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denis23
1 month ago
14

You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2 that has a worn label mak

ing it impossible to read. You know the concentration is below 1.0 M as you can make out "0.xxx" at the beginning of the label. In order to determine the concentration, you decide to precipitate out the lead in the solution as PbSO4.
If you added 1.0 mL of the unknown Pb(NO3)2 to a test tube, what is the amount of H2SO4 in mL you will need to add to be sure the H2SO4 is the excess reagent?
Chemistry
1 answer:
castortr0y [3K]1 month ago
5 0

Answer:

Minimum volume of H₂SO₄ to ensure excess = 0.0556 mL

Explanation:

Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃

In this reaction, we consider that the specified maximum concentration of Pb(NO₃) is 0.999M, and to guarantee an excess of the other reactant, we will calculate assuming Pb(NO₃) has a concentration slightly above that, established at 1.0M.

We recognize that Concentration in mol/L = (number of moles)/(volume in L)

Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole

In accordance with the reaction,

1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄

Thus, 0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄

Consequently, the volume of H₂SO₄ needed to ensure excess is 0.001 mole of H₂SO₄

Knowing the concentration of commercial H₂SO₄ is typically 18.0M, we use this value in our calculations.

Volume of H₂SO₄ = (amount of H₂SO₄ required for excess)/(concentration of H₂SO₄)

Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.

QED!!!

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