-4
$\begingroup$

please, does anyone know, how to properly (with calculation procedure) calcutate $\mathrm{pH}$ of an $\ce{CH3COOH}$, when you know only:


-$\ce{CH3COOH}$ is 8% (water solution)

-density of $\ce{CH3COOH}$ - $\pu{1.01 g cm-3}$

-$K_\mathrm{a}$ - $1.74 \times 10^{-3}$


I know, how to calculate pH and I tried to do it myself. But...

I have done this:

enter image description here - V is multiplied by 0,08 for 8% solution.

and then calculated pH from concentration.

pH = - log [H3O+]

But still, I am getting wrong numbers...


EDIT:

I found this:

enter image description here

and calculated $\mathrm{pH}$ 2,71, but this is still way too far from 2,9...

$\endgroup$
2
  • 1
    $\begingroup$ It wasn't clear that what percentage of the solution. It could be $w/w$ or $w/v$ or etc. The density of the solution is given for a reason. Did you use it? $\endgroup$ – Mathew Mahindaratne Mar 15 at 19:23
  • 1
    $\begingroup$ $\pu{K_a}$ is not $1.74·10^{-3}$ as you state. It is $1.74·10^{-5}$ $\endgroup$ – Maurice Mar 15 at 20:38
2
$\begingroup$

I have asked OP to verify the solution concentration but didn't get the answer. Thus, I assume it is $8\% \ (w/w)$. Thus, if you assume $[\ce{HA}] = c$ then:

$$ c = 8\% \ (w/w) = \frac{\pu{8 g}\text{ of HA}}{\pu{100 g}\text{ of sol}} \times \frac{\pu{1.0 mol}\text{ of HA}}{\pu{60.05 g}\text{ of HA}} \times \frac{\pu{1.01 g}\text{ of sol}}{\pu{1.0 mL}\text{ of sol}} \times \frac{\pu{10^3 mL}\text{ of sol}}{\pu{1.0 L}\text{ of sol}}\\ = \pu{1.35 mol L-1}$$

Acetic acid ionization according to:

$$\ce{HA + H2O <=> H3O+ + A-}$$

If $\alpha$ amount of $[\ce{HA}]$ is ionized at equilibrium, concentrations at equilibrium would be $[\ce{HA}] = c - \alpha$, and $[\ce{A-}] = [\ce{H3O+}] = \alpha$. Thus:

$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]} = \frac{\alpha^2}{c- \alpha} \tag1$$

Since $c = \pu{1.35 mol L-1}$ and $K_\mathrm{a} = 1.77 \times 10^{-5}$, $c - \alpha \approx c$, the equation $(1)$ can be simplified here to get $\mathrm{pH}$. Take $\log$ on each side of the equation:

$$\alpha^2 = K_\mathrm{a} \times c \ \Rightarrow \ 2 \log \alpha = \log K_\mathrm{a} + \log c$$

Since $\alpha = [\ce{H3O+}]$:

$$2\times \mathrm{pH} = \mathrm{p}K_\mathrm{a} - \log c = 4.75 - 0.13$$

Thus, $\mathrm{pH} = 2.31$.


Late addition: OP insists that his given answer for $\mathrm{pH}$ of the solution is $2.9$. Since I suspected OP's given data for the problem, I assumed the concentration of the acetic acid solution must be $0.8\% \ (w/w)$ instead of $8\% \ (w/w)$. When calculating with that value, you get $c = \pu{0.135 mol L-1}$. Hence,

$$\mathrm{pH} = \frac{1}{2}\left(\mathrm{p}K_\mathrm{a} - \log c \right) = \frac{1}{2}\left(4.75 + 0.87\right) = 2.81$$

$\endgroup$
6
  • $\begingroup$ Sorry @Mathew Mahindarante this cannot be the solution. Thus pH of the 8% CH3COOH is 2,9, how I mentioned in question... $\endgroup$ – HASHTAG Mar 15 at 21:44
  • 3
    $\begingroup$ @HASHTAG: You forgot to divide by 2. $\endgroup$ – Mathew Mahindaratne Mar 16 at 5:51
  • 1
    $\begingroup$ @HASHTAG: Also, you forgot to correct your data. For example, percentage of HAc. Its $K_\mathrm{a}$ is not correct either. Hence, I assume your suggested answer might not be correct. My answer is based on $8\% \ (w/w)$ concentrated HAc solution with known $\mathrm{p}K_\mathrm{a}$ of 4.75 (I just assume the given density is correct). $\endgroup$ – Mathew Mahindaratne Mar 16 at 5:59
  • 1
    $\begingroup$ @MathewMahindaratne Unit symbols should never be appended with "of something". If you want to show what physical quantity does a particular value refer to, use an equation in symbolic form, and then plug in the numbers, say, next line for the ease of comparison. $\endgroup$ – andselisk Mar 16 at 7:19
  • 4
    $\begingroup$ IMHO, the task is errorneous and the input data do not match the expected result, if calculation is done correctly. What you, @HASHTAG , ask for is what incorrect calculation fits the incorrect, but expected result. $\endgroup$ – Poutnik Mar 16 at 11:09

Not the answer you're looking for? Browse other questions tagged or ask your own question.