Doubt from Ionic Equilibrium

[Viram_2019]

K_{a_1} = 4.3 \times 10^{-7}
K_{a_2} = 4.8 \times 10^{-11} for carbonic acid.

[Viram_2019]

@Azimuddin_Sheikh @pratyaksh_tyagi @Supreeta_Sen @Satyendra_Srivastva @Yash_Srivastava2

Given answer is 63.3 \ mL and I'm getting 68 \ mL by standard way of solving.

[Viram_2019]

@Yash2 @Chirag_Hegde @Siddhant_Mudholkar @Mayank_Chowdhary
Try this.

[Supreeta_2019]

Bro even I'm getting 68
Around 68.55 something
@Viraam_Rao

[Viram_2019]

@Supreeta_Sen
Yup. I think I got the same.
Don't understand where its going wrong.

[Supreeta_2019]

Let's wait for the others then
Me neither, can't understand what's wrong

[Chirag_2019]

What's your approach?

[Viram_2019]

@Chirag_Hegde
Finding the final ratio of bicarbonate to carbonate using the hydrolysis constant.

[Chirag_2019]

Can you upload the working Viraam?
I seem to be missing something
Btw did you account for the dilution due to h2so4?

[Viram_2019]

@Chirag_Hegde
Ohh wait. I didn't account for that. Let me try that then ill post whatever I did so far.

@Chirag_Hegde
Do we have to account for dilution? Im just finding the ratio of moles as the volume is same for both of them.

[Viram_2019]

@Chirag_Hegde
This is what I did initially

First of all we calculate h for the hydrolysis of carbonate. h = 0.2
[OH^-] = \cfrac{K_h [CO_3^{-2}]}{[HCO_3^-]}
Initial pOH = 3.08
Final pOH = 4.08
4.08 = pK_h + \log(\cfrac{Ch + x}{C(1-h) - x})
where x is the number of moles of sulphuric acid that is added.

Don't think there a need to account for dilution here and the final volume in the denominator is same for both concentrations.

[Soumyadeep_2019]

@Viraam_Rao
But bro isnt the last expression poh due to hydrolysis of na2co3 only. wont there be h2co3 formed after reaction which will add to the ph??
Sorry if I am wrong please explain me.

[Viram_2019]

@Soumyadeep
Yeah. Haven't accounted for that.
Can you try that as well.?

[Soumyadeep_2019]

@Viraam_Rao Bro just check what you get after taking that.
sure I will try too.
I think Poh will become 2.08 after that not 4.08

[Viram_2019]

@Soumyadeep
Have you done a similar problem like this bro? I haven't done one where we have to consider both of them so my working is just getting too complicated. Do you have an example problem ?

However, since answers aren't matching with what I initially did, its most likely that we have to consider h2co3 as you said.

@Soumyadeep
Send working pls.

@Soumyadeep
Wait is this somewhat like simultaneous solubility?

[Soumyadeep_2019]

Nah bro i havent done such problems yet. And after i saw this problem i went into deep thoughts of what I have done these years.
Gotta revise Ionic.

[Viram_2019]

@Soumyadeep
Same here bro. I feel like I have done may equilibrium problems and then such problems come my way and im like

How did you get pH = 2.08 though?

[Soumyadeep_2019]

@Viraam_Rao
No bro wrong resoning, now I think it will be 4.08 rather, LOL whats happening with me.

[Viram_2019]

Considering both I get this
K_{h1 }K_{h2} = \cfrac{[OH^-]^2[H_2CO_3]}{[CO_3^{-2}]}
And the respective concentrations are:
[CO_3^{-2}] = C(1-\alpha _1)
[HCO_3^- ]= C\alpha_1(1-\alpha_2)
[H_2CO_3] = C\alpha_1\alpha_2

@Soumyadeep

[Siddhant_2019]

Do you have an example problem ?

I saw similar question like this will find tomorrow