1. Calculate the concentration of IO−3 in a 5.52 mM Pb(NO3)2 solution saturated with Pb(IO3)2 . The 𝐾sp of Pb(IO3)2 is 2.5×10−13 . Assume that Pb(IO3)2 is a negligible source of Pb2+ compared to Pb(NO3)2 .
2. A different solution contains dissolved NaIO3 . What is the concentration of NaIO3 if adding excess Pb(IO3)2(s) produces a Pb2+ concentration of 5.40×10−6 M ?
................Pb(IO3)2 ==> Pb^2+ + 2IO3^-
I.....................solid...........0..............0
C....................solid...........x..............2x
E.....................solid...........x..............2x
Ksp = 2.5E-13 = (Pb^2+)(IO3^-)^2
The problem tells you that (Pb^2+) is 5,52E-3 M, you want to calculate (IO3^-) in this solution AND THAT we can ignore the 2x contribution from the solubility of Pb(IO3)2. So
Ksp = 2.5E-13 = (Pb^2+)(IO3^-)^2
2.5E-13 = (5.52E-3)(IO3^-)^2
Solve for (IO3^-) = ?
2. A separate post.
1. To calculate the concentration of IO^3- in the saturated solution, we need to determine the solubility of Pb(IO3)2 using the given Ksp value.
The solubility product expression for Pb(IO3)2 is:
Ksp = [Pb2+][IO3-]^2
Since we assume Pb(IO3)2 is a negligible source of Pb2+ compared to Pb(NO3)2, the concentration of Pb2+ from Pb(IO3)2 can be considered as zero. Thus, the solubility product expression can be simplified as:
Ksp = [IO3-]^2
Given that Ksp = 2.5 × 10^-13, we can solve for [IO3-]:
2.5 × 10^-13 = [IO3-]^2
Taking the square root of both sides, we get:
[IO3-] = sqrt(2.5 × 10^-13)
[IO3-] ≈ 5.0 × 10^-7 M
Therefore, the concentration of IO3- in the saturated solution is approximately 5.0 × 10^-7 M.
2. When excess Pb(IO3)2(s) is added to the solution containing NaIO3, a reaction occurs to form Pb2+ ions.
The balanced chemical equation for the reaction is:
Pb(IO3)2(aq) + 2 NaIO3(aq) → Pb(IO3)2(s) + 2 Na+(aq) + 2 IO3-(aq)
According to the stoichiometry of the balanced equation, the concentration of Pb2+ ions formed is equal to the concentration of IO3- ions in solution. Therefore, the concentration of NaIO3 can be calculated based on the concentration of Pb2+ ions.
Given that the concentration of Pb2+ is 5.40 × 10^-6 M, we can assume this concentration is also the concentration of IO3-. Hence, the concentration of NaIO3 is also 5.40 × 10^-6 M.
Therefore, the concentration of NaIO3 in the solution is 5.40 × 10^-6 M.
To calculate the concentration of IO−3 in a solution and the concentration of NaIO3 in a different solution, you need to understand the concept of solubility product constant (Ksp) and how it relates to the concentrations of ions in solution.
1. Calculate the concentration of IO−3 in a Pb(NO3)2 solution saturated with Pb(IO3)2:
First, write the balanced equation for the dissolution of Pb(IO3)2:
Pb(IO3)2(s) ⇌ Pb2+(aq) + 2IO−3(aq)
The solubility product constant expression for Pb(IO3)2 is:
Ksp = [Pb2+][IO−3]^2
Assuming Pb(IO3)2 is a negligible source of Pb2+ compared to Pb(NO3)2, we can assume that the concentration of Pb2+ is equal to the concentration of Pb(NO3)2 in the solution.
Let's assume the concentration of Pb(NO3)2 is 5.52 mM. This means the concentration of Pb2+ is also 5.52 mM.
Now, let's substitute the known values into the solubility product constant expression:
2.5×10−13 = (5.52 mM)([IO−3]^2)
Rearranging the equation and solving for [IO−3]:
[IO−3]^2 = (2.5×10−13) / (5.52 mM)
[IO−3]^2 ≈ 4.53 × 10−12
Taking the square root of both sides to find [IO−3]:
[IO−3] ≈ 6.73 × 10−6 M
Therefore, the concentration of IO−3 in the Pb(NO3)2 solution saturated with Pb(IO3)2 is approximately 6.73 × 10−6 M.
2. Calculate the concentration of NaIO3 in a solution:
Since adding excess Pb(IO3)2(s) produces a Pb2+ concentration of 5.40 × 10−6 M, we can assume that all the Pb(IO3)2 has reacted to form Pb2+ ions.
The balanced equation for the reaction of Pb(IO3)2 and NaIO3 is:
Pb(IO3)2(s) + 2NaIO3(aq) → 2Na+ + Pb2+ + 2IO−3
From the equation, we can see that for every 1 mol of Pb(IO3)2 that reacts, we get 2 mol of IO−3. Since we know the concentration of Pb2+ is 5.40 × 10−6 M, and assuming all the Pb(IO3)2 has reacted:
[IO−3] = 2 × (Pb2+ concentration)
[IO−3] ≈ 2 × (5.40 × 10−6 M)
[IO−3] ≈ 1.08 × 10−5 M
Therefore, the concentration of NaIO3 in the solution is approximately 1.08 × 10−5 M.
................Pb(IO3)2 ==> Pb^2+ + 2IO3^-
I.....................solid...........0..............0
C....................solid...........x..............2x
E.....................solid...........x..............2x
The problem tells you that you have a solution of NaIO3 (remember it's completely ionized; i.e., 100%) so the amount of Pb(IO3)2 SOLID that dissolves will have (Pb^2+) = 5.4E-6 M. So if we ignore any extra IO3^- that might come from Pb(IO3)2 (whatever amount it will be small), then
Ksp = 2.5E-13 = (Pb^2+)(IO3^-)^2
Plug in the (Pb^2+) and solve for (IO3^-).
Post your work if you get stuck.