Question

Question 72What is the final concentration of HCI in a solution prepared by addition of 930.0 mL of 8.77 M HCI to 468.0 mL of 3.22 M HCI?Assume volumes are additive.O 6.00 MO 0.00858 MO 12.0 MO 6.91 MO 5.08 MB

Answer 1

6.91 M

Step-by-step explanation

Given:

Concentration of HCl in solution 1 = 8.77 M

Volume of HCl solution 1 = 930.0 mL = 0.930 L

Concentration of HCl in solution 2 = 3.22 M

Volume of HCl solution 1 = 468.0 mL = 0.468 L

What to find:

The final concentration of HCI in a solution prepared by addition of 930.0 mL of 8.77 M HCI to 468.0 mL of 3.22 M HCI.

Step-by-step solution:

Step 1: Determine the number of moles of HCl in solutions 1 and 2.

Moles of HCl = Molarity x Volume in L

Moles of HCl in solution 1 = 8.77 M x 0.930 L = 8.1561 mol

Moles of HCl in solution 2 = 3.22 M x 0.468 L = 1.50696 mol

Step 2: Determine the total number of moles of HCl and volume in the final solution.

Total moles of HCl in final solution = Moles of HCl in solution 1 + Moles of HCl in solution 2

Total moles of HCl in final solution = 8.1561 mol + 1.50696 mol = 9.66306 mol

Total Volume in the final solution = Volume of solution 1 + Volume of solution 2

Total Volume in the final solution = 0.930 L + 0.468 L = 1.398 L

Step 3: Calculated the final concentration of the solution.

`Final\text{ }concentration=\frac{Total\text{ }moles\text{ }of\text{ }HCl}{Total\text{ }volume}=\frac{9.66306\text{ }mol}{1.398\text{ }L}=6.91\text{ }M`

Hence, the final concentration of HCI in a solution is 6.91 M