$\ce{CH3COOH->CH3COO- + H+}$
\begin{array}{c|c c c} \mathbf{Initial} & \mathrm{1~mol} & \mathrm{0~mol} & \mathrm{0 ~mol} \\\hline \mathbf{Final} & \mathrm{1-\alpha ~mol}&\mathrm{\alpha ~mol}&\mathrm{\alpha~ mol} \end{array} Where $\alpha$ is the degree of dissociation.

Therefore, $i = 1-\alpha+\alpha+\alpha =1+\alpha$, and $\alpha=(i-1)$.
This is what is written in my book. And I understand this.

$\ce{CH3COOH->CH3COO- + H+}$ \begin{array}{c|c c c} \mathbf{Initial} & \mathrm{c~ mol}&\mathrm{0~mol}&\mathrm{0~ mol}\\\hline \mathbf{Final} & \mathrm{c(1-\alpha)~ mol}&\mathrm{c\alpha~ mol}&\mathrm{c\alpha~ mol} \end{array} Where $\alpha$ is the degree of dissociation.

Therefore, $$i = \frac{c\alpha+c\alpha+c(1-\alpha)}{c} =1+\alpha$$ (same as in case 1), and $\alpha=(i-1)$.

$$K_a=\frac{c^2\alpha^2}{c(1-\alpha)} = \frac{c\alpha^2}{(1-\alpha)}$$ This is also written in my book, and I understand this, too.

But, my question is why can we not use the former method to calculate the dissociation constant, $K_a$, which will give me: $$K_a =\frac{ \alpha^2}{(1-\alpha)}$$ as opposed to what I got in the second (and the correct) case, which was: $$K_a = \frac{c\alpha^2}{(1-\alpha)}$$ I can see the difference of a $c$ in the two equations, but can someone explain why do we get the difference?


There is absolutely no difference in your two equations but you forgot the units which makes it seem like there is a difference.

Indeed it is in both cases $$k_a=\frac{c~\alpha^2}{1-\alpha}$$ since what happens if your concentration would be $c=1~\mathrm{mol~L^{-1}}\approx1~\color{\red}{\text{forgotten unit}}$?


How do I know that the c is the concentration?

Using the acid dissociation constant $$k_a=\frac{c(\ce{A-})~c(\ce{H3O+})}{c(\ce{HA})}$$ we have a good basis to obtain what we search. Now by including the degree of dissociation we can define equations for the undissolved acid molecules $c(\ce{HA})$ and the concentration of the conjugate base in solution $c(\ce{A-})$ $$\begin{align}c(\ce{HA})=&~(1-\alpha)~c_0(\ce{HA})\\c(\ce{H3O+})=c(\ce{A-})=&~\alpha~c_0(\ce{HA})\end{align}$$ and combine it with the equation for the acid dissociation constant to obtain $$k_a=\frac{(\alpha~c_0(\ce{HA}))^2}{(1-\alpha)~c_0(\ce{HA})}=\frac{c_0(\ce{HA})~\alpha^2}{1-\alpha}$$ and this is exactly your equation which tells us that the $c$ is indeed a concentration ... namely the concentration of your solution.

  • $\begingroup$ The concentration of the solution can be expressed in many ways viz. molarity, molality, mole fraction etc. So, can c have any of those units (or no unit at all for mole fraction), or necessarily that of molarity? I ask this because many questions in my book have been solved using molality, and not molarity. $\endgroup$
    – agdhruv
    Jun 30 '15 at 21:19
  • 2
    $\begingroup$ You can simply convert them. Molality $b_x=\frac{n_x}{m_L}$ (moles of x per kilogram solvent) and concentration $c_x=\frac{n_x}{V_L}$ (moles per liter) which gives you through $n=c_x~V_L=b_x~m_L$ how to convert from molality to molarity: $c_x=\frac{b_x~m_L}{V_L}$ $\endgroup$ Jun 30 '15 at 21:32

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