Be sure to answer all parts. One way to remove lead ion from water is to add a s

Question

Be sure to answer all parts.

One way to remove lead ion from water is to add a source of iodide ion so that lead iodide will precipitate out of solution:

Pb2+(aq) + 2I(aq) PbI2(s)

(a) What volume of a 1.0 M KI solution must be added to 300.0 mL of a solution that is 0.19 M in Pb2+ ion to precipitate all the lead ion?

L I

(b) What mass of PbI2 should precipitate?

g PbI2 Be sure to answer all parts.

One way to remove lead ion from water is to add a source of iodide ion so that lead iodide will precipitate out of solution:

Pb2+(aq) + 2I(aq) PbI2(s)

(a) What volume of a 1.0 M KI solution must be added to 300.0 mL of a solution that is 0.19 M in Pb2+ ion to precipitate all the lead ion?

L I

(b) What mass of PbI2 should precipitate?

g PbI2

Explanation / Answer

M1V1 = M2V2

M1 = Molarity of KI = 1M

V1 = Volume of KI = ?

M2 = Molarity of Pb2+ = 0.19 M

V2 = Volume of Pb2+ = 300 mL = 0.3 L

M1V1 = M2V2

1 * V1 = 0.19 * 0.3

V1 = 0.057 L I-

Pb(NO3)2 + Kl ----> Pbl2 + KNO3

1st balance the equation
Pb(NO3)2 + 2 Kl ----> Pbl2 + 2 KNO3


According to the balanced equation above, 1 mole of Pb(NO3)2 will completely react with 2 moles of KI to produce 1 mole of PbI2 and 2 moles of KNO3.

2nd
Determine the moles of KI

Molarity = No. of moles / L = 1 mole / 0.057 L = 0.057 moles

3rd
The mole ratio of Pb(NO3)2 to KI = 1 : 2

Set up a proportion:
1 / 2 = Pb(NO3)2 / 0.057

Pb(NO3)2 = ½ * 0.057= 0.0285 moles of lead nitrate

Mass of lead nitrate = No. of moles * molar mass
Molar mass = 207.2 + 2 * (14 + 48) = 331.2 grams

Mass of lead nitrate = 0.0285 * 331.2 = 9.4392 grams

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