Question

My Question is the first picture. Others have tried to answer but I don't follow there use of variables/ algebra. I want to know how my instructor went from :

--> 2 [HA] = [A-] to 66.6% both algebraically and scientifically

Please don't respond to this question if you are unwilling to explain !!!!!! STOP using letters to justify your algebra! I don't understand how the variables are equal or even equilavent. Please continue this question step-by -step without letters.


5.1 = 48 + los e 1083 = 2 = 14 J = 7 - 66.6 %. Ionized A-] ( - 11/27 (HA) 1 2 [HA] - [A-

Do NOT use short hand! this is unaccepable:

NOW, HT HA - Initial cone - c final con coca +A o o ca cx ? Degree of alissociation 9 er av = 2 3) Q = 2 (1-) 3 g ? 2-28 2  

. T-Mobile 19:52 1 64% + Expert Q&A Do Expert Answer kiran kumar R 19 minutes later Let a be the dissociation of the weak aci

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Answer #1

We assume that only 2 forms of acid exist at equilibrium, that is HA & A-.

Thus, there total concentration becomes 100. Hope,this helps in clearing your doubt. feel free to ask for any further queries regarding the solution.

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