# 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.

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 $\alpha$.
$$\alpha = \frac{\text{amount of substance of the reactant dissociated}}{\text{amount of substance of the reactant present initially}}$$
• 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 $\text{(amount of substance initially present) - (amount of substance dissociated)}$ while for B and C it is just $\text{amount of substance of A dissociated)}$.

Writing our equation again, \begin{array}{lcccc} & \ce{A &-> &B &+ &C}\\ \text{Initial amount of substance}& a && 0 && 0\\ \text{Final amount of substance}& a - a\cdot(\mathrm{DOD}) && a\cdot(\mathrm{DOD}) && a\cdot(\mathrm{DOD})\\ \end{array}

• @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 unhappy with SX 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 unhappy with SX May 31 '12 at 9:08

x is the extent of the reaction and α is the DoD. if the initial amount of the only reactant is a than
α=x/a and x=aα α is useful to dissociation type reactions, where reactant is only one and the coefficient is 1. such as PCl5-> PCl3 + Cl2 N2O4 -> 2NO2 NH3 -> (1/2)N2 +(3/2)H2

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,