Suppose 50.00 mL of 2.0 E -5 M Fe(NO3)3 is added to 50.00ml of 2.0 E -4M KIO3. Which of the following statements is true?

For Fe(IO3)3, Ksp= 1.0 E-14

a-precipate forms because Qc<Ksp
b-No precipate forms because Qc>Ksp
c-Nothing happended
d-No precipate forms because Qc<Ksp

I chose answer a...

A ppt can't form if Ksp is not exceeded. a can't be right if Qc<Ksp since Ksp is not exceeded.

To determine which statement is true, we need to compare the reaction quotient (Qc) with the solubility product constant (Ksp).

First, let's write the balanced chemical equation for the reaction between Fe(NO3)3 and KIO3:

2 Fe(NO3)3 + 3 KIO3 → Fe(IO3)3 + 3 KNO3

From the balanced equation, we can see that the reaction produces Fe(IO3)3 as a precipitate.

Now, let's calculate the initial concentrations of the reactants:

For Fe(NO3)3:
Initial concentration = 2.0 E -5 M

For KIO3:
Initial concentration = 2.0 E -4 M

Next, we need to calculate the reaction quotient (Qc), which is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients.

Since we have equal volumes for both reactants, the final concentrations of products and reactants will be the same.

For Fe(IO3)3, the concentration is x M (unknown because it's the amount that will precipitate).

For Fe(NO3)3:
Concentration = initial concentration = 2.0 E -5 M

For KIO3:
Concentration = initial concentration = 2.0 E -4 M

Using the balanced equation, we can see that the stoichiometric coefficients are 2 for Fe(NO3)3 and Fe(IO3)3. Therefore:

Qc = [Fe(IO3)3]^2 / [Fe(NO3)3]^2

Now let's substitute the concentrations into the equation:

Qc = (x)^2 / (2.0 E -5)^2

To determine the value of x, we need to consider the solubility product constant (Ksp). In this case, Ksp = 1.0 E-14.

If Qc < Ksp, it means that the reaction has not yet reached equilibrium, and a precipitate will form. If Qc > Ksp, it means that the reaction has surpassed equilibrium and no precipitate will form.

So, to determine the right answer, we need to compare Qc with Ksp.

Comparing Qc with Ksp:

If Qc < Ksp, then a precipitate will form. In this case, answer choice (a) is correct.

Therefore, you have chosen the correct answer, answer choice (a).