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lbvjy [14]
3 years ago
11

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% sta

rch 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
Chemistry
1 answer:
STALIN [3.7K]3 years ago
5 0

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.

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