Potassium iodide reacts with lead(ii) nitrate in the following precipitation reaction:
2ki(aq)+pb(no3)2(aq)→2kno3(aq)+pbi2 (s)

what minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 155.0 ml of a 0.122 m lead(ii) nitrate solution?

0.1891 L minimum volume of 0.200 M potassium iodide solution is required.

Explanation:

Considering:

Or,

Given :

For lead(II) nitrate solution :

Molarity = 0.122 M

Volume = 155.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 155.0×10⁻³ L

Thus, moles of lead(II) nitrate :

Moles of lead(II) nitrate  = 0.01891 moles

According to the given reaction:

1 mole of lead(II) nitrate reacts with 2 moles of potassium iodide

0.01891 mole of lead(II) nitrate reacts with 2*0.01891 moles of potassium iodide

Moles of potassium iodide = 0.03782 moles

Also,

Molarity = 0.200 M

Volume = ?

0.1891 L minimum volume of 0.200 M potassium iodide solution is required.

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