# How do I calculate the degree of dissociation in equilibrium?

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

The solution is:

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

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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@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

x is subtracted from reactants , because the reaction is forward, so it will change in to ions, or we can say the concentration of reactants will decrease therefore we write minus x, and the acid you mention here is weak acid so it will not dissociate completely, so we donot know directly how much it will dissociate thus we use a variable x ,to find the value of x we make equation ,eg c-x=x+x then kc =x2/c-x ,then we solve for x , and get the amount which dissociated,

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