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:
[tex]CH_{3}COOH\rightleftharpoons H^+ + CH_{3}COO^-[/tex]
If you write [tex](0,1-x)[/tex], [tex]{x}[/tex], and [tex]{x}[/tex], respectively, right below acetic acid and hydrogen, and acetate ions, then you can calculate the equilibrium constant (sorry for Latex-based inability):
[tex]K_d=\frac{x^2}{(0,1-x)}[/tex]. Here, you've stated that [tex]K_d[/tex] is [tex]1,8.10^{-5}[/tex], the ratio between the initial concentration and this constant makes ten-thousandfold difference; this can easily be tolerated and the [tex]x[/tex] in [tex](0,1-x)[/tex] may be safely omitted.
So we have this finally:
[tex]1,8.10^{-5}=\frac{x^2}{(0,1)}[/tex], and we find [tex]x[/tex] as [tex]0,0013[/tex], whose pH is
ca. [tex]2,87[/tex].
The pH of a 0.01 M ethanoic acid solution is 3.4. What is the ionisation constant of the acid at this temperature?
About your second question, we'll need to decode the data backwards. [tex]pH{3,4}[/tex] is equal to [tex]10^{-3,4}[/tex], meaning that the conjugate base, acetate, has the same value. The ratio between [tex]\frac{x^2}{0,01-x}[/tex] gives the exact dissociation constant. However, [tex]x^2[/tex] gives [tex]10^{-6,8}[/tex], and the difference between this and the initial concentration reaches as much as
ca. 158.500; so we may omit the [tex]x[/tex] in [tex](0,01-x)[/tex] for the sake of simplicity.
When we omit the it, we obtain [tex]10^{-4,8}[/tex], that is the sufficiently close dissociation constant of acetic acid to the real value.
Regards, chem_tr