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pH calculation (1 Viewer)

skillstriker

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Calculate the pH of a 0.1 M soln of carbonic acid if only 8% ionises. I keep getting pH = 1.8 but the answer is pH = 2.1 (they didn't double the hydrogen ion concentration). Is the answer wrong or are you not meant to double the hydrogen ion concentration?
 

Parvee

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Degree of Ionisation = [H+]/concentration
Therefore [H+]= Degree of Ionisation x Concentration
= 0.08 x 0.1
=0.008

pH=-log([H+])
=-log(0.008)
=2.1
 

skillstriker

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Carbonic acid (H2CO3) is diprotic though. H2CO3 <--> 2H(+) + CO3(2-). So shouldn't you double the [H+]? i.e. [H+] = 0.08 x 0.1 x 2
 
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madharris

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carbonic acid (h2co3) is diprotic though. h2co3 <--> 2h(+) + co3(2-). So shouldn't you double the [h+]? I.e. [h+] = 0.08 x 0.1 x 2
I always thought it was

H2CO3 <--> H+ + HCO3-


Howcome my capital letters aren't working
EDIT: nevermind! it worked :)
 
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RealiseNothing

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Does that mean that no HCO3- ionisation occurs?
I should have been more clear, my post was at this post:

Carbonic acid (H2CO3) is diprotic though. H2CO3 <--> 2H(+) + CO3(2-). So shouldn't you double the [H+]? i.e. [H+] = 0.08 x 0.1 x 2
I was actually just agreeing with madharris with the "^^^" then went on to say why the OP's equation is wrong lol.
 

RealiseNothing

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Does that mean that no HCO3- ionisation occurs?
I think that the degree of ionisation (%) for HCO3- will be uber low and would occur
To clarify this anyway, the amount of HCO3- in the solution will depend on which reaction the equlibrium favours. If heaps of H2CO3 was suddenly added, then the equilibrium would shift right and the concentration of the HCO3- would be high.

In terms of the question, only 8% ionises, so it's a very low amount of HCO3- concentration in the solution.
 

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