Question

Practice Problem 1: The following reagents are combined in a test tube: o 2.00 mL of 0.20 M potassium iodide o 1.00 mL of 1% starch o 0.50 mL of 0.20 M ammonium persulfate o 0.50 mL of 0.012 M sodium thiosulfate o 2.00 mL of 0.20 M potassium nitrate o 2.00 mL of 0.20 M ammonium sulfate What is the concentration of iodide in the solution in the test tube

Answer 1

0.05 M is the concentration of iodide in the solution in the test tube.

Step-by-step explanation:

Molarity of KI = 0.20 M

Volume of KI solution = 2.00 mL = 0.002 L ( 1 mL = 0.001 L)

Moles of KI = n

`Molarity=\frac{\text{Moles of solute }}{\text{Volume of solution in L}}`

`0.20 M=(n)/(0.002 L)`

`n=0.20 M* 0.002 mL =0.0004 mol`

Volume of solutions formed by mixing all the solutions in a test tube = V

V = 2.00 mL+1.00 mL+0.50 mL+0.50 mL+2.00 mL+2.00 mL = 8.00 ml

V = 8.00 mL = 0.008 L

Concentration of KI in the test tube = [KI]

`[KI]=(0.0004 mol)/(0.008 L)=0.05 M`

`KI(aq)\rightarrow K^+(aq)+I^-(aq)`

`[I^-]=[KI]=0.05 M`

0.05 M is the concentration of iodide in the solution in the test tube.