Chemistry question. How to determine whether or not a precipitate will form...?

Question

Will a precipitate form if 5mL of.03M NaOh is added to 100mL of 0.125M Fe(No3)2???

So I know that a precipitate will form...because you have to compare Ksp to P but I'm not sure how to solve the problem, so can someone please explain it..?

Here's another problem that I have the solution to but I'm not sure where some of the results came from:

Will a precipitate fom if 15mL of .0100M KOH is added to 475 mL of 0.0075M Pb(NO3)2?

So first you write...

P=[Pb^2+][OH^-]²

[Pb^2+](initial)= (475 x 10^-3) x (.0075)/(490 x 10^-3) =0.0072M

(NOTE*** Where does the 490 x 10^-3 come from?)

(..and why do you have to write 475 x 10^-3)

[OH-](initial) = (15 x 10^-3 x 0.01)/(490 x 10^-3) = 0.0003M

P= (.0072)(.0003)²

P= (6.97 x 10^-10) ≥Ksp

So therefore precipitation occurs....

Thanks!

Answer 1

I assume that you understand scientific notation, metric conversions and how precipitates are formed (when looking at a chemical reaction).

NaOH + Fe(NO3)2  Fe(OH)2 + Na(NO3)

Fe(OH)2 (iron hydroxide) forms.

Ksp = [Fe^2+][OH^-]²

SOLUTION:

Convert milliliters to liters because M (molarity) is moles per a liter (NOT moles per a milliliter). NOTE: We use 10^-3 because there are 1000 milliliters in one liter; and no, you do not have to write the 10^-3 so long as you put the decimal point in the right place.

[1] 5mL = 5mL * 10^-3 = .005L (or 5.0 * 10^-3L).

[2] 100mL = 100mL * 10^-3 = .100L (or 1.0 * 10^-1L).

We’re adding .005L of NaOH to .100L of Fe(NO3)2, so we’ll want to know the total volume of solution:

[3A] Volume of solution = .005L + .100L

[3B] Volume of solution = .105L.

Now we want to know how much Fe^2+ ion is in the solution:

[4A] [Fe^2+] = (volume of Fe(NO3)2) * (Molarity of Fe(NO3)2 / volume of solution)

[4B] [Fe^2+] = (.100L) * (.125M / .105L)

[4C] [Fe^2+] = .1190M.

Next we’ll find how much OH^- ions are in solution the same way we found the Fe^2+ concentration:

[5A] [OH^-]² = (volume of NaOH) * (Molarity of NaOH / volume of solution)

[5B] [OH^-]² = (.005L) * (.030M / .105L)

[5C] [OH^-]² = .0014M.

Now we find the sum of P but DON’T forget that there are 2 OH- for every 1 Fe.

[6A] P = .1190M + (.0014M)²

[6B] P = .1190M

The Ksp for iron hydroxide = 7.9 * 10^-16

[7A] P > Ksp.

[7B] .1190 > 7.9 * 10^-16