When 72.0mL of a 0.110M lead(II) nitrate solution is mixed with 111.5mL of a 0.1

Question

When 72.0mL of a 0.110M lead(II) nitrate solution is mixed with 111.5mL of a 0.195M potassium iodide solution, a yellow-orange precipitate of lead(II) iodide is formed.

Part A

What mass (in grams) of lead(II) iodide is formed, assuming the reaction goes to completion?

Part B

What is the molarity of Pb2+ in the resulting solution?

Part C

What is the molarity of K+ in the resulting solution?

Part D

What is the molarity of NO?3 in the resulting solution?

Part E

What is the molarity of I? in the resulting solution?

Explanation / Answer

The balanced reaction will be -

Pb(NO3)2 + 2KI ---> PbI2 + 2KNO3

Formulae to be used -

1. moles = Molarity x Volume

2. mass (in gms) = moles x molecular weight

Initially, moles of Pb(NO3)2 in solution = 0.110M x 0.72ml = 7.92m.moles (we are calculating in millimoles since volume is given in ml)

And moles of KI = 0.195M x 111.5ml = 21.7425m.moles

According to the balanced reaction -

1 mole Pb(NO3)2 reacts with 2 moles KI to give 1 mole PbI2 and 2 moles KNO3

In shorthand,

1 mole Pb(NO3)2 ~ 2 moles KI ~ 1 mole PbI2 ~ 2 moles KNO3

=> 7.92 m.moles of Pb(NO3)2 will react with 15.84 m.moles(2x7.92) of KI

But KI present is 21.7425 m.moles and hence is in excess. => Pb(NO3)2 is the Limiting Reagant.

Therefore, on completion of reaction, using molar ratios from balanced reaction -

All of Pb(NO3)2 will be used up.

5.9025 m.moles of KI (21.7425 - 15.84) will be left in the solution

7.92 m.moles of PbI2 will be formed

15.84 m.moles of KNO3 will be formed

New volume of the solution after mixing = 72ml + 111.5ml = 183.5ml

PART A) Mass of PbI2 formed = 3.651 gms

mass = moles x Molecular weight

=> mass of PbI2 = 7.92 m.moles x 461.01 gms = 3.651 gms

PART B) Molarity of Pb2+ = 0M

Assuming that all of the lead is converted into PbI2 and precipitated, therefore, no Pb2+ ions will remain in solution.

PART C) Molarity of K+ = 0.118M

K+ comes from

1. KI which is left behind in solution unreacted (5.9025m.moles)

2. KNO3 formed (15.84m.moles)

Since both KI and KNO3 ionize completely, therefore -

m.moles of K+ = m.moles of K+ (KI left unreacted) + m.moles of K+ (KNO3)

=> m.moles of K+ = (5.9025 m.moles + 15.84 m.moles) = 21.7425 m.moles

Molarity = moles/Volume(lts) or m.moles/Volume(ml)

=> Molarity K+ = 21.7425/183.5 = 0.118M

PART D) Molarity NO3- = 0.086M

NO3- comes from KNO3 only (as all Pb(NO3)2 is reacted completely

=> m.moles NO3- = m.moles of KNO3 in solution = 15.84 m.moles

=> Molarity NO3- = m.moles/vol(ml) = 15.84/183.5 = 0.086M

PART E) Molarity of I- = 0.032M

I- comes only from KI left unreacted in solution (since PbI2 is precipitated completely)

=> m.moles of I- = m.moles of KI left unreacted = 5.9025 m.moles

=> Molarity of I- = m.moles/vol(ml) = 5.9025/183.5 = 0.032M

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