It is necessary to add iodide ions to precipitate
the lead(II) ions from 250 mL of 0.076 M
Pb(NO3)2(aq). What minimum iodide ion
concentration is required for the onset of PbI2
precipitation? The solubility product of PbI2
is 1.4 × 10−8
.
Answer in units of mol/L.
What mass of KI must be added for PbI2 to
form?
Answer in units of g.
To determine the minimum iodide ion concentration required for the precipitation of lead(II) ions, we can use the solubility product constant (Ksp) of PbI2.
The balanced chemical equation for the precipitation reaction is:
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
From the equation, we can see that the stoichiometric ratio between lead(II) ions (Pb2+) and iodide ions (I-) is 1:2.
Given:
Volume of Pb(NO3)2 solution = 250 mL = 0.250 L
Molarity of Pb(NO3)2 = 0.076 M
Solubility product constant (Ksp) of PbI2 = 1.4 × 10^-8
Step 1: Calculate the number of moles of Pb(NO3)2:
moles of Pb(NO3)2 = Molarity × Volume
moles of Pb(NO3)2 = 0.076 M × 0.250 L
moles of Pb(NO3)2 = 0.019 mol
Step 2: Determine the minimum number of moles of I- ions required:
Since the stoichiometric ratio between Pb2+ and I- ions is 1:2, the number of moles of I- ions required is twice the number of moles of Pb(NO3)2.
moles of I- ions required = 2 × moles of Pb(NO3)2
moles of I- ions required = 2 × 0.019 mol
moles of I- ions required = 0.038 mol
Step 3: Convert moles of I- ions to concentration (mol/L):
Concentration of I- ions = moles of I- ions / Volume
Concentration of I- ions = 0.038 mol / 0.250 L
Concentration of I- ions = 0.152 mol/L
Therefore, the minimum iodide ion concentration required for the precipitation of PbI2 is 0.152 mol/L.
To calculate the mass of KI required for the formation of PbI2, we need to consider that the molar ratio between KI and PbI2 is 2:1 from the balanced chemical equation.
Given:
Molar mass of KI = 166.0026 g/mol (from periodic table)
Step 4: Calculate the number of moles of KI required:
Since the stoichiometric ratio between KI and PbI2 is 2:1, the number of moles of KI required is half the number of moles of Pb(NO3)2.
moles of KI required = 0.5 × moles of Pb(NO3)2
moles of KI required = 0.5 × 0.019 mol
moles of KI required = 0.0095 mol
Step 5: Convert moles of KI to mass (g):
Mass of KI = moles of KI × molar mass of KI
Mass of KI = 0.0095 mol × 166.0026 g/mol
Mass of KI ≈ 1.573 g
Therefore, approximately 1.573 grams of KI must be added for PbI2 to form.
To determine the minimum iodide ion concentration required for the onset of PbI2 precipitation, we can use the solubility product expression of PbI2.
The solubility product expression for the dissociation of PbI2 is:
PbI2 ⇌ Pb2+ + 2I-
The solubility product constant, Ksp, is given as 1.4 × 10^(-8). Ksp is equal to the concentration of the products (Pb2+ and I-) when the system reaches saturation.
In this case, we are given the initial concentration of lead(II) ions, [Pb(NO3)2], which is 0.076 M. Since lead(II) ions dissociate into one Pb2+ ion, the concentration of Pb2+ is also 0.076 M.
Let's assume the concentration of iodide ions as x M.
Using the solubility product expression, we can write the equation:
Ksp = [Pb2+][I-]^2
Substituting the given values:
1.4 × 10^(-8) = (0.076)(x)^2
Simplifying the equation, we get:
x^2 = (1.4 × 10^(-8))/(0.076)
x^2 = 1.84 × 10^(-7)
Taking the square root of both sides:
x ≈ 4.29 × 10^(-4) M
Therefore, the minimum iodide ion concentration required for the onset of PbI2 precipitation is approximately 4.29 × 10^(-4) mol/L.
Now, to calculate the mass of KI required for PbI2 to form, we need to consider the stoichiometry of the reaction. From the balanced equation:
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
We can see that for every mole of PbI2 formed, we need 2 moles of KI.
The molar mass of KI is approximately 166 g/mol.
Using the formula:
Mass of KI = (Concentration of KI) × (Volume of solution) × (Molar mass of KI)
Substituting the values:
Mass of KI = (4.29 × 10^(-4) mol/L) × (0.250 L) × (166 g/mol)
Mass of KI ≈ 0.018 g
Therefore, to form PbI2, approximately 0.018 g of KI needs to be added.