Question

In acid dilution problem, describe how you created 500 ml of a 5.5 M HCl solution from the 11.6 M stock solution. Show your calculations and give details about the equipment used, which solutions you mixed together, and amounts of each.

Answer 1

Answer: 237 ml of stock solution is taken in a 500 ml volumetric flask and diluted with water until the 500 ml mark to obtain 5.5 M diluted solution

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock $HCl$ solution = 11.6 M

$V_1$ = volume of stock $HCl$ solution = ?

$M_2$ = molarity of dilute $HCl$ solution = 5.5 M

$V_2$ = volume of dilute $HCl$ solution = 500 ml

$11.6\times V_1=5.5\times 500$

$V_1=237ml$

Thus 237 ml of stock solution is taken in a 500 ml volumetric flask and diluted with water until the 500 ml mark to obtain 5.5 M diluted solution