# Ionisation constant

1. Sep 19, 2004

### josephcollins

Hi peeps, could someone help me out:

Define the term dissociation constant of an aid. The value for ethanoic acid is 1.8*10^-5 moldm-3. Calculate the pH of a 0.1M solution of ethanoic acid.

The pH of a 0.01 M ethanoic acid solution is 3.4. What is the ionisation constant of the acid at this temperature?

2. Sep 19, 2004

### chem_tr

My first LATEX experience, sorry for rookie post

Hello, you are doing great help as I am also preparing for PhD proficiency exam on Chemistry

I will not consult any books about the definition and use mine instead. Dissociation constant for an acid is the ratio of the product of final concentration of hydrogen ions and the conjugate base over the remaining (if the acid is sufficiently weak, practically initial then) concentration of the acid solution after the equilibrium is reached.

Let me solve the problem with this information:

$$CH_{3}COOH\rightleftharpoons H^+ + CH_{3}COO^-$$

If you write $$(0,1-x)$$, $${x}$$, and $${x}$$, respectively, right below acetic acid and hydrogen, and acetate ions, then you can calculate the equilibrium constant (sorry for Latex-based inability):

$$K_d=\frac{x^2}{(0,1-x)}$$. Here, you've stated that $$K_d$$ is $$1,8.10^{-5}$$, the ratio between the initial concentration and this constant makes ten-thousandfold difference; this can easily be tolerated and the $$x$$ in $$(0,1-x)$$ may be safely omitted.

So we have this finally:

$$1,8.10^{-5}=\frac{x^2}{(0,1)}$$, and we find $$x$$ as $$0,0013$$, whose pH is ca. $$2,87$$.

About your second question, we'll need to decode the data backwards. $$pH{3,4}$$ is equal to $$10^{-3,4}$$, meaning that the conjugate base, acetate, has the same value. The ratio between $$\frac{x^2}{0,01-x}$$ gives the exact dissociation constant. However, $$x^2$$ gives $$10^{-6,8}$$, and the difference between this and the initial concentration reaches as much as ca. 158.500; so we may omit the $$x$$ in $$(0,01-x)$$ for the sake of simplicity.

When we omit the it, we obtain $$10^{-4,8}$$, that is the sufficiently close dissociation constant of acetic acid to the real value.

Regards, chem_tr

Last edited: Sep 19, 2004