Calc Ionisation Const of Ethanoic Acid in 0.1M Soln

[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?

 

[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. $$2,87$$.

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. $$pH{3,4}$$ 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

 

[ravenprp]

The dissociation constant, also known as the ionization constant, of an acid is a measure of its strength. It is the equilibrium constant for the dissociation reaction of an acid in water, where the acid donates a proton (H+) to water to form its conjugate base.

In this case, the dissociation constant of ethanoic acid (also known as acetic acid) is 1.8*10^-5 moldm-3. This means that at equilibrium, only a small fraction of the acid molecules have dissociated into ions in the solution.

To calculate the pH of a 0.1M solution of ethanoic acid, we can use the dissociation constant and the equation for the ionization of a weak acid: Ka = [H+][A-]/[HA], where Ka is the dissociation constant, [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

Rearranging the equation, we get [H+] = √(Ka*[HA]). Plugging in the values, we get [H+] = √(1.8*10^-5 * 0.1) = 1.34*10^-3 moldm-3. Taking the negative logarithm of this value, we get the pH of the solution to be 2.87.

Now, to find the ionization constant at a pH of 3.4, we can use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Rearranging the equation, we get pKa = pH - log([A-]/[HA]). Plugging in the values, we get pKa = 3.4 - log(0.01/0.1) = 3.4 + 1 = 4.4.

Therefore, the ionization constant at a pH of 3.4 is 10^-4.4 = 3.98*10^-5 moldm-3. This value is close to the given dissociation constant of 1.8*10^-5 moldm-3, indicating that the acid is mostly dissociated at this pH.

I hope this helps! Let me know if you have any other questions.