propanoic acid ch3ch2cooh is a weak acid which has a value for dissociation content of ka = 6.3 x 10^-6 calculate the pH of an aqueous solution of propanoic acid containing 37 g of propanoic acid per litre.
i'm stumbled on this one.. i calculated that the concentration of the propanoic acid is .635 mol/L but i don't know what to do after.. do i just do -log(.635)?? thanks for any help!
someone? anyone? help!
Not quite. Let's call propionic acid (Propanoic acid) HPr to avoid all that typing.
HPr ==> H^+ + Pr^-
K = (H^+)(Pr^-)/(HPr)
Now, we have 37 g HPr/L which is 37/74 = 0.50 mole/L = 0.50 M.
Set up an ICE chart.
(H^+) = x
(Pr^-) = x
(HPr) = 0.5-x
Substitute into Ka expression and solv for (H^+). My answer is 0.00177 M. Then pH = -log(1.77 x 10^-3) and round to 3 s.f. for about 2.75 but you need to confirm all of that.
thank you very much. i did 47/74 instead of 37/74. I knew the steps to answer the question but the confusing part was forming the balanced equation because i didn't know how to do it and the question didn't offer one. May I ask how you got that? how do you know that H+ when balanced has the same amount of moles as ch3ch2cooh?
First, you need to remember that ch2ch2cooh doesn't mean a thing to a chemist.
CH3CH2COOH is the formula for propanoic acid. It is the last H on the right hand side of the formula that is the acidic hydrogen. All of the others are glued onto their respective C atoms and they aren't going anywhere. In organic chemistry, the -COOH group is acidic because the H can ionize. Most of the COOH compounds are relatively weak acids. So we write RCOOH to show any organic acid. I just used the right hand H as H and stuck Pr on for the R(the rest of the molecule) because it is so much easier to write HPr than
CH3CH2COOH ==> CH3CH2COO^- + H^+.
Thank you sooo much!
To calculate the pH of an aqueous solution of propanoic acid, you need to consider that it is a weak acid that partially dissociates in water. Propanoic acid dissociates into propanoate ions (CH3CH2COO-) and hydrogen ions (H+):
CH3CH2COOH ⇌ CH3CH2COO- + H+
The dissociation constant (Ka) represents the equilibrium expression for this dissociation:
Ka = [CH3CH2COO-][H+]/[CH3CH2COOH]
Given that the value for Ka is 6.3 x 10^-6 and you have calculated the concentration of the propanoic acid to be 0.635 mol/L, you can proceed to solve for the concentration of the hydrogen ions (H+).
Since the initial concentration of CH3CH2COO- is negligible compared to the initial concentration of the propanoic acid, you can assume that [CH3CH2COO-] ≈ 0. Therefore, you can simplify the equation to:
Ka = [H+]/[CH3CH2COOH]
Now, you can rearrange the equation to solve for [H+]:
[H+] = Ka × [CH3CH2COOH]
[H+] = (6.3 x 10^-6) × (0.635 mol/L)
[H+] ≈ 3.995 x 10^-6 mol/L
Now that you have the concentration of hydrogen ions in mol/L, you can use the definition of pH to calculate the pH:
pH = -log[H+]
pH = -log(3.995 x 10^-6)
pH ≈ 5.40
Therefore, the pH of the aqueous solution of propanoic acid with a concentration of 37 g/L is approximately 5.40.