When 10.0 mL of 0.012 M Pb(No3)2 is mixed with 10.0 mL of 0.030M KI, a yellow precipitate of PbI2(s) forms.
a. Calculate the molarity of [Pb2+]
b. calculate the initial molarity of [I-]
c. On measuring the equilibrium concentration of [I-] it cam out to be 8.0 x 10^-3 M. Calculate the molarity of [I-] precipitated.
d. Calculate the concentration of [Pb2+] precipitated.
e. calculate the equilibrium concentration of [Pb2+]
f. Calculate the Ksp of PbI2.
i need help with all of it to solve it step by step
To solve this problem step-by-step, let's start by identifying the balanced chemical equation for the reaction between Pb(NO3)2 and KI:
Pb(NO3)2 (aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
a. Calculate the molarity of [Pb2+]:
Since 10.0 mL of 0.012 M Pb(NO3)2 is mixed, the moles of Pb(NO3)2 present can be calculated as follows:
moles of Pb(NO3)2 = volume (L) × concentration (M)
moles of Pb(NO3)2 = (10.0 mL ÷ 1000 mL/L) × 0.012 M
moles of Pb(NO3)2 = 0.00012 mol
Since the coefficient of Pb(NO3)2 is 1 in the balanced chemical equation, the molarity of [Pb2+] is equal to the moles of Pb(NO3)2:
molarity of [Pb2+] = moles of Pb(NO3)2 ÷ volume (L)
molarity of [Pb2+] = 0.00012 mol ÷ (10 mL ÷ 1000 mL/L)
molarity of [Pb2+] = 0.012 M
Therefore, the molarity of [Pb2+] is 0.012 M.
b. Calculate the initial molarity of [I-]:
Since 10.0 mL of 0.030 M KI is mixed, the moles of KI can be calculated as follows:
moles of KI = volume (L) × concentration (M)
moles of KI = (10.0 mL ÷ 1000 mL/L) × 0.030 M
moles of KI = 0.0003 mol
Since the coefficient of KI is 2 in the balanced chemical equation, the initial moles of [I-] is twice the moles of KI:
moles of [I-] = 2 × moles of KI
moles of [I-] = 2 × 0.0003 mol
moles of [I-] = 0.0006 mol
Now, calculate the initial molarity of [I-]:
initial molarity of [I-] = moles of [I-] ÷ volume (L)
initial molarity of [I-] = 0.0006 mol ÷ (10 mL ÷ 1000 mL/L)
initial molarity of [I-] = 0.06 M
Therefore, the initial molarity of [I-] is 0.06 M.
c. Calculate the molarity of [I-] precipitated:
Considering the stoichiometry of the balanced chemical equation, 1 mole of PbI2 is formed for every 2 moles of KI reacted. Since the initial moles of [I-] is 0.0006 mol:
moles of [I-] precipitated = moles of PbI2 formed (1:2 stoichiometry)
moles of [I-] precipitated = 0.0006 mol ÷ 2
moles of [I-] precipitated = 0.0003 mol
Now, calculate the molarity of [I-] precipitated:
molarity of [I-] precipitated = moles of [I-] precipitated ÷ volume (L)
molarity of [I-] precipitated = 0.0003 mol ÷ (10 mL ÷ 1000 mL/L)
molarity of [I-] precipitated = 0.03 M
Therefore, the molarity of [I-] precipitated is 0.03 M.
d. Calculate the concentration of [Pb2+] precipitated:
Again, considering the stoichiometry of the balanced chemical equation, 1 mole of PbI2 is formed for every 1 mole of Pb(NO3)2 reacted. Since the initial moles of Pb(NO3)2 is 0.00012 mol:
moles of Pb2+ precipitated = moles of PbI2 formed (1:1 stoichiometry)
moles of Pb2+ precipitated = 0.00012 mol
Now, calculate the concentration of [Pb2+] precipitated:
concentration of [Pb2+] precipitated = moles of Pb2+ precipitated ÷ volume (L)
concentration of [Pb2+] precipitated = 0.00012 mol ÷ (10 mL ÷ 1000 mL/L)
concentration of [Pb2+] precipitated = 0.012 M
Therefore, the concentration of [Pb2+] precipitated is 0.012 M.
