Question

In the laboratory a student combines 47.8 mL of a 0.321 M aluminum nitrate solution with 21.8 mL of a 0.366 M aluminum iodide solution.

What is the final concentration of aluminum cation ?
M

Answer 1

Answer: The final concentration of aluminum cation is 0.335 M.

Step-by-step explanation:

Given:
`V_(1)` = 47.8 mL (1 mL = 0.001 L) = 0.0478 L

`M_(1)` = 0.321 M,
`V_(2)` = 21.8 mL = 0.0218 L,
`M_(2)` = 0.366 M

As concentration of a substance is the moles of solute divided by volume of solution.

Hence, concentration of aluminum cation is calculated as follows.

`[Al^(3+)] = (M_(1)V_(1) + M_(2)V_(2))/(V_(1) + V_(2))`

Substitute the values into above formula as follows.

`[Al^(3+)] = (M_(1)V_(1) + M_(2)V_(2))/(V_(1) + V_(2))\\= (0.321 M * 0.0478 L + 0.366 M * 0.0218 L)/(0.0478 L + 0.0218 L)\\= (0.0153438 + 0.0079788)/(0.0696)\\= 0.335 M`

Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.