Prepare 30mL of a stock KI soln by mixing 10.0mL of the 0.03M KI in 0.2M KNO3 with 20.0mL of the .2M KNO3
Determine the concentration of the iodide ion.
What does it mean by the 0.03M KI in 0.2M KNO3? Is this just bad wording of the question?
Am I just mixing the 10.0mL of .03M KI with 20.0mL .2M KNO3?
In which case my answer would be :
.03 moles KI/1L * 0.01L=0.0003 moles KI
.2 moles KNO3/1L * .02L=.004 moles KNO3
0.0003 moles I-/.03L=.01M I-
Is .01M I- correct?
0.01 M in KI is correct. It isn't bad wording; all of the KNO3 stuff is put in there to confuse you.
In the given question, the description "0.03M KI in 0.2M KNO3" means that the 0.03M KI solution is dissolved in a solvent containing 0.2M KNO3. So yes, you are correct in understanding that you need to mix 10.0mL of the 0.03M KI solution with 20.0mL of the 0.2M KNO3 solution.
Now, let's calculate the concentration of the iodide ion (I-) in the final solution:
First, find the moles of KI:
moles of KI = concentration (M) × volume (L)
moles of KI = 0.03 M × 0.01 L
moles of KI = 0.0003 moles
Next, find the final volume of the solution:
final volume = volume of KI + volume of KNO3
final volume = 10.0 mL + 20.0 mL
final volume = 30.0 mL = 0.03 L
Lastly, calculate the concentration of the iodide ion (I-):
concentration of I- = moles of I- / final volume
concentration of I- = 0.0003 moles / 0.03 L
concentration of I- ≈ 0.01 M
So, based on your calculations, the concentration of the iodide ion (I-) in the final solution is approximately 0.01 M. Therefore, your answer of 0.01M I- is correct.