Ask AI: prepare 0.25 hcl acid from 32% hcl acid

Preparing a 0.25 Molar (M) hydrochloric acid (HCl) solution requires the use of dilution, which is the process of reducing the concentration of a solute in a solution. Unfortunately, the term "32% HCl" alone doesn't offer a clear enough picture of the original concentration needed to complete this dilution because it doesn't specify whether the percentage is by mass or by volume. Assuming that 32% HCl is by weight and the density of the solution is 1.16 g/mL, we can use the following steps for making a 1L solution of 0.25 M HCl: 1. Calculate the molarity of the 32% HCL: If it is 32% w/w, then a 100 g solution contains 32 g HCl. The molar mass of HCl is about 36.5 g/mol, so this is about 0.88 moles. If the density is 1.16 g/ml, then 100 g is about 86.2 ml, so the molarity is 0.88 moles / 0.0862 L which equals to about 10.2 M. 2. Use the dilution formula to find out the volume of 32% HCL needed: (M1)(V1)=(M2)(V2) where: M1= initial molarity V1= initial volume M2=final molarity V2= final volume. You want to make 1 liter of 0.25 M HCl from a stock of 10.2 M HCl. Filling in these values: (10.2 M)(V1)=(0.25 M)(1 L) Rearranging terms gives: V1 = [(0.25 M) * (1 L)]/10.2 M ≈ 0.0245 L or 24.5 mL 3. Measure and pour 24.5 mL of the 32% HCL into a volumetric flask. 4. Add enough distilled water to the flask to make the total volume 1 L. Remember to pour acid into water and not water into acid! Please keep in mind, all of these steps should be performed while wearing appropriate personal protective equipment, including gloves and safety glasses, due to the hazardous nature of concentrated acids.