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I need help with a question thx (1 Viewer)

pencilspanker

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Hey so I've just got a few questions for my understanding would appreciate if anyone could help;

So they give you the Ksp of BaC2O4 to be 2.0 x 10-7. Then what you do is you write out the ionic product for its ions so [Ba2+][C2O42-] = 2.0 x 10-7

First of all, I don't actually understand what the fuck this means, is it saying that there is an equilibrium that exists within the solution between the solid precipitate barium oxalate and its ions???

Also, after that, I know that you are meant to find the concentration of the barium ions but i dont understand what the concentration correlates to, is it the concentration of barium ions left in solution after precipitation?

haha adding to this, because you already 'know' the concentration of C2O42- as they say she adds 100mL of 0.0020 mol L-1 Na2C2O4 what i dont understand is to why the concentration of C2O42- is 0.0020mol/L. Shouldn't the concentration actually be 100ml/200ml x 0.0020 mol/ L because we are actually putting the 100mL of that solution into another 100mL so u have a total of 200mL of stuff. Therefore the concetration is actually less of what was before when looking at the new solution as a whole.

Im sorry for this mess, chemistry is my worst subject and im having trouble with it heapsssss. THANK U FOR ANY HELP I REALLLLY FREAKING APPRECIATE IT.
 
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fingerscrossed2019

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My understanding is somewhat limited but I’ll have a go.
1. Yes, equilibrium in the barium oxalate continually dissolving and forming its precipitate.
2. Concentration: the concentration you are finding is the concentration of barium ions at the given ksp. These are dissolved in solution. The ksp indicates the position of equilibrium for the reaction, and hence indicates the concentration of barium ions in solution at equilibrium (not in precipitate). I don’t understand the reasoning behind the choice of C2O4 as the concentration substituted in line 2 so any help would be appreciated greatly.
3. The concentration of C2O4 is the same, as although you double the volume you also double the amount of C2O4 as it is present in the sodium oxalate solution.

That’s my guesses. Any help re the choice of C2O4 in line 2 from anyone would be awesome.
 

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