Posted by Riana on Friday, December 14, 2012 at 10:23pm.
FeCl3(aq) ==> Fe^3+(aq) + 3Cl^-(aq)
NaOH(aq) ==> Na^+(aq) + OH^-(aq)
mols FeCl3 = grams/molar mass
mols NaOH = grams/molar mass
(FeCl3) = mols/total L
(NaOH) = mols/total L.
The reaction is
FeCl3(aq) + 3NaOH(aq) ==> 3NaCl(aq) + Fe(OH)3(s)
The remainder of this problem is a limiting reagent, a Ksp, and a common ion effect all rolled into one.
formula is 2FeCl3 + 6 NaOH -> 2Fe(OH)3 + 6NaCl
I wrote the balanced equation; yours is balanced also but you didn't use the lowest set of numbers possible as I did.
it says determine the concentration of four components ions.
I wrote what I did to get you started. You take it from where I stopped.
i don't know how to find the concentration of ion.....just show me only one of them so i can carry on with the rest.
plz can you help me?
FeCl3 + 3NaOH ==> Fe(OH)3 + 3NaCl
0.0285...0.0625
mols FeCl3 = 4.62/162.2 = 0.0285
mols NaOH = 2.50/40 = 0.0625
You need to confirm the molar masses and these values. It's late here and bed time for me.
Fe(OH)3 is a ppt. Which is the limiting reagent?
If we use all of the FeCl3 we could form 0.0285 mol Fe(OH)3.
If we use all of the NaOH, we could form 0.0625/3 = about 0.0.0208.
Obviously both answers can't be right; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent.
Therefore we will form about 0.0208 mols Fe(OH)3 and NaOH is the limiting reagent. All of the OH ion will be used.
How much of the FeCl3 will be used? That is 0.0208 mol FeCl3. How much Fe is left? That is 0.0285-0.0208 = 0.0077
Total volume is 250 mL + 200 mL = 450 mL = 0.450 L.
(Cl^-) = mols/L = (3*0.0285/0.450 L) = ?
(Na^+) = mols/L = (3*0.0625/0.450 L) = ?
(Fe^3+) = (0.0077/0.450)= ?
Have you had anything about the solubility product. If you have, then
(Fe^3+)(OH^-) = Ksp = 4E-38 but you need to confirm that and use the value in your text. This number is listed variously in different texts. Substitute (Fe^3) from above and Ksp and solve for OH-. That will give you the molarity of the OH^-.
Check all of the carefully; it's late.
I got Cl ion concentration 0.342mol/L
The question says write two equation so we will write 2fecl3 -> 2fe+3 + 6cl1-
And 6naoh -> 6na1+ + 6oh1-
Is that the way we have to write ???
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