl solution being 11.64 M and the volume of the HCl solution being 1.75 L, we calculate that 1,711.08 grams of sodium bicarbonate would be needed to neutralize the spill of HCl solution.