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)
What volume of a 1.0 M KI solution must be added to 320.0 mL of a solution that is 0.22 M in Pb2+ ion to precipitate all the lead ion?
What mass of PbI2 should precipitate?
Lead(II) iodide is PbI2. If you have 0.070 mols Pb wouldn't that need twice that or 0.140 mols I^-?
grams PbI2 pptd will be mols PbI2 x molar mass PbI2 = ?
.320L * 0.22M = 0.070 moles of Pb
So, you will need 0.70 moles of I
x L * 1.0 M = 0.70 moles
Looks like 700 mL needed
To determine the volume of 1.0 M KI solution needed, we can use the stoichiometry of the balanced equation:
Pb2+(aq) + 2I−(aq) → PbI2(s)
From the equation, we can see that one mole of Pb2+ reacts with two moles of I- to produce one mole of PbI2.
First, let's calculate the number of moles of Pb2+ in the 320.0 mL of the 0.22 M Pb2+ solution:
moles of Pb2+ = volume (L) x concentration (M)
= 0.320 L x 0.22 M
= 0.0704 moles Pb2+
To precipitate all the Pb2+, we need to add an equal number of moles of I- ions. Since the stoichiometry of the balanced equation is 1:2 (Pb2+:I-), we need twice as many moles of I-:
moles of I- = 2 x moles of Pb2+
= 2 x 0.0704 moles
= 0.1408 moles I-
Now, let's calculate the volume of 1.0 M KI solution needed to provide 0.1408 moles of I-:
volume (L) = moles / concentration (M)
= 0.1408 moles / 1.0 M
= 0.1408 L
Since the concentration is given in moles per liter, the volume needed is equal to the number of moles.
Therefore, 0.1408 L (or 140.8 mL) of the 1.0 M KI solution must be added to precipitate all the lead ion.
To determine the mass of PbI2 that will precipitate, we need to use the stoichiometry relationship in the balanced equation:
1 mole of PbI2 = 1 mole of Pb2+ + 2 moles of I-
Since we already calculated that we have 0.0704 moles of Pb2+, we can use this to find the amount of PbI2 formed:
moles of PbI2 = 0.0704 moles Pb2+
To convert moles of PbI2 to mass, we need to know the molar mass of PbI2. The atomic masses are:
Pb: 207.2 g/mol
I: 126.9 g/mol
So, the molar mass of PbI2 is:
207.2 g/mol + (2 x 126.9 g/mol) = 459 g/mol
Now, let's calculate the mass of PbI2 that should precipitate:
mass of PbI2 = moles of PbI2 x molar mass of PbI2
= 0.0704 moles x 459 g/mol
= 32.2736 grams
Therefore, approximately 32.27 grams of PbI2 should precipitate.
To determine the volume of a 1.0 M KI solution needed to precipitate all the lead ions and the mass of PbI2 that will form, we can use stoichiometry and the concept of molarity.
Let's break down the problem step by step:
Step 1: Calculate the amount of Pb2+ ions present in the 320.0 mL of the 0.22 M Pb2+ ion solution.
To calculate the amount of Pb2+ ions, we can use the formula:
Amount (in moles) = Concentration (in M) × Volume (in L)
Given:
Concentration of Pb2+ ion solution = 0.22 M
Volume of Pb2+ ion solution = 320.0 mL = 320.0 mL ÷ 1000 = 0.320 L
Using the formula:
Amount of Pb2+ ions = 0.22 M × 0.320 L
Step 2: Determine the stoichiometric ratio between Pb2+ ions and I- ions.
From the balanced chemical equation:
Pb2+(aq) + 2I−(aq) → PbI2(s)
The ratio between Pb2+ and I- is 1:2. This means that for every Pb2+ ion, we need 2 I- ions.
Step 3: Calculate the number of moles of I- ions required to precipitate all the Pb2+ ions.
Since we have determined the amount of Pb2+ ions in Step 1, we can use the 1:2 stoichiometric ratio to find the moles of I- ions needed.
Moles of I- ions needed = (Amount of Pb2+ ions) × 2
Step 4: Convert the moles of I- ions to volume (in L) using the concentration of the KI solution.
Given:
Concentration of KI solution = 1.0 M
Using the formula:
Volume (in L) = (Moles of I- ions needed) ÷ (Concentration of KI solution)
Step 5: Calculate the mass of PbI2 that will form.
To calculate the mass of PbI2, we need to convert the moles of I- ions to moles of PbI2 using the 1:1 stoichiometric ratio from the balanced chemical equation. Then, we can use the molar mass of PbI2 to calculate the mass.
Given:
Molar mass of PbI2 = (207.2 g/mol) + 2(126.9 g/mol) = 459 g/mol
Mass (in g) = (Moles of I- ions needed) × (Molar mass of PbI2)
By following these steps, you should be able to solve the given problem and determine both the volume of KI solution required and the mass of PbI2 that will precipitate.