. Ksp for Fe(IO3)3 is 10-14. Two solutions, one being iron(III) nitrate and the other being sodium iodate, were mixed. At the instant of mixing, [Fe3+] = 10-4M and [IO3-] = 10-5M. What happens?
My answer is that a precipitate forms because Qsp > Ksp.
Would you agree?
Chemistry(Please check) - DrBob222, Sunday, April 8, 2012 at 5:02pm
No. Calculate Qsp again. And don't forget to cube IO3^- this time.
How do you calculate Qsp again? Would it be the concentration of Fe^3+ divided by IO3^-^?
You calculate Qsp the same way you do Ksp; you just didn't cube the IO3^-.
Fe(IO3)3 ==> Fe^3+ + 3IO3^-
Ksp = (Fe^3+)(IO3^-)^3 or
Qsp = (Fe^3+)(IO3^-)^3 = [1*10^-4][1*10^-5]^3 = 1E-19
Well, calculating Qsp is no joking matter! Qsp stands for "reaction quotient" in the context of solubility product calculations. To calculate Qsp, you simply raise the concentrations of the ions in the solution to the power of their respective stoichiometric coefficients.
In this case, Qsp = [Fe3+] * [IO3-]^3. So, Qsp is the product of the concentration of Fe3+ and the concentration of IO3- raised to the power of 3.
To calculate Qsp, you need to multiply the concentrations of the ions raised to the power of their stoichiometric coefficients. In this case, the stoichiometric coefficient for Fe(IO3)3 is 1:1 for both Fe3+ and IO3-.
So, the calculation for Qsp would be:
Qsp = [Fe3+] * ([IO3-])^3
Plugging in the values given:
[Fe3+] = 10^-4 M (given)
[IO3-] = 10^-5 M (given)
Qsp = (10^-4) * (10^-5)^3
= 10^-4 * 10^-15
= 10^-19
Therefore, Qsp is 10^-19.
To calculate Qsp, you need to multiply the concentrations of the ions in the solution, each raised to the power of their stoichiometric coefficient in the balanced equation.
In this case, the balanced equation for the reaction between Fe(IO3)3 and NaIO3 is:
Fe(IO3)3 (aq) + 3NaIO3 (aq) → Fe(IO3)3 (s) + 3Na+ (aq) + 3IO3- (aq)
The stoichiometric coefficients are 1 for Fe(IO3)3 and 3 for IO3-, so you need to cube the IO3- concentration when calculating Qsp.
Therefore, to calculate Qsp, you would multiply the concentration of Fe3+ by the concentration of IO3-, cubed. In this case, it would be:
Qsp = [Fe3+] x [IO3-]^3
Using the given concentrations of [Fe3+] = 10-4M and [IO3-] = 10-5M, the calculation would be:
Qsp = (10-4M) x (10-5M)^3 = 10-4M x 10-15M = 10-19
So, Qsp = 10-19.
Since Qsp is much smaller than the Ksp value of 10-14, which means that Qsp < Ksp, a precipitate will not form in this case.