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

The final concentration of aluminum cation is 0.335 M.

Explanation:

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

= 0.321 M,       = 21.8 mL = 0.0218 L,       = 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.

Substitute the values into above formula as follows.

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

dynamic equilibrium is the answer.