# How to calculate pH of an CH3COOH solution? [closed]

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:

- 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:

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

• 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? – Mathew Mahindaratne Mar 15 at 19:23
• $\pu{K_a}$ is not $1.74·10^{-3}$ as you state. It is $1.74·10^{-5}$ – Maurice Mar 15 at 20:38

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,

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

• Sorry @Mathew Mahindarante this cannot be the solution. Thus pH of the 8% CH3COOH is 2,9, how I mentioned in question... – HASHTAG Mar 15 at 21:44
• @HASHTAG: You forgot to divide by 2. – Mathew Mahindaratne Mar 16 at 5:51
• @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). – Mathew Mahindaratne Mar 16 at 5:59
• @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. – andselisk Mar 16 at 7:19
• 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. – Poutnik Mar 16 at 11:09