Calculate the concentration of I- in an aqueous solution prepared in a 250 ml volumetric flask from 50.0 ml of a 0.840 mol/L aluminum iodide solution and 160.0 ml of a 0.500 mol/L tin(IV) iodide solution.
To calculate the concentration of I- in the final solution, we need to determine the total moles of iodine in the solution and divide it by the final volume of the solution.
Step 1: Calculate the moles of I- from aluminum iodide (AlI3).
The concentration of the aluminum iodide solution is given as 0.840 mol/L. Therefore, the moles of I- in 50.0 ml of the solution can be calculated as follows:
Moles of I- from AlI3 = concentration * volume
= 0.840 mol/L * 0.050 L
= 0.042 mol
Step 2: Calculate the moles of I- from tin(IV) iodide (SnI4).
The concentration of the tin(IV) iodide solution is given as 0.500 mol/L. The moles of I- in 160.0 ml of the solution can be calculated as follows:
Moles of I- from SnI4 = concentration * volume
= 0.500 mol/L * 0.160 L
= 0.08 mol
Step 3: Calculate the total moles of I- in the solution.
Add up the moles of I- from both solutions:
Total moles of I- = Moles of I- from AlI3 + Moles of I- from SnI4
= 0.042 mol + 0.08 mol
= 0.122 mol
Step 4: Calculate the concentration of I- in the final solution.
Concentration of I- = Total moles of I- / Final volume of the solution
The final volume of the solution is given as 250 ml. Convert it to liters:
Final volume = 250 ml = 250 / 1000 L
= 0.250 L
Concentration of I- = 0.122 mol / 0.250 L
= 0.488 mol/L
Therefore, the concentration of I- in the final solution is 0.488 mol/L.