ionic
How much solid Na2S2O3 should be added to 1.0 L of water so that 0.0005 mol Cd (OH2) could just barely dissolve?K1 and K2 for S2O3-- complexation with Cd++ are 8.3 × 10^3 and 2.5 × 10^2, respectively. Ksp (Cd(OH)2 = 4.5 × 10–15
ques in case photo is not clear
34 Replies
@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.woahhhhh very cool question i have no idea how to do this 😄
equilibrium??
Can you post a screenshot of the original q please?
ah phone pe dikha nahi lol
got it
this one is kinda direct
K1 and K2 are significantly greater than Ksp
so assume that they go to completion
Does this help?
This is a NSEC+ level question.
I'll explain it only if you're preparing for chem olympiads.
I might have just an idea how to do this.
huh
this seems direct to me :O
lolwelcome shaurya
not rly
yeah sure
can you show me your working so far?
Cd x Cd x2-
those are the two complexes
x = (s2o3)2-
completely the reactions go forward dirn me. so any amount of thiosulphate ions will directly dissolve the Cd2+ in solution.
i tried writing a few equilibria but it wasnt really of much use
can you share that here?
you are on the right track
just write three equations
gotta sum up two of them to get Cd (X)2 2- conc = Cd2+ conc in solution.
cos the reactions go to completion
BUT X2- conc can be figured from there, just take root :)
sorry its a bit messy
s is solubility of salt
a is initial moles of s2o3 2-
x moles break from cd2+ and further y from cds2o3
i do have a doubt
its mentioned 0.005 mol just dissolve
does that imply solubility of salt is 0.005mol/L
so this is where the basics help yea, 0.005 mol/L is the amount of Cd2+ ions since the reaction goes to completion. Add up both of those reactions above.
What you have done would work if K1 K2 were kinda small. Since they are both large, this approxmation works
So K1 K2 = Cd (X2) 2-/(X^2)Cd2+ ... (i)
Cd2+ btw without all this would be 1 x 10^-5, that part is fine?
So you are increasing the solubility by 10 times technically
[Cd (X2) 2-] = 0.0005
So new [OH-] = 2s = 0.001, from Ksp, you can get [Cd2+] = 4.5 x10^(-9) ... (ii)
[X] = 0.2 (substituting (ii) in (i)
Note that I haven't gone into the x y etc.
Do that when necessary, but FIRST write this down, helps you see things for what they are. Sometimes variables can mess things up in your head!
@hardcoreisdead
So this is it
I plugged this in desmos and this gives a=0.2324
Molar mass of na2s2o3 is 158.11
This gives 36.74 grams as the ans
Ans is 31.7 tho
Is this because of approximations??
@iTeachChem
Yes
Did you follow this logic?
not really , i just made my own flow
exam mein aise kaise chalega 😦
its impossible to solve this without calci
i assume this is above mains level and adv wouldnt give such calc so its olympiad specific only??
Bro I just solved
Without calci XD
Try toh kar
No calci required
oh ill try your method then
sorry but i dont understand where am i going wrong
like where does my method and approximations go wrong
shouldnt that be ksp = 4s^3
s being solubility ig i kinda get it [oh-] = 10^-3 using this i get [cd2+] 4.5 * 10^-9 (same as yours ) now since almost all cd2+ is used so cds2o3 will be same as cd2+ initially now all of that cds2o3 is used up to make cd(x2) ( x being thiosulphate ion) [cd(x2) = 0.0005 or 5*10^-4]
s being solubility ig i kinda get it [oh-] = 10^-3 using this i get [cd2+] 4.5 * 10^-9 (same as yours ) now since almost all cd2+ is used so cds2o3 will be same as cd2+ initially now all of that cds2o3 is used up to make cd(x2) ( x being thiosulphate ion) [cd(x2) = 0.0005 or 5*10^-4]
1. is [x] and 2. molar mass of na2s2o3 * moles
so basically the same ?
@iTeachChem
Yo travelling.
Check my earlier messages once. Have explained everything :)
If you have followed the process don’t worry about the end result.
ohh , no problem
check the final ans whenever u get some time
i believe i have followed it as intended yet the ans is slightly off
Haan that's okay. Don't worry about that.
If you are confident process samajh me aa gaya then move on :)
aight, thank you so much
leaving this thread open so u can confirm ans later on
Nope please close :p
You know my agenda
I don’t encourage test taking ke liye mugging
Process over everything else yo
i am not mugging tho
ans confirm ho jata toh ill know my process is 100% correct
yes 0.231 it is!
@hardcoreisdead we good ?
Thank you so much sir
Yep
+solved @iTeachChem
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