<u>Answer:</u> The concentration of iodide ion in the resulting solution is 0.225 M
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
.......(1)
Molarity of solution = 0.30 M
Volume of solution = 6.00 mL
Putting values in equation 1, we get:
1 mole of KI produces 1 mole of ions and 1 mole of ions
Moles of iodide ions in KI solution = (1 × 0.0018) = 0.0018 moles
Molarity of solution = 0.45 M
Volume of solution = 2.00 mL
Putting values in equation 1, we get:
1 mole of NaI produces 1 mole of ions and 1 mole of ions
Moles of iodide ions in NaI solution = (1 × 0.0009) = 0.0009 moles
Now, calculating the chloride ions in the solution by using equation 1, we get:
Total moles of iodide ions = [0.0018 + 0.0009] = 0.0027 moles
Total volume of base solution = [6 + 4 + 2] = 12 mL
Putting values in equation 1, we get:
Hence, the concentration of iodide ion in the resulting solution is 0.225 M