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

Answer:

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

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

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

$0.20 M=\frac{n}{0.002 L}$

$n=0.20 M\times 0.002 mL =0.0004 mol$

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]

$[KI]=\frac{0.0004 mol}{0.008 L}=0.05 M$

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

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

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