e. Calculate the equilibrium concentration of [Pb2+]:
Since we have the concentration of [Pb2+] precipitated, we need to subtract it from the initial molarity of [Pb2+] calculated in part a:
equilibrium concentration of [Pb2+] = initial molarity of [Pb2+] - concentration of [Pb2+] precipitated
equilibrium concentration of [Pb2+] = 0.012 M - 0.012 M
equilibrium concentration of [Pb2+] = 0 M
Therefore, the equilibrium concentration of [Pb2+] is 0 M.
f. Calculate the Ksp of PbI2:
The solubility product expression for PbI2 is:
Ksp = [Pb2+] × [I-]^2
Substituting the equilibrium concentration of [Pb2+] (0 M) and the initial molarity of [I-] (0.06 M) squared:
Ksp = 0 × (0.06 M)^2
Ksp = 0
Therefore, the Ksp of PbI2 is 0.
To solve these calculations, we will need to employ concepts from stoichiometry, molarity, and solubility. Let's go through each question step by step:
a. To calculate the molarity of [Pb2+], we need to determine the moles of Pb2+ ions present in the solution.
- The volume of the solution is 10.0 mL, which is equivalent to 0.0100 L.
- The concentration of Pb(No3)2 solution is 0.012 M.
- Thus, the moles of Pb2+ ions can be calculated by multiplying the volume of the solution by its concentration:
Moles of Pb2+ = 0.0100 L * 0.012 M = 0.00012 mol
- Finally, we can convert the moles back into molarity by dividing the moles by the volume of the solution in liters:
Molarity of [Pb2+] = Moles of Pb2+ / Volume of solution (in L)
b. Similar to part a, we need to calculate the initial molarity of [I-] in the solution.
- The volume of the solution is 10.0 mL, which is equivalent to 0.0100 L.
- The concentration of KI solution is 0.030 M.
- As KI dissociates into one potassium ion (K+) and one iodide ion (I-), the initial moles of I- can be calculated by multiplying the volume of the solution by its concentration:
Moles of I- = 0.0100 L * 0.030 M = 0.00030 mol
- Finally, we can convert the moles back into molarity by dividing the moles by the volume of the solution in liters:
Molarity of [I-] = Moles of I- / Volume of solution (in L)
c. The equilibrium concentration of [I-] is given as 8.0 x 10^-3 M. To calculate the molarity of [I-] precipitated, we need to subtract this equilibrium concentration from the initial molarity of [I-] calculated in part b.
d. To calculate the concentration of [Pb2+] precipitated, we need to determine the stoichiometric ratio between Pb(No3)2 and PbI2. From the balanced equation, we can see that for every mole of Pb(No3)2, one mole of PbI2 is formed. Therefore, the concentration of [Pb2+] precipitated is equal to the concentration of Pb(No3)2 used, which is given as 0.012 M.
e. To calculate the equilibrium concentration of [Pb2+], we can use the stoichiometric ratio mentioned in part d. As every mole of Pb(No3)2 produces one mole of Pb2+, the equilibrium concentration of [Pb2+] is equal to the original molarity of Pb(No3)2, which is given as 0.012 M.
f. The solubility product constant (Ksp) represents the equilibrium concentration of the ions present in the solid precipitate, PbI2. According to the balanced equation, it can be written as:
Ksp = [Pb2+] * [I-]^2
Substitute the equilibrium concentration of [Pb2+] from part e and the equilibrium concentration of [I-] from part c to calculate Ksp.
By following these steps, you can calculate the different concentrations and the solubility product constant for PbI2.