$\ce{CH3COOH->CH3COO- + H+}$
\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?


1 Answer 1


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, 2015 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, 2015 at 21:32

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