Answer: The final concentration of aluminum cation is 0.335 M.
Explanation:
Given: [tex]V_{1}[/tex] = 47.8 mL (1 mL = 0.001 L) = 0.0478 L
[tex]M_{1}[/tex] = 0.321 M, [tex]V_{2}[/tex] = 21.8 mL = 0.0218 L, [tex]M_{2}[/tex] = 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.
[tex][Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}[/tex]
Substitute the values into above formula as follows.
[tex][Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M[/tex]
Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.
