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Boedecjt01 Boedecjt01
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7 years ago
   I have tried using Molarity= Moles of Solute/Liters of Solution, but i cannot figure the answer out. If you can only get the first two questions, that is okay, thank you!

1) Calculate the amount of potassium ferrocyanide trihydrate [K4Fe(CN)6•3H2O] needed to prepare 250 ml of 0.025 M solution. 

2) Calculate the amount of zinc sulfate heptahydrate [ZnSO4•7H2O] needed to prepare 250 ml of 0.050 M solution. 

3) Assuming that the actual molarities of your stock solutions are equal to the target molarities given above, calculate what titration volume of zinc sulfate would be expected for titrating  a 25.00 ml aliquot of the ferrocyanide solution for each of the five  possible reactions presented below:

   K4Fe(CN)6(aq)  +  ZnSO4(aq) -----> K2ZnFe(CN)6(s)  +  K2SO4(aq)

   K4Fe(CN)6(aq)  +  2 ZnSO4(aq) ----> Zn2Fe(CN)6(s)  +  2 K2SO4(aq)

   2 K4Fe(CN)6(aq)  +  ZnSO4(aq) ----> K6Zn[Fe(CN)6]2(s)  +  K2SO4(aq)

   2 K4Fe(CN)6(aq)  +  3 ZnSO4(aq) ----> K2Zn3[Fe(CN)6]2(s)  +  3 K2SO4(aq)

   3 K4Fe(CN)6(aq)  +  2 ZnSO4(aq) ----> K8Zn2[Fe(CN)6]3(s)  +  2 K2SO4(aq)
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wrote...
Educator
7 years ago
For #1 and #2, see if this helps before I continue
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Boedecjt01 Author
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7 years ago
Where did the 3 come from on the outside of the bracket in problem 1?
wrote...
Educator
7 years ago
I thought it contained it... opps.
wrote...
7 years ago
The MWt of [K4Fe(CN)6•3H2O] (C6H6FeK3N6O3) = 383.3 g

250 mL of 0.025 M K4Fe(CN)6•3H2O contains (250/1000)×0.025 = 6.25×10^-3 mole

Hence amount needed = (6.25×10^-3)×383.3 = 2.40 g (should only be 2 sig figs)
Source  https://biology-forums.com/index.php?action=downloads;sa=view;down=14860
Boedecjt01 Author
wrote...
7 years ago
I have tried using Molarity= Moles of Solute/Liters of Solution, but i cannot figure the answer out. I have the first two, but the third question is confusing.
wrote...
Educator
7 years ago
I'm unsure how to tackle it too Frowning Face
wrote...
5 years ago
Thank you
wrote...
2 years ago
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SMR
wrote...
2 years ago
thank you
wrote...
2 years ago
3. Molarity of ZnSO4. 7H2O = 0.050 M

Volume = 250 mL = 0.250 L

Moles of ZnSO4. 7H2O = Molarity * Volume = 0.050 M * 0.250 L = 0.0125 moles

Molar mass of ZnSO4. 7H2O = 287.5496 g/mol

Mass of  ZnSO4. 7H2O = Moles * Molar mass = 0.0125 moles * 287.5496 g/mol = 3.59 g
wrote...
2 years ago
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