<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>
Step 1: Calculate [H⁺] in the dilute solution
We will use the following expresion.
pH = -log [H⁺]
[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M
Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.
Step 2: Calculate the volume of the concentrated HCl solution
We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.
Considering the reaction stoichiometry and Avogadro's Number, the mass of barium chloride produced is 21.6 grams.
<h2>Balanced reaction</h2>
The balanced reaction is
2 HCl + Ba → BaCl₂ + H₂
<h2>Moles of HCl that react</h2>
Avogadro's Number is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.
Then you can apply the following rule of three: if 6.023×10²³ molecules are contained in 1 mole of HCl, then 1.25×10²³ molecules are contained in how many moles of HCl?
amount of moles of HCl= (1.25×10²³ molecules × 1 mole)÷ 6.023×10²³ atoms
<u><em>amount of moles of HCl= 0.2075 moles</em></u>
Then, 0.2075 moles of HCl react.
<h2>Reaction stoichiometry</h2>
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
HCl: 2 moles
Ba: 1 mole
BaCl₂: 1 mole
H₂: 1 mole
<h2>Mass of barium chloride produced</h2>
Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl produce 1 mole of BaCl₂, 0.2075 moles of HCl will produce how many moles of BaCl₂?
<u><em>amount of moles of BaCl₂= 0.10375 moles</em></u>
Being the molar mass of BaCl₂ 208.24 g/mole, then the mass of barium chloride produced is calculated as:
Finally, the mass of barium chloride produced is 21.6 grams.