The ksp for Ce(IO3)3 is 3.2E-10. What is Ce^3+ concentration in a solution prepared by mixing 60ml of 0.025M Ce^3+ with 40ml of 0.045M IO3^-?
millimols Ce^3+ = mL x M = 60 x 0.025 = 1.5
mmols IO3^- = 40 x 0.045 = 1.8
Qsp = (Ce^3+)(IO3^-)^3 = approx 9E-8 but you can do the more exact number. Note that M = mmols/mL. which means Ksp will be exceeded and a ppt will form. Now we must find the limiting reagent.
Ce^3+ + 3IO3^- ==> Ce(IO3)3
1.5.....1.8
Therefore, IO3^- is the limiting reagent and all of it will be used to form a ppt. The (Ce^3+) will be what is left behind. Can you take it from here.
I understand the solution above but if the limiting reagent is 1.8 that means to get the [Ce^3+]= 1.5-1.8= 0.3mols/ 0.10L wherein 0.1L is the total volume. Is this correct? Thanks.
To find the concentration of Ce^3+ in the solution, we need to consider the equilibrium reaction of Ce(IO3)3 dissociating into Ce^3+ and IO3^- ions.
The balanced equation for this reaction is:
Ce(IO3)3 ⇌ Ce^3+ + 3IO3^-
The equilibrium constant (Ksp) for the reaction is given as 3.2E-10.
Now, let's calculate the moles of Ce^3+ and IO3^- in the given volumes of solutions:
Moles of Ce^3+ = volume (in liters) x concentration (in moles/liter)
= 0.060 L x 0.025 mol/L
= 0.0015 mol
Moles of IO3^- = volume (in liters) x concentration (in moles/liter)
= 0.040 L x 0.045 mol/L
= 0.0018 mol
Since the reaction stoichiometry is 1:1 between Ce(IO3)3 and Ce^3+, and 1:3 between Ce(IO3)3 and IO3^-, the mole ratio of Ce^3+ to Ce(IO3)3 is 1:1. Therefore, the moles of Ce^3+ will be the same as the moles of Ce(IO3)3.
Now, let's assume that 'x' is the concentration of Ce^3+ in the final equilibrium solution. Since the solution is prepared by mixing two solutions, the final volume of the solution will be the sum of the initial volumes, which is 0.06 L + 0.04 L = 0.1 L.
Using the equilibrium expression, Ksp = [Ce^3+][IO3^-]^3, and substituting the values, we get:
3.2E-10 = x * (0.0018)^3
Solving for 'x', we have:
x = (3.2E-10) / (0.0018)^3
x ≈ 4.39E-4 mol/L
Therefore, the concentration of Ce^3+ in the solution is approximately 4.39E-4 mol/L.
To find the concentration of Ce^3+ in the solution, we need to calculate the concentration of the Ce^3+ ions in each solution and then determine the amount of Ce(IO3)3 that will dissolve.
First, let's calculate the initial moles of Ce^3+ in each solution by using the formula:
moles = concentration (in M) x volume (in L)
For the 60 ml of 0.025 M Ce^3+ solution:
moles of Ce^3+ = 0.025 M x (60 ml / 1000 ml/L) = 0.0015 moles of Ce^3+
And for the 40 ml of 0.045 M IO3^- solution:
moles of IO3^- = 0.045 M x (40 ml / 1000 ml/L) = 0.0018 moles of IO3^-
For every mole of Ce^3+ that reacts with IO3^-, one mole of Ce(IO3)3 is formed. So, the limiting reactant is the one with fewer moles, which in this case, is Ce^3+.
Now, let's determine the amount of Ce(IO3)3 that will dissolve. Since Ksp is the solubility product constant, it represents the product of Ce^3+ and IO3^- concentrations in an equilibrium solution.
Ksp = [Ce^3+]^3 x [IO3^-]
We are given that Ksp for Ce(IO3)3 is 3.2 x 10^-10. We can rearrange the equation to solve for [Ce^3+]:
[Ce^3+]^3 = Ksp / [IO3^-]
Now substitute the given values:
[Ce^3+]^3 = (3.2 x 10^-10) / 0.0018
[Ce^3+]^3 = 1.78 x 10^-7
Taking the cubic root of both sides, we get:
[Ce^3+] = (1.78 x 10^-7)^(1/3)
[Ce^3+] ≈ 0.006 M
Therefore, the concentration of Ce^3+ in the final solution is approximately 0.006 M.