# Degree of dissociation in equilibrium problem

In my textbook, for calculating the percentage dissociation of $HF$ for the given equation: $$\ce{HF + H2O <-> H3O+ + F-}$$

The solution is:

Initial Concentrations $$[\ce{HF}] = 0.08M, \: \ce{[H3O+]} = 0, \:\ce{[F- ]}= 0$$ Equilibrium concentrations $$[\ce{HF}] = 0.08M - x, \: \ce{[H3O+]} = x, \:\ce{[F- ]}= x$$

I am not able to undestand why $x$ is subtracted from 0.08 and not $cx$ [ where $x$ is the degree of dissociation]

I tried solving the same problem taking $cx$ but not able to get the solution, can anybody explain the difference to me? I have tried asking a similar question earlier too but it is really hard for me to get my head around this concept.

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From where do you get $\ce{HOCl}$? It is not there anywhere in the original equation –  Ashu May 26 '12 at 17:43
1) [no official consensus on this yet], [homework] is tagged according to the spirit of the post. Whether or not it actually is homework is irrelevant. 2) What Ashu said. Your equations have some issue. –  ManishEarth May 26 '12 at 17:50
@ashu sorry i type HOCL by mistake instead of HF , corrected it now . –  Bunny Rabbit May 30 '12 at 15:47
@BunnyRabbit do you need more clarification? –  Ashu May 30 '12 at 16:43
@BunnyRabbit: if you ask this as a follow-up, please link the original question. –  cbeleites May 31 '12 at 8:56

Let us first define the terms needed here.

• Degree of dissociation (DOD)
Degree of dissociation is the fraction of a mole of the reactant that underwent dissociation. It is represented by $\alpha$.

• Number of moles dissociated
It is defined as the product of the initial concentration of the reactant and the degree of dissociation

Now suppose you have a reaction like this
$$\ce{A->B + C}$$

The initial state of A is always the concentration of A (should be given in the question) while initial moles of B and C are zero (if anything else is not specified). The final state of A is always defined as (number of moles initially present) - (Number of moles dissociated) while for B and C it is just ( number of moles of A dissociated)

Writing our equation again,

                 A--------------> B         +     C
Initial moles   a                 0               0
Final moles     a - a*(DOD)       a*(DOD)        a*(DOD)


This should be quite simple . If you need more clarification feel free to ask.

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Tip: use double dollars ($$.....$$) to make the equation get its own line. Tildes(squigglies) are OK, but they're a pain to use. –  ManishEarth May 26 '12 at 19:39
@Manishearth Oh, Thanks –  Ashu May 27 '12 at 4:30
@BunnyRabbit: In addition to this nice answer from Ashu, I still think (as for the original question) that you should think of your units. As the $x$ in your equation has the unit of a concentration it cannot be the degree of dissociation which is dimenionsless. –  cbeleites May 31 '12 at 9:04
@BunnyRabbit: Note that the no of moles here are proportional to the respective concentration: $c = \frac{n}{V}$ and $V$ is constant in your problem. –  cbeleites May 31 '12 at 9:08