Experiment 110

Objective

To prepare a 250 mL solution of 0.1 M hydrochloric acid (HCl) from a 12 M concentrated stock solution.

Introduction

Hydrochloric acid (HCl) is commonly diluted from a concentrated form to prepare standard solutions. Using the dilution formula V1S1​=V2​S2 ​, we calculate the required volume of 12 M HCl to achieve a 0.1 M solution in 250 mL.

V2 = 250 x 0.1 / 12 = 2.08 mL of 12 M HCI

Adding 2.08 mL of 12 M HCl to distilled water will produce a 0.1 M solution in a final volume of 250 mL.

Materials

Chemicals

●      Concentrated HCl (12 M)

●      Distilled water

Apparatus

●      250 mL volumetric flask

●      Funnel

●      10 mL pipette

Reagents

●      2.08 mL of 12 M HCl in 250 mL distilled water

Procedure

1.    Wash and dry the 250 mL flask thoroughly.

2.    Use a pipette to measure 2.08 mL of 12 M HCl.

3.    Carefully add the HCl to the flask.

4.    Add distilled water to the flask until the total volume reaches 250 mL.

5.    Mix the solution thoroughly to ensure even distribution of HCl.

Observation

After combining 2.08 mL of 12 M HCl with distilled water to reach 250 mL, the resulting solution concentration is 0.1 M.

Questions

1.    What is a primary standard substance?

2.    What is a primary standard solution?

3.    How is a 0.1 M HCl solution defined?

4.    What is the molecular weight of HCl?

5.    What are solute and solvent in this solution